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Areas Of Bounded Regions

Class 12th Mathematics RD Sharma Volume 2 Solution
Exercise 21.1
  1. Using integration, find the area of the region bounded between the line x = 2…
  2. Using integration, find the area of the region bounded by the line y - 1 = x,…
  3. Find the area the region bounded by the parabola y^2 = 4ax and the line x = a.…
  4. Find the area lying above the x - axis and under the parabola y = 4x - x^2 .…
  5. Draw a rough sketch to indicate the bounded between the curve y^2 = 4x and the…
  6. Make a rough sketch of the graph of the function y = 4 - x^2 , 0 ≤ x ≤ 2 and…
  7. Sketch the graph of y = root x+1 in [0,4] and determine the area of the region…
  8. Find the area under the curve y = root 6x+4 above x - axis from x = 0 to x = 2.…
  9. Draw the rough sketch of y^2 + 1 = x, x ≤ 2. Find the area enclosed by the…
  10. Draw a rough sketch of the graph of the curve x^2/4 + y^2/9 = 1 and evaluate…
  11. Sketch the region {(x,y):9x^2 + 4y^2 = 36} and the find the area of the region…
  12. Draw a rough sketch of the graph of the function y=2√1-x^2 ,x[0,1] and…
  13. Determine the area under the y = root a^2 - x^2 included between the lines x =…
  14. Using integration, find the area of the region bounded by the line 2y = 5x +…
  15. Using definite integrals, find the area of circle x^2 + y^2 = a^2…
  16. Using integration, find the area of the region bounded by the following…
  17. Sketch the graph y = |x - 5|. Evaluate integrate _0^1 |x-5|dx . What does this…
  18. Sketch the graph y = |x + 3|. Evaluate integrate _-5^0x+3|dx . What does this…
  19. Sketch the graph y = |x + 1|. Evaluate integrate _-4^2 |x+1|dx . What does the…
  20. Draw a rough sketch of the curve xy -3x - 2y - 10 = 0, x - axis and the lines…
  21. Draw a rough sketch of the curve y = pi /2 + 2sin^2x and find the area between…
  22. Draw a rough sketch of the curve y = pi /2 + 2sin^2x and find the area between…
  23. Find the area bounded by the curve y = cosx, x - axis and the ordinates x = 0…
  24. Show that the areas under the curves y = sin x and y = sin 2x between x = 0…
  25. Compare the areas under the curves y = cos^2 x and y = sin^2 x between x = 0…
  26. Find the area bounded by the ellipse x^2/a^2 + y^2/b^2 = 1 and the ordinates x…
  27. Find the area of the minor segment of the circle x^2 + y^2 = a^2 cut off by…
  28. Find the area of the region bounded by the curve x = at^2 , y = 2at between…
  29. Find the area enclosed by the curve x = 3 cost, y = 2 sint
Exercise 21.2
  1. Find the area of the region in the first quadrant bounded by the parabola y =…
  2. Find the area of the region bounded by x^2 = 16y, y = 1, y = 4 and the y - axis…
  3. Find the area of the region bounded by x^2 = 4ay and its latus rectum.…
  4. Find the area of the region bounded by x^2 + 16y = 0 and its latus rectum.…
  5. Find the area of the region bounded by the curve ay^2 = x^3 , they y-axis and…
Exercise 21.3
  1. Calculate the area of the region bounded by the parabolas y^2 = 6x and x^2 =6y.…
  2. Find the area of the region common to the parabolas 4y^2 = 9x and 3 x^2 =16y.…
  3. Find the area of the region bounded by y = √x and y = x
  4. Find the area bounded by the curve y = 4 - x^2 and the lined y = 0, y = 3.…
  5. Find the area of the region (x , y) : x^2/a^2 + y^2/b^2 less than equal to 1…
  6. Using integration find the area of the region bounded by the triangle whose…
  7. Using integration, find the area of the region bounded by the triangle ABC…
  8. Using integration, find the area of the triangular region, the equations of…
  9. Find the area of the region {(x, y) : y^2 ≤8x, x^2 + y^2 ≤ 9}
  10. Find the area of the region common to the circle x^2 + y^2 = 16 and the…
  11. Find the area of the region between circles x^2 + y^2 = 4 and (x - 2)^2 + y^2…
  12. Find the area of the region included between the parabola y^2 = x and the line…
  13. Draw a rough sketch of the region {(x, y) : y^2 ≤ 3x, 3x^2 + 3y^2 ≤ 16} and…
  14. Draw a rough sketch of the region {(x,y) : y^2 ≤ 5x, 5x^2 + 5y^2 ≤ 36} and…
  15. Draw a rough sketch and find the area of the region bounded by the two…
  16. Find the area included between the parabolasy^2 = 4ax and x^2 = 4by.…
  17. Prove that the area in the first quadrant enclosed by the axis, the line x =…
  18. Find the area of the region bounded by by y = √x and x = 2y + 3 in the first…
  19. Find the area common to the circle x^2 + y^2 = 16 a^2 and the parabola y^2 =…
  20. Find the area, lying above x - axis and included between the circle circle x^2…
  21. Find the area enclosed by the parabolas y = 5x^2 and y = 2x^2 + 9.…
  22. Prove that the area common to the two parabolas y = 2x^2 and y = x^2 + 4 is…
  23. Using integration, find the area of the region bounded by the triangle whose…
  24. Find the area of the region bounded by y = √x and y = x.
  25. Find the area of the region in the first quadrant enclosed by the x - axis,…
  26. Find the area of the region bounded by the parabola y^2 = 2x + 1 and the line…
  27. Find the area of the region bounded by the curves y = x - 1 and (y - 1)^2 = 4…
  28. Find the area enclosed by the curve y = - x^2 and the straight line x + y + 2…
  29. Find the area enclosed by the curve Y = 2 - x^2 and the straight line x + y =…
  30. Using the method of integration, find the area of the region bounded by the…
  31. Sketch the region bounded by the curves y = x^2 + 2, y = x, x = 0 and x = 1.…
  32. Find the area bounded by the curves x = y^2 and x = 3 - 2 y^2 .
  33. Using integration, find the area of the triangle ABC coordinates of whose…
  34. Using integration find the area of the region {(x,y)|x - 1| ≤ y ≤ √5 - x^2 }.…
  35. Find the area of the region bounded by y - |x - 1| and y = 1.
  36. Find the area of the region bounded by y = x and circle x^2 + y^2 = 32 in the…
  37. Find the area of the circle x^2 + y^2 = 16 which is exterior the parabola y^2…
  38. Find the area of the region enclosed by the parabola x^2 = y and the line y =…
  39. Make a sketch of the region{(x,y): 0 ≤ y ≤ x^2 + 3; 0 ≤ y ≤ 2x + 3; 0 ≤ x ≤ 3}…
  40. Find the area of the region bounded by the curve y = √1 - x^2 , line y = x and…
  41. Find the area bounded by the lines y = 4x + 5, y = 5 - x and 4y = x + 5.…
  42. Find the area of the region enclosed between the two curves x^2 + y^2 = 9 and…
  43. Find the area of the region {(x,y): x^2 + y^2 ≤ 4, x + y ≥ 2}
  44. Using integration, find the area of the following region (x , y) : x^2/9 +…
  45. Using integration find the area of the region bounded by the curve y = root…
  46. Find the area enclosed by the curves y = |x - 1| and y = - |x - 1| + 1.…
  47. Find the area enclosed by the curves 3x^2 + 5y = 32 and y = |x - 2|.…
  48. Find the area enclosed by the parabolas y = 4x - x^2 and y = x^2 - x.…
  49. In what ratio does the x-axis divide the area of the region bounded by the…
  50. Find the area of the figure bounded by the curves y = |x - 1| and y = 3 - |x|.…
  51. If the area bounded by the parabola y^2 = 4ax and the line y = mx is a^2 /12…
  52. If the area enclosed by the parabolas y^2 = 16ax and x^2 = 16ay, a0 is 1024/3…
Exercise 21.4
  1. Find the area of the region between the parabola x = 4y - y^2 and the line x =…
  2. Find the area bounded by the parabola x = 8 + 2y - y^2 ; the y - axis and the…
  3. Find the area bounded by the parabola y^2 = 4x and the line y = 2x - 4. (i) By…
  4. Find the area of the region bounded by the parabola y^2 = 2x and the…
Mcq
  1. If the area above the x-axis, bounded by the curves y = 2kx and x = 0, and x = 2 is…
  2. The area of the region bounded by the parabola (y – 2)2 =x – 1, the tangent to it at…
  3. The area included between the parabolas y2 = 4x and x2 = 4y is (in square units)…
  4. The area bounded by the curve y = loge x and x-axis and the straight line x = e is…
  5. The area bounded by y = 2 – x2 and x + y = 0 is
  6. The area bounded by the parabola x = 4 – y2 and y-axis, in square units, is…
  7. If An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x =…
  8. The area of the region formed by x2 + y2 – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is…
  9. The area enclosed between the curves y =loge (x + e), x = loge ( {1}/{y} ) and the…
  10. The area bounded by the curves y = sin x between the ordinates x = 0, x = π and the…
  11. The area bounded by the parabola y2 = 4ax and x2 = 4ay is
  12. The area bounded by the curve y = x4 – 2x3 + x2 + 3 with x-axis and ordinates…
  13. The area bounded by the parabola y2 = 4ax, latus rectum and x-axis is…
  14. The area of the region { ( x , y ) :x^{2} + y^{2}less than equal to 1 leqx+y } is…
  15. The area common to the parabola y = 2x2 and y = x2 + 4 is
  16. The area of the region bounded by the parabola y = x2 + 1 and the straight line x + y…
  17. The ratio of the areas between the curves y = cosx and y = cos 2x and x-axis from x =…
  18. The area between x-axis and curve y = cos x when 0 ≤ x ≤ 2π is
  19. Area bounded by parabola y2 = x and straight line 2y = x is
  20. The area bounded by the curve y = 4x – x2 and the x-axis is
  21. Area enclosed between the curve y2 (2a – x) = x3 and the line x = 2a above x-axis is…
  22. The area of the region (in square units) bounded by the curve x2 = 4y, line x = 2 and…
  23. The area bounded by the curve y = f(x), x-axis, and the ordinates x = 1 and x = b is…
  24. The area bounded by the curve y2 = 8x and x2 = 8y is
  25. The area bounded by the parabola y2 = 8x, the x-axis and the latusrectum is…
  26. Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is…
  27. The area bounded by the curve y = x |x| and the ordinates x = –1 and x = 1 is given by…
  28. The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ π/2 is…
  29. The area of the circle x2 + y2 = 16 enterior to the parabola y2 = 6x is…
  30. Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is…
  31. Area lying between the curves y2 = 4x and y = 2x is
  32. Area lying in first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0…
  33. Area of the region bounded by the cure y2 = 4x, y-axis and the line y = 3, is…

Exercise 21.1
Question 1.

Using integration, find the area of the region bounded between the line x = 2 and the parabola y2 = 8x.


Answer:

Given equations are:

x = 2 ...... (1)


And y2 = 8x ...... (2)


Equation (1) represents a line parallel to y - axis at a distance of 2 units and equation (2) represents a parabola with vertex at origin and x - axis as its axis, A rough sketch is given as below: -



We have to find the area of shaded region.


Required area


= shaded region OBAO


= 2 (shaded region OBCO) (as it is symmetrical about the x - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,2) and the value of y varies)


(as )



On integrating we get,




On applying the limits, we get,




Hence the area of the region bounded between the line x = 2 and the parabola y2 = 8x is equal to square units.



Question 2.

Using integration, find the area of the region bounded by the line y – 1 = x, the x – axis and the ordinates x = – 2 and x = 3.


Answer:

Given equations are:

y – 1 = x (is a line that meets at axes at (0,1) and ( – 1,0))


x = – 2 (is line parallel to y – axis at a distance of 2 units to the left)


x = 3 (is line parallel to y - axis at a distance of 3 units to the right)


A rough sketch is given as below: -



We have to find the area bounded by these three lines with the x - axis, i.e., area of the shaded region.


Required area


= shaded region ABCA + shaded region ADEA


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between ( – 1,3) for the region ABCA and it is between ( – 2, – 1) for the region ADEA and the value of y varies)


(as y – 1 = x ⇒ y = x + 1)


(as x0 = 1)


On integrating we get,



(Combining terms with same limits)


On applying the limits, we get







Hence the area of the region bounded by the line y – 1 = x, the x – axis and the ordinates x = – 2 and x = 3 is equal to square units.



Question 3.

Find the area the region bounded by the parabola y2 = 4ax and the line x = a.


Answer:

Given equations are:

x = a ...... (1)


And y2 = 4ax ...... (2)


Equation (1) represents a line parallel to the y - axis at a distance of units and equation (2) represents a parabola with vertex at origin and x - axis as its axis; A rough sketch is given as below: -



We have to find the area of the shaded region.


Required area


= shaded region OBAO


= 2 (shaded region OBCO) (as it is symmetrical about the x - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,a) and the value of y varies)


(as )



On integrating we get,




On applying the limits, we get,




Hence the area of the region bounded between the line x = a and the parabola y2 = 4ax is equal to square units.



Question 4.

Find the area lying above the x - axis and under the parabola y = 4x – x2.


Answer:

Given equations are:

x – axis ...... (1)


And y = 4x – x2 ...... (2)


⇒ y + 4 = – (x2 – 4x – 4) (adding 4 on both sides)


⇒ – (y + 4) = (x – 2)2


equation (2) represents a downward parabola with vertex at (2,4) and passing through (0,0) and (4,0) on the x – axis, A rough sketch is given as below: –



We have to find the area of the shaded region.


Required area


= shaded region OABO (as it is symmetrical about the x - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,4) and the value of y varies)


(as y = 4x – x2)



On integrating we get,



On applying the limits, we get,





Hence the area lying above the x - axis and under the parabola y = 4x – x2 is equal to square units.



Question 5.

Draw a rough sketch to indicate the bounded between the curve y2 = 4x and the line x = 3. Also, find the area of this region


Answer:

Given equations are:

x = 3 ...... (1)


And y2 = 4x ...... (2)


Equation (1) represents a line parallel to the y - axis at a distance of 3 units and equation (2) represents a parabola with vertex at origin and x - axis as its axis; A rough sketch is given as below: -



We have to find the area of shaded region.


Required area


= shaded region OBCAO


= 2 (shaded region OBCO) (as it is symmetrical about the x - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,3) and the value of y varies)


(as )



On integrating we get,




On applying the limits, we get,




Hence the area of the region bounded between the line x = 3 and the parabola y2 = 4x is equal to square units.



Question 6.

Make a rough sketch of the graph of the function y = 4 – x2, 0 ≤ x ≤ 2 and determine the area enclosed by the curve, the x - axis and the lines x = 0 and x = 2.


Answer:

Given equations are:

x – axis ...... (1)


x = 0 ...... (2)


x = 2 ...... (3)


And y = 4 – x2, 0 ≤ x ≤ 2 ...... (4)


⇒ y = – (x2 – 4) ⇒ x2 = – (y – 4)


equation (4) represents a downward parabola with vertex at (0,4) and passing through (2,0) and ( – 2,0) on x – axis, equation (3) represents a line parallel to y – axis at a distance of 2 units and equation (2) represents y - axis.


A rough sketch is given as below: -



We have to find the area of the shaded region.


Required area


= shaded region OABO


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,2) and the value of y varies)


(as y = 4 – x2 )


(as x0 = 1)


On integrating we get,



On applying the limits, we get,





Hence the area enclosed by the curve, the x - axis and the lines x = 0 and x = 2 is equal to square units.



Question 7.

Sketch the graph of in [0,4] and determine the area of the region enclosed by the curve, the x - axis and the lines x = 0, x = 4


Answer:

Given equations are:

x – axis ...... (1)


x = 0 ...... (2)


x = 4 ...... (3)


And , 0 ≤ x ≤ 4 ...... (4)


equation (4) represents a half parabola with vertex at ( – 1,0) and passing through (4,0) on x – axis, equation (3) represents a line parallel to y - axis at a distance of 4 units and equation (2) represents y - axis.


A rough sketch is given as below: -



We have to find the area of the shaded region.


Required area


= shaded region AOBC


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,4) and the value of y varies)


(as )



Substitute


So the above equation becomes,



On integrating we get,



On applying the limits we get,





Hence the area of the region enclosed by the curve, the x - axis and the lines x = 0, x = 4 is equal to square units.



Question 8.

Find the area under the curve above x - axis from x = 0 to x = 2. Draw a sketch of curve also.


Answer:

Given equations are:

x – axis ...... (1)


x = 0 ...... (2)


x = 2 ...... (3)


And ...... (4)


equation (4) represents a half parabola with vertex at and passing through (2,0) on x - axis, equation (3) represents a line parallel to y - axis at a distance of 2 units and equation (2) represents y - axis.


A rough sketch is given as below: -



We have to find the area of shaded region.


Required area


= shaded region OABC


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,2) and the value of y varies)


(as )



Substitute


So the above equation becomes,



On integrating we get,



On applying the limits we get,





Hence the area under the curve above x - axis from x = 0 to x = 2 is equal to square units.



Question 9.

Draw the rough sketch of y2 + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2


Answer:

Given equations are:

x = 2 ...... (1)


And y2 + 1 = x, x ≤ 2 ...... (2)


equation (2) represents a parabola with vertex at (1,0) and passing through (2,0) on x - axis, equation (1) represents a line parallel to y - axis at a distance of 2 units.


A rough sketch is given as below: -



We have to find the area of shaded region.


Required area


= shaded region ABCA


= 2 (shaded region ACDA) ( as it is symmetrical about the x - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (1,2) and the value of y varies)


(as )



Substitute


So the above equation becomes,



On integrating we get,



On applying the limits we get,




Hence the area enclosed by the curve and the line x = 2 is equal to square units.



Question 10.

Draw a rough sketch of the graph of the curve and evaluate the area of the region under the curve and above the x - axis


Answer:

Given equations are:

...... (1)


And x - axis ...... (2)


equation (1) represents an eclipse that is symmetrical about the x - axis and also about the y - axis, with center at origin and passes through (±2, 0) and (0, ±3).


A rough sketch is given as below: -



We have to find the area of shaded region.


Required area


= shaded region ABCA


= 2 (shaded region OBCO) ( as it is symmetrical about the x - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,2) and the value of y varies)


(as )




Substitute


So the above equation becomes,




We know,


So the above equation becomes,




Apply reduction formula:



On integrating we get,




Undo the substituting, we get





On applying the limits we get,





Hence the area of the region under the given curve and above the x - axis is equal to square units.



Question 11.

Sketch the region {(x,y):9x2 + 4y2 = 36} and the find the area of the region enclosed by it, using integration


Answer:

Given equation:

9x2 + 4y2 = 36 ...... (1)


equation (1) represents an eclipse that is symmetrical about the x - axis and also about the y - axis, with center at origin and passes through (±2, 0) and (0, ±3).


A rough sketch is given as below: -



We have to find the area of the shaded region.


Required area


= shaded region ABCDA


= 4 (shaded region OBCO) ( as it is symmetrical about the x - axis as well as y - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,2) and the value of y varies)


(as )




Substitute


So the above equation becomes,




We know,


So the above equation becomes,




Apply reduction formula:



On integrating we get,




Undo the substituting, we get





On applying the limits we get,





Hence the area of the region enclosed by it is equal to square units.



Question 12.

Draw a rough sketch of the graph of the function y=2√1–x2,x[0,1] and evaluate the are enclosed between the curve and the x–axis.


Answer:

Given equation:

...... (1)



equation (1) represents a half eclipse that is symmetrical about the x - axis and also about the y - axis with center at origin and passes through (±1, 0) and (0, ±2). And x∈[0,1] is represented by region between y - axis and line x = 1.


A rough sketch is given as below: -



We have to find the area of shaded region.


Required area


= (shaded region OBCO)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,1) and the value of y varies)


(as )



Substitute


So the above equation becomes,




We know,


So the above equation becomes,




Apply reduction formula:



On integrating we get,




Undo the substituting, we get





On applying the limits we get,





Hence the area enclosed between the curve and the x - axis is equal to square units.



Question 13.

Determine the area under the included between the lines x = 0 and x = 1


Answer:

Given equations are :

...... (1)



x = 0 (y – axis)


x = 1 (represents a line parallel to y - axis at a distance 1 to the right)


equation (1) represents a half eclipse that is symmetrical about the x - axis and also about the y - axis with center at origin.


A rough sketch is given as below: -



We have to find the area of shaded region.


Required area


= (shaded region OABCO)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,1) and the value of y varies)


(as )



Substitute


So the above equation becomes,




We know,


So the above equation becomes,




Apply reduction formula:



On integrating we get,




Undo the substituting, we get





On applying the limits we get,





Hence the area under the included between the lines x = 0 and x = 1 is equal to square units.



Question 14.

Using integration, find the area of the region bounded by the line 2y = 5x + 7, x - axis the lines x = 2 and x = 8.


Answer:

Given equations are:

2y = 5x + 7 ...... (1)


x = 2 ...... (2)


x = 8 ...... (3)


Equation (1) represents line passing through and . Equation (2), (3) shows line parallel to y - axis passing through (2,0), (8,0) respectively.


A rough sketch of curves is as below:



We have to find the area of shaded region.


Required area


= (shaded region ABCDA)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (2,8) and the value of y varies)


(as )




Now integrating by applying power rule, we get



Now applying the limits we get





Hence the area of the region bounded by the line 2y = 5x + 7, x - axis the lines x = 2 and x = 8 is equal to 96 square units.



Question 15.

Using definite integrals, find the area of circle x2 + y2 = a2


Answer:

Given equations are :

x2 + y2 = a2 ...... (1)


Equation (1) represents a circle with centre (0,0) and radius a, so it meets the axes at (±a,0), (0,±a). A rough sketch of the curve is given below: -



We have to find the area of shaded region.


Required area


= (shaded region ABCDA)


= 4(shaded region OBCO) (as the circle is symmetrical about the x - axis as well as the y - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,a) and the value of y varies)


(as )


Substitute


So the above equation becomes,




We know,


So the above equation becomes,




Apply reduction formula:



On integrating we get,




Undo the substituting, we get





On applying the limits we get,





Hence the area of circle x2 + y2 = a2 is equal to square units.



Question 16.

Using integration, find the area of the region bounded by the following curves, after making a rough sketch: y = 1 + |x + 1|, x = - 2, x = 3, y = 0.


Answer:

Given equations are:

y = 1 + |x + 1|


y = 1 + x + 1, if x + 1 ≥ 0


y2 = 2 + x …… (1), if x ≥ - 1


And y = 1 – (x + 1), if x + 1 < 0


y = 1 – x – 2, if x < - 1


y1 = - x …… (2), if x < - 1


x = - 2 ...... (3)


x = 3 ...... (4)


y = 0 ...... (5)


So, equation (1) is straight line that passes thorough (0,2) and ( - 1,1). Equation (2) is a line passing through ( - 1,1) and ( - 2,2) and it is enclosed by line x = - 2 and x = 3 which are lines parallel to y – axis and pass through ( - 2,0) and (3,0) respectively, y = 0 is x – axis. So, a rough sketch of the curves is gives as: -



We have to find the area of shaded region.


Required area


= (shaded region ABCDEA)


= shaded region ABCFA + Shaded region FCDEFC


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between ( - 2, - 1) in first shaded region and x is between ( - 1,3) for the second shaded region)


(from equation (1) and (2))



Now integrating by applying power rule, we get



Now applying the limits we get





Hence the area of the region bounded by the curves, y = 1 + |x + 1|, x = - 2, x = 3, y = 0 is equal to square units.



Question 17.

Sketch the graph y = |x - 5|. Evaluate . What does this value of the integral represent on the graph?


Answer:

Given equations are:

y = |x - 5|


y1 = x - 5, if x - 5 ≥ 0


y1 = x - 5 …… (1), if x ≥ 5


And y2 = - (x - 5), if x - 5 < 0


y2 = - (x - 5) …… (2), if x < 5


So, equation (1) is straight line that passes thorough (5,0). Equation (2) is a line passing through (5,0) and (0,5). So, the graph of which is as follows:




(As when x is between (0,1) the given equation becomes y = - (x - 5) as shown in equation (2) shown ass shaded region in the above graph)


(from equation (2))



Now integrating by applying power rule, we get



Now applying the limits we get





Hence the value of represents the area of the shaded region OABC (as shown in the graph) and is equal to square units.



Question 18.

Sketch the graph y = |x + 3|. Evaluate . What does this integral represent on the graph?


Answer:

Given equations are:

y = |x + 3|


y1 = x + 3, if x + 3 ≥ 0


y1 = x + 3 …… (1), if x ≥ - 3


And y2 = - (x + 3), if x + 3 < 0


y2 = - (x + 3) …… (2), if x < - 3


So, equation (1) is straight line that passes thorough ( - 3,0) and (0,3). Equation (2) is a line passing through ( - 3,0). So, the graph of which is as follows:




(As x is between ( - 6, - 3) in first shaded region equation becomes as y2 and when x is between ( - 3,0) for the second shaded region equation becomes y1)


(from equation (2))



Now integrating by applying power rule, we get



Now applying the limits we get





Hence the value of represents the area of the shaded region (as shown in the graph) and is equal to 9 square units.



Question 19.

Sketch the graph y = |x + 1|. Evaluate . What does the value of this integral represent on the graph?


Answer:

Given equations are:

y = |x + 1|


y1 = x + 1, if x + 1 ≥ 0


y1 = x + 1 …… (1), if x ≥ - 1


And y2 = - (x + 1), if x + 1 < 0


y2 = - (x + 1) …… (2), if x < - 1


So, equation (1) is straight line that passes thorough ( - 1,0) and (0,1). Equation (2) is a line passing through ( - 1,0). So, the graph of which is as follows:




(As x is between ( - 4, - 1) in first shaded region equation becomes as y2 and when x is between ( - 1,2) for the second shaded region equation becomes y1)


(from equation (2))



Now integrating by applying power rule, we get



Now applying the limits we get





Hence the value of represents the area of the shaded region (as shown in the graph) and is equal to 9 square units.



Question 20.

Draw a rough sketch of the curve xy –3x – 2y – 10 = 0, x - axis and the lines x = 3, x = 4.


Answer:

Given equations are:

xy –3x – 2y – 10 = 0 …..(i)


y (x - 2) = 3x + 10


…..(ii)


x - axis …..(iii)


x = 3 ……(iv)


x = 4 …..(v)


A rough sketch of the curves is given below: -



We have to find the area of shaded region.


Required area


= (shaded region ABCDA)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (3,4) and the value of y varies)


(from equation(ii))


Substitute u = x−2 ⟶ dx = du






Now on integrating we get



Undo substitution, we get




On applying the limits we get





Hence the area of the region bounded by the curves, xy –3x – 2y – 10 = 0, x - axis and the lines x = 3, x = 4 is equal to square units.



Question 21.

Draw a rough sketch of the curve and find the area between x - axis, the curve and the ordinates x = 0, x = π.


Answer:

Given equations are:

…..(i)


x - axis …..(ii)


x = 0 ……(iii)


x = …..(iv)


A table for values of is: -



A rough sketch of the curves is given below: -



We have to find the area of shaded region.


Required area


= (shaded region ABCDOA)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,) and the value of y varies)


(as )



Apply reduction formula:



On integrating we get,




On applying the limits we get




Hence the area between x - axis, the curve and the ordinates x = 0, x = π is equal to square units.



Question 22.

Draw a rough sketch of the curve and find the area between x - axis, the curve and the ordinates x = 0, x = π.


Answer:

Given equations are:

…..(i)


x - axis …..(ii)


x = 0 ……(iii)


x = …..(iv)


A table for values of is: -



A rough sketch of the curves is given below: -



We have to find the area of shaded region.


Required area


= (shaded region ABCDOA)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,) and the value of y varies)


(as )



Apply reduction formula:



On integrating we get,




On applying the limits we get




Hence the area between x - axis, the curve and the ordinates x = 0, x = π is equal to square units.



Question 23.

Find the area bounded by the curve y = cosx, x - axis and the ordinates x = 0 and x = 2π.


Answer:

Given equations are:

y = cos x …..(i)


x - axis …..(ii)


x = 0 ……(iii)


x = …..(iv)


A table for values of y = cos x is: -



A rough sketch of the curves is given below: -



We have to find the area of shaded region.


Required area


= (shaded region ABOA + shaded region BCDB + shaded region DEFD)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,) and the value of y varies)


(as y = cos x)


On integrating we get,



On applying the limits we get



= 1 - 0 + | - 1 - 1| + 0 - ( - 1) = 4


Hence the area bounded by the curve y = cosx, x - axis and the ordinates x = 0 and x = 2π is equal to 4 square units.



Question 24.

Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and are in the ration 2:3.


Answer:

Given equations are:

y = sin x …..(i)


y = sin 2x …..(ii)


x = 0 ……(iii)


x = …..(iv)


A table for values of y = sin x and y = sin 2x is: -



A rough sketch of the curves is given below: -



The area under the curve y = sin x , x = 0 and is


A1 = (area of the region OPBCA)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between and the value of y varies)


(as y = sin x)


On integrating we get,



On applying the limits we get




The area under the curve y = sin 2x , x = 0 and is


A2 = (area of the region OABCO)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between and the value of y varies)


(as y = sin 2x)


On integrating we get,



On applying the limits we get




So the ratio of the areas under the curves y = sin x and y = sin 2x between x = 0 and are



Hence showed



Question 25.

Compare the areas under the curves y = cos2x and y = sin2 x between x = 0 and x = π.


Answer:

Given equations are:

y = cos2x …..(i)


y = sin2x …..(ii)


x = 0 ……(iii)


x = …..(iv)


A table for values of y = cos2x and y = sin2x is: -



A rough sketch of the curves is given below: -



The area under the curve y = cos2x, x = 0 and x = is


A1 = (area of the region OABCO + area of the region CEFGC)


A1 = 2(area of the region CEFGC)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between and the value of y varies)


Apply reduction formula:



On integrating we get,




On applying the limits we get




The area under the curve y = cos2x, x = 0 and x = is


A2 = (area of the region OBDGEO)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between and the value of y varies)


Apply reduction formula:



On integrating we get,




On applying the limits we get




Hence A1 = A2


Therefore the areas under the curves y = cos2x and y = sin2 x between x = 0 and x = π are equal.



Question 26.

Find the area bounded by the ellipse and the ordinates x = ae and x = 0, where b2 = a2 (1 - e2) and e<1.


Answer:

Given equations are:

...... (1)


And x = ae, x = 0 ...... (2)


equation (1) represents an eclipse that is symmetrical about the x - axis and also about the y - axis, with center at origin and passes through (±a, 0) and (0, ±a).


A rough sketch is given as below: -



We have to find the area of shaded region.


Required area


= shaded region ABCDA


= 2 (shaded region OABO) ( as it is symmetrical about the x - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between (0,ae) and the value of y varies)


(as )




Substitute


So the above equation becomes,




We know,


So the above equation becomes,




Apply reduction formula:



On integrating we get,




Undo the substituting, we get




On applying the limits we get,




Hence the area bounded by the ellipse and the ordinates x = ae and x = 0, where b2 = a2 (1 - e2) and e<1 is equal to square units.



Question 27.

Find the area of the minor segment of the circle x2 + y2 = a2 cut off by the line .


Answer:

Given equations are :

x2 + y2 = a2 ...... (1)


..... (2)


Equation (1) represents a circle with centre (0,0) and radius a, so it meets the axes at (±a,0), (0,±a).


Equation (2) represents a line parallel to y axis.


A rough sketch of the circle is given below: -



We have to find the area of shaded region.


Required area


= (shaded region BCDB)


= 2(shaded region ABCA) (as the circle is symmetrical about the x - axis as well as the y - axis)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between and the value of y varies)


(as )


Substitute


So the above equation becomes,




We know,


So the above equation becomes,




Apply reduction formula:



On integrating we get,




Undo the substituting, we get





On applying the limits we get,








Hence the area of the minor segment of the circle x2 + y2 = a2 cut off by the line is equal to square units.



Question 28.

Find the area of the region bounded by the curve x = at2, y = 2at between the ordinates corresponding t = 1 and t = 2


Answer:

Given equations are:

x = at2 ...... (1)


y = 2at ..... (2)


t = 1 ..... (3)


t = 2 ..... (4)


Equation (1) and (2) represents the parametric equation of the parabola.


Eliminating the parameter t, we get



This represents the Cartesian equation of the parabola opening towards the positive x - axis with focus at (a,0).


A rough sketch of the circle is given below: -



When t = 1, x = a


When t = 2, x = 4a


We have to find the area of shaded region.


Required area


= (shaded region ABCDEF)


= 2(shaded region BCDEB)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between and the value of y varies, here y is Cartesian equation of the parabola)


(as )



On integrating we get,


(by applying power rule)


On applying the limits we get,





Hence the area of the region bounded by the curve x = at2, y = 2at between the ordinates corresponding t = 1 and t = 2 is equal to square units.



Question 29.

Find the area enclosed by the curve x = 3 cost, y = 2 sint


Answer:

Given equations are x = 3 cost, y = 2 sint

These are the parametric equation of the eclipse.


Eliminating the parameter t, we get




Squaring and adding equation (i) and (ii), we get


(as sin2t + cos2t = 1)


This is Cartesian equation of the eclipse.


A rough sketch of the circle is given below: -



We have to find the area of shaded region.


Required area


= (shaded region ABCDA)


= 4(shaded region OBCO)


(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)


(As x is between and the value of y varies, here y is Cartesian equation of the eclipse)


(as )




Substitute


So the above equation becomes,




We know,


So the above equation becomes,




Apply reduction formula:



On integrating we get,




Undo the substituting, we get




On applying the limits we get,




Hence the area enclosed by the curve x = 3 cost, y = 2 sint is equal to square units.




Exercise 21.2
Question 1.

Find the area of the region in the first quadrant bounded by the parabola y = 4x2 and the lines x = 0, y = 1 and y = 4.


Answer:


To find the area under two or more than two curves, the first crucial step is to find the INTERSECTION POINTS of the curves.





The coordinates



⟹ y = 4x2, y = 4


⟹ 4 = 4x2


⟹ x = + 1


Required Area can be calculated by breaking the problem into two parts.


I. Calculate Area under the curve A and Line C


II. Subtract the area enclosed by curve A and Line B from the above area.


Therefore, the areas are:


I. = Area enclosed by line C and curve A




II. = Area enclosed by curve A and Line B.





Now the required area under the curves:





Question 2.

Find the area of the region bounded by x2 = 16y, y = 1, y = 4 and the y – axis in the first quadrant.


Answer:


To find the area under two or more than two curves, the first crucial step is to find the INTERSECTION POINTS of the curves.





Between Curve A and Line C




Between Curve A and Line B





Required Area can be calculated by breaking the problem into two parts.


I. Calculate Area under the curve A and Line B.


II. Subtract the area enclosed by curve A and Line C from the above area.


I. = Area under B and A




II. = Area under C and A




The required area under the curve




Question 3.

Find the area of the region bounded by x2 = 4ay and its latus rectum.


Answer:


This is a simple problem of the area under the two curves.


Step 1: Find the latus rectum and its intersection points with the parabola.




Comparing it with the standard form of a parabola


Y = 4 A x2


Where (0, A) is the coordinate of the focus of the parabola. And the latus rectum passes through this point and is perpendicular to the axis of symmetry.


Therefore, the equation of latus rectum is y = a.


Comparing equation (1) and (2)




The equation of the latus rectum:



Intersection points





Step 2: Integrating the expression to find the area enclosed by the curves.


Since the latus rectum is above the parabola in the cartesian plane, the expression will be:







Question 4.

Find the area of the region bounded by x2 + 16y = 0 and its latus rectum.


Answer:


This is a simple problem of the area under the two curves.


Step 1: Find the latus rectum and its intersection points with the parabola.


As following the above questions procedure:


We find the equation of latus rectum as



And the intersection points:



Step 2: Integrating the expression to find the area enclosed by the curves.





Question 5.

Find the area of the region bounded by the curve ay2 = x3, they y-axis and the lines y = a and y = 2a.


Answer:


Similarly as problem 1,


Area of the bounded region = Area under (y = 2a) and (ay2 = x3) – Area under (y = a) and (ay2 = x3)


Area under (y = 2) and (ay2 = x3 ) =





Area under (y = a) and (ay2 = x3)


=




Required area:



Simplify further and you will get the answer.




Exercise 21.3
Question 1.

Calculate the area of the region bounded by the parabolas y2 = 6x and x2 =6y.


Answer:

The given equations are,


y2 = 6x



And x2 = 6y.



When y = 0 then x = 0,


Or


Putting x value on y2 = 6x,


y2 = 6


Or 6


Or


Or y = 6


When y = 0 then x = 0,


And When y = 6 then x = 6,


On solving these two equations, we get point of intersections.


The points are O (0,0) and A(6,6). These are shown in the graph below



Now the bounded area is the required area to be calculated, Hence,


Bounded Area, A = [Area between the curve (i) and x axis from 0 to 6] - [Area between the curve (ii) and x axis from 0 to 6]




On integrating the above definite integration,



Area of the region bounded by the parabolas y2 = 6x and x2 = 6y is 12sq. units.



Question 2.

Find the area of the region common to the parabolas 4y2 = 9x and 3 x2 =16y.


Answer:

The given equations are,


4y2 = 9x



And 3x2 = 16y.



Equating equation (i) and (ii)



Or


Or x = 4


When we put x = 4 in equation (i) then y = 3,


When we put x = 0 in equation (i) then y = 0,


On solving these two equations, we get the point of intersections.


The points are O (0, 0) and A(4,3). These are shown in the graph below



Now the bounded area is the required area to be calculated, Hence,


Bounded Area, A = [Area between the curve (i) and x axis from 0 to 4] - [Area between the curve (ii) and x axis from 0 to 4]




On integrating the above definite integration,


The required area =






Area of the region common to the parabolas 4y2 = 9x and 3 x2 = 16y is 4 sq. units



Question 3.

Find the area of the region bounded by y = √x and y = x


Answer:

The given equations are,



And y = x ...(ii)


Solving equation (i) and (ii)


= x = y


Or = y


Or y(y – 1) = 0


So, y = 0 or y = 1 and x = 0 or x = 1


On solving these two equations, we get the points of intersection.


The points are O (0, 0) and A(1,1). These are shown in the graph below



Now the bounded area is the required area to be calculated,


Hence, Bounded Area, A = [Area between the curve (i) and x axis from 0 to 1] - [Area between the curve (ii) and x axis from 0 to 1]




On integrating the above definite integration,







Area of the region bounded by and Y = X is.



Question 4.

Find the area bounded by the curve y = 4 – x2 and the lined y = 0, y = 3.


Answer:

The given equations are,


Y = 4 – x2 ...(i)


Y = 0 ...(ii)


And y = 3 ...(iii)


Equation (i) represents a parabola with vertex (0,4) and passes through (0,2),(0,02)


Equation (ii) is x - axis and cutting the parabola at C (2, 0)and D( – 2,0)


Equation (iii) is a line parallel to x - axis cutting the parabola at A(3,1)and B( – 3,1)


On solving these equations, we get point of intersections.


The points of intersections of a parabola with the other two lines are A(3,1), B( – 3,1), C(2,0) and D( – 2,0). These are shown in the graph below



Now the bounded area is the required area to be calculated,


Hence, Bounded Area, A = 2 times [Area between the equation(i) and y axis from y = 0 to y = 3]



On integrating the above definite integration,





The area bounded by the curve y = 4 – x2 and the lined y = 0, y = 3 is .



Question 5.

Find the area of the region .


Answer:

There are two equations involved in the question,


Represents an ellipse, symmetrical about both axis and cutting x - axis


at B (a,0) and ( – a,0)


Represents the area inside the ellipse



Represents a straight line cutting x - axis at B(a,0)



Represents the area above the straight line.


Form the given these two equations; we get the point of intersections. The points are B(a,0) and A(0,b). These are shown in the graph below



The common area is the smaller area of an ellipse.


A = [Area between the curve (i) and x axis from 0 to a] – [Area between the curve (ii) and x axis from 0 to a]








SO, the required area is square units.



Question 6.

Using integration find the area of the region bounded by the triangle whose vertices are (2,1), (3,4) and (5,2).


Answer:

Here we have to find the area of the triangle whose triangle are A(2,1), B(3,4) and C(5,2) as shown below.



The equation of AB,





Y = 3x – 5 ...(i)


The equation of BC,




...(ii)


The equation of AC,






Now the required area (A) =


[(Area between line AB and x - axis) - (Area between line AC and x - axis) from x = 2 to x = 3]


+ [(Area between line BC and x - axis) - (Area between line AC and x - axis) from x = 3 to x = 5]











The area of the region bounded by the triangle whose vertices are (2,1), (3,4) and (5,2) is 4 sq. units



Question 7.

Using integration, find the area of the region bounded by the triangle ABC whose vertices A, B, C are ( – 1,1), (0,5) and (3,2) respectively.


Answer:

We have to find the area of the triangle whose vertices are A( – 1,1), B (0,5), C(3,2) as shown below.



The equation of AB,






Y = 4x + 5 ...(i)


The equation of BC,





Y = 5 – x ...(ii)


The equation of AC,






Now the required area(A) =


[(Area between line AB and x - axis) - (Area between line AC and x - axis) from x = - 1 to x = 0]


+ [(Area between line BC and x - axis) - (Area between line AC and x - axis) from x = 0 to x = 3]


Say,








And,







So the enclosed area of the triangle is.



Question 8.

Using integration, find the area of the triangular region, the equations of whose sides are y = 2x + 1, y = 3x + 1 and x = 4.


Answer:

To find the area of the triangular region bounded by


y = 2x + 1 (Say, line AB) ...(i)


y = 3x + 1 (Say, line BC) ...(ii)


y = 4 (Say, line AC) ...(iii)


The sketch of the curves are drawn below,



Equation (i) represents a line passing through points B(0,1) and ,


Equation (ii) represents a line passing through (0,1) and .


Equation (ii) represents a line parallel to y - axis passing through (4, 0).


Solving equation (i) and (ii) gives point B (0, 1).


Solving equation (ii) and (iii) gives point C (4, 13).


Solving equation (i) and (iii) gives point A (4, 9).


So, the Required area, A = (Region ABCA) = [Area between line BC and x - axis from x = 0 to x = 4] - [Area between line AB and x - axis from x = 0 to x = 4]






= 8 sq.units


So the Required area is 8 sq. units.



Question 9.

Find the area of the region {(x, y) : y2≤8x, x2 + y2≤ 9}


Answer:

To find area {(x, y): y2≤ 8x, x2 + y2≤9}

y2 = 8x ...(i)


x2 + y2 = 9 ...(ii)


On solving the equation (i) and (ii),


Or, x2 + 8x = 9


Or, x2 + 8x – 9 = 0


Or, (x + 9)(x – 1) = 0


Or, x = – 9 or x = 1


And when x = 1 then y = ±2√2


Equation (i) represents a parabola with vertex (0,0) and axis as x – axis, equation (ii) represents a circle with centre (0,0) and radius 3 units, so it meets area at (±3, 0), (0,±3).


Point of intersection of parabola and circle is (1,2√2) and (1, – 2√2).


The sketch of the curves is as below:



Or, required area = 2(region ODCO + region DBCD)







Hence, the required area is sq. units.



Question 10.

Find the area of the region common to the circle x2 + y2 = 16 and the parabola y2 = 6x.


Answer:

There are two equations,

x2 + y2 = 16 ...(i)


y2 = 6x ...(ii)


From (i) and (ii)


X2 + 6x = 16


Or, X2 + 6x – 16 = 0


Or, (x + 8)(x – 2) = 0


Or, x = – 8 or x = 2


And when x = 2 then y = ±2√3


Equation (i) represents a circle with centre (0,0) and radius 4 units, so it meets x - axis at (±4,0) and equation (ii) represents a parabola with vertex (0,) and axis as x - axis


Points of intersection of parabola and circle are (2,2√3) and (2, – 2√3).


The sketch of the two curves are drawn below,



The shaded region represents the required area.


Required area = Region OBCAO


Required area = 2 (region ODAO + region DCAD)










So, the required area is sq units.



Question 11.

Find the area of the region between circles x2 + y2 = 4 and (x – 2)2 + y2 = 4.


Answer:

The given equations are,


x2 + y2 = 4 ...(i)


(x – 2)2 + y2 = 4 ...(ii)


Equation (i) is a circle with centre O at origin and radius 2.


Equation (ii) is a circle with centre C (2,0) and radius 2.


On solving these two equations, we have


(x – 2)2 + y2 = x2 + y2


Or x2 – 4x + 4 + y2 = x2 + y2


Or x = 1 which gives y ± √3


Thus, the points of intersection of the given circles are A (1, √3) and A’ (1, – √3) as show in the graph below



Now the bounded area is the required area to be calculated, Hence,


Required area of the enclosed region OACA’O between circle


A = [area of the region ODCAO]


= 2 [area of the region ODAO + area of the region DCAD]










The area of the region between circles x2 + y2 = 4 and (x – 2)2 + y2 = 4 is



Question 12.

Find the area of the region included between the parabola y2 = x and the line x + y = 2.


Answer:

To find region enclosed by


y2 = x ...(i)


And x + y = 2 ...(ii)


From equation (i) and (ii),


y2 + y – 2 = 0


Or, (y + 2) (y – 1) = 0


Or, y = – 2, 1


x = 4,1


Equation represents a parabola with vertex at origin and its axis as x - axis


Equation represents a line passing through (2,0) and (0,2)


On solving these two equations, we get point of intersections. The points of intersection of line and parabola are (1,1) and (4, – 2) These are shown in the graph below



Shaded region represents the required area. We slice it in rectangles of width Δy and length = (x1 – x2).


Area of rectangle = (x1 – x2)Δy.


Required area of Region AOBA









The area of the region included between the parabola y2 = x and the line x + y = 2.


is



Question 13.

Draw a rough sketch of the region {(x, y) : y2 ≤ 3x, 3x2 + 3y2 ≤ 16} and find the area enclosed by the region using method of integration.


Answer:

To find area of region{(x, y) : y2 ≤ 3x, 3x2 + 3y2 ≤ 16}

The given equations are,


y2 = 3x ...(i)


And 3x2 + 3y2 = 16


...(ii)


Equation (i) represents a parabola with vertex (0,0) and axis as x - axis,


Equation (ii) represents a circle with centre (0,0) and radius 4/√3 and meet at A and B A rough sketch of the region is drawn below



Required area = Region OCBAO


= 2(area of Region)


= 2(area of Region OBAO)


= 2(area of Region ODAO + area of Region DBAD)







The area enclosed by the region is



Question 14.

Draw a rough sketch of the region {(x,y) : y2 ≤ 5x, 5x2 + 5y2 ≤ 36} and find the area enclosed by the region using the method of integration.


Answer:

To find the area enclosed by the region{(x,y) : y2 ≤ 5x, 5x2 + 5y2 ≤ 36}

The given equations are,


y2 = 5x ...(i)


And 5x2 + 5y2 = 36 ...(ii)



Substituting the value of y2 from (i) into (ii)


5x2 + 25x = 36


5x2 + 25x – 36 = 0


x =


Equation (i) represents a parabola with vertex (0, 0) and axis as x - axis.


Equation (ii) represents a circle with centre (0, 0) and radius 6/√5 and meets axes at and . X ordinate of the point of intersection of circle and parabola is A where a = .


A rough sketch of curves is: -



Required area = Region OCBAO


= 2 (Region OBAO)


= 2 (Region ODAO + Region DBAD)







The area enclosed by the region is sq. Units



Question 15.

Draw a rough sketch and find the area of the region bounded by the two parabolas y2 = 4x and x2 = 4y by using methods of integration.


Answer:

To find the area bounded by

y2 = 4x


...(i)


And x2 = 4y


...(ii)


On solving the equation (i) and (ii),


= 4x


Or, x4 – 64x = 0


Or, x(x3 – 64) = 0


Or, x = 0, 4


Then y = 0,4


Equation (i) represents a parabola with vertex (0,0) and axis as x – axis. Equation (ii) represents a parabola with vertex (0,0) and axis as y - axis.


Points of intersection of the parabola are (0,0) and (4,4).


A rough sketch is given as: -



Now the bounded area is the required area to be calculated, Hence,


Bounded Area, A = [Area between the curve (i) and x axis from 0 to 4] – [Area between the curve (ii) and x axis from 0 to 4]




On integrating the above definite integration,







Area of the region bounded by the parabolasy2 = 4x and x2 = 4y is sq. units.



Question 16.

Find the area included between the parabolasy2 = 4ax and x2 = 4by.


Answer:

To find area enclosed by

Y2 = 4ax


...(i)


And X2 = 4by


...(ii)


On solving the equation (i) and (ii),



Or, x4 – 64ab2x = 0


Or, x(x3 – 64ab2) = 0


Or, x = 0 and x =


Then y = 0 and y =


Equation (i) represents a parabola with vertex (0,0) and axis as x–axis,


Equation (ii) represents a parabola with vertex (0,0) and axis as x - axis,


Points of intersection of parabolas are O (0,0) and


These are shown in the graph below: -



The shaded region is required area, and it is sliced into rectangles of width and length (y1 – y2)ΔX.


This approximation rectangle slides from x = 0 to , so


Required area = Region OQCPO








The area included between the parabolasy2 = 4ax and x2 = 4by is



Question 17.

Prove that the area in the first quadrant enclosed by the axis, the line x = √3y and the circle x2 + y2 = 4 is π/3.


Answer:

To find an area in the first quadrant enclosed by the x – axis,

x = √3y


x2 + y2 = 4


Or


Or


Or


And


Equation (i) represents a line passing through (0,0), ( – √3, – 1), (√3,1).


Equation (ii) represents a circle centre (0,0) and passing through (±2,0), (0,±2).


Points of intersection of line and circle are ( – √3, – 1) and (√3,1).


These are shown in the graph below: -



Required enclosed area = Region OABO


= Region OCBO + Region ABCA








Hence proved that the area in the first quadrant enclosed by the axis, the line and the circle is π/3.



Question 18.

Find the area of the region bounded by by y = √x and x = 2y + 3 in the first quadrant and x - axis.


Answer:

To find an area in the first quadrant enclosed by the x - axis

y = √x ...(i)


x = 2y + 3 ...(ii)


On solving the equation (i) and (ii),



y2 = 2y + 3


Or, y2 – 2y – 3 = 0


Or, (y – 3)(y + 1) = 0


Or, y = 3 or y = – 1


These are shown in the graph below: -



Required area of the bounded region



The area of the region bounded by by y = √x and x = 2y + 3 is



Question 19.

Find the area common to the circle x2 + y2 = 16 a2 and the parabola y2 = 6ax.

OR

Find the area of the region {(x,y):y2 ≤ 6ax} and {(x,y):x2 + y2 ≤ 16a2}.


Answer:

To find area given equations are

y2 = 6ax ...(i)


x2 + y2 = 16 a2 ...(ii)


On solving Equation (i) and (ii)


Or x2 + (6ax)2 = 16a2


Or x2 + (6ax)2 – 16a2 = 0


Or (x + 8a) (x – 2a) = 0


Or x = 2a or x = – 8a is not possible solution.


Then y2 = 6a(2a) = 12a2 = 2√3a


Equation (i) represents a parabola with vertex (0,0) and axis as x - axis.


Equation (ii) represents a with centre (0,0) and meets axes (±4a,0), (0,±4a).


Point of intersection of circle and parabola are (2a,2√3a), (2a, – 2√3a).


These are shown in the graph below: -



Required area = 2[Region ODCO + Region BCDB]









The area common to the circlex2 + y2 = 16 a2 and the parabola y2 = 6ax is



Question 20.

Find the area, lying above x - axis and included between the circle circle x2 + y2 = 8x and the parabola y2 = 4x.


Answer:

To find area lying above x - axis and included in the circle

x2 + y2 = 8x ...(i)


Or (x – 4)2 + y2 = 16


And ...(ii)


On solving the equation (i) and (ii),


x2 + y2 = 8x


Or x2 – 4x = 0


Or x(x – 4) = 0


Or x = 0 and x = 4


When x = 0, y = 0


When x = 4, y = ±4


Equation (i) represents a circle with centre (4,0) and meets axes at (0,0) and (8,0).


Equation (ii) represent a parabola with vertex (0,0) and axis as x - axis.


They intersect at (4, – 4) and (4,4).


These are shown in the graph below: -



Required area = Region OABO


Required area = Region ODBO + Region DABD …(1)


Region ODBO



Region OBDO = 32/3 sq. units …(2)


Region DABD




Region DABD = 4π sq. units …(3)


Using (1),(2) and (3), We get


Required area =


= 4 sq. units


The area lying above the x - axis and included between the circle x2 + y2 = 8x and the parabola y2 = 4x is 4 sq. Units



Question 21.

Find the area enclosed by the parabolas y = 5x2 and y = 2x2 + 9.


Answer:

To find the area enclosed by

y = 5x2 ...(i)


and y = 2x2 + 9 ...(ii)


On solving the equation (i) and (ii),


5x2 = 2x2 + 9


Or 3x2 = 9


Or x =


Or 15


Equation (i) represents a parabola with vertex O (0, 0) and axis as y - axis .


Equation (ii) represents a parabola with vertex C (0, 9) and axis as the y - axis.


Points of intersection of parabolas are A (√3, 15) and B( – √3,15)


These are shown in the graph below: -



Required area = Region AOBCA


= 2(Region AOCA)







The area enclosed by the parabolas y = 5x2 and y = 2x2 + 9 is



Question 22.

Prove that the area common to the two parabolas y = 2x2 and y = x2 + 4 is 32/3 sq. Units.


Answer:

To find the area enclosed by,

y = 2x2 ...(i)


And y = x2 + 4 ...(ii)


On solving the equation (i) and (ii),


2x2 = x2 + 4


Or x2 = 4


Or x =


y = 8


Equation (1) represents a parabola with vertex (0,0) and axis as y - axis.


Equation (2) represents a parabola with vertex (0,4) and axis as the y - axis.


Points of intersection of parabolas are A(2,8) and B( – 2,8).


These are shown in the graph below: -



Required area = Region AOBCA


= 2(Region AOCA)







Hence, proved that the area common to the two parabolas y = 2x2 and y = x2 + 4 is 32/3 sq. Units.



Question 23.

Using integration, find the area of the region bounded by the triangle whose vertices are

(i) ( – 1, 2), (1, 5) and (3, 4)

(ii) ( – 2, 1), (0, 4) and (2, 3)


Answer:

Equation of side AB,


Or


Or 3x + 3 = 2y – 4


Or 2y – 3x = 7


...(i)


Similarly, the equation of side BC,



Or


Or – x + 1 = 2y – 10


Or 2y = 11 – x


(ii)


And equation of side AC,



Or


Or


Or x + 1 = 2y – 4


...(iii)


These are shown in the graph below: -



Area of required region


= Area of EABFE + Area of BFGCB – Area of AEGCA









= 4 sq. units


The area of the region bounded by the triangle is 4 sq. units.



Question 24.

Find the area of the region bounded by y = √x and y = x.


Answer:

To find the area bounded by

y = √x ...(i)


y = x ...(ii)


On solving the equation (i) and (ii),


Or x2 = x


Or, x(x – 1) = 0


Or, x = 0 or x = 1


Then y = 0 or y = 1


Equation (i) represent a parabola with vertex (0,0) and axis as x - axis


Equation (ii) represents a line passing through origin.


Points of intersection are (0,0) and (1,1).


These are shown in the graph below: -



Area of bounded region






The area of the region bounded by y = √x and y = x is



Question 25.

Find the area of the region in the first quadrant enclosed by the x - axis, the line y = √3x and the circle x2 + y2 = 16.


Answer:

To find the area enclosed by

y = √3x ...(i)


x2 + y2 = 16 ...(ii)


On solving the equation (i) and (ii),


Or x2 + = 16


Or 4x2 = 16


Or x2 = 4


Or x =


∴ Y = ±


Equation (i) represents a parabola with vertex (0,0) and axis as x - axis.


Equation (ii) represent axis a circle with centre (4,0) and meets axes at (0,0) and (4,0).


They intersect at A (2,) and C ( – 2, – ).


These are shown in the graph below: -



Area of the region OAB = Area OAC + Area ACB









The area of the region in the first quadrant enclosed by x - axis, the line y = √3x and the circle x2 + y2 = 16 is



Question 26.

Find the area of the region bounded by the parabola y2 = 2x + 1 and the line x – y – 1 = 0.


Answer:

To find area bounded by

y2 = 2x + 1 ...(i)


X – y – 1 = 0. ...(ii)


On solving the equation (i) and (ii),


X – y = 1


Or y2 = 2(y – 1) + 1


Or y2 = 2y – 1


Or (y + 1)(y – 3) = 0


Or y = 3 or – 1


∴ x = 4,0


Equation (i) is a parabola with vertex and passes through (0, 1), A (0, – 1)


Equation (ii) is a line passing through (1, 0) and (0, – 1).


Points of intersection of parabola and line are B (4, 3) and A (0, – 1)


These are shown in the graph below: -



Required area = Region ABCDA









Area of the region bounded by the parabola y2 = 2x + 1 and the line x – y – 1 = 0is .



Question 27.

Find the area of the region bounded by the curves y = x – 1 and (y – 1)2 = 4 (x + 1).


Answer:

To find region bounded by curves

y = x – 1 ...(i)


(y – 1)2 = 4 (x + 1) ...(ii)


On solving the equation (i) and (ii),


Or (x – 1 – 1)2 = 4 (x + 1)


Or (x – 2)2 = 4 (x + 1)


Or x2 + 4 – 4x = 4x + 4


Or x2 – 8x = 0


Or x = 0 or 8


∴y = – 1 or7


Equation (i) represents a line passing through (1,0) and (0, – 1)


Equation (ii) represents a parabola with vertex ( – 1,1) passes through (0,3),(0, – 1),.


Their points of intersection A(0, – 1) and B(8,7).


These are shown in the graph below: –



It slides from y = – 1 to y = 7,


So, required area = Region ABCDA









The area of the region bounded by the curves y = x – 1 and (y – 1)2 = 4 (x + 1) is



Question 28.

Find the area enclosed by the curve y = – x2 and the straight line x + y + 2 = 0


Answer:

To find region enclosed by

y = – x2 ...(i)


x + y + 2 = 0 ...(ii)


On solving the equation (i) and (ii),


x – x2 + 2 = 0


Or x = 2 or – 1


∴y = – 4, – 1


Equation (i) represents a parabola opening towards the negative y - axis.


Equation (ii) represents a line passing through ( – 2,0) and (0, – 2).


Their points of intersection A( – 1, – 1) and B(2, – 4).


These are shown in the graph below: -



Area of the bounded region






The area enclosed by the curve y = – x2 and the straight line x + y + 2 = 0 is



Question 29.

Find the area enclosed by the curve Y = 2 – x2 and the straight line x + y = 0.


Answer:

To find region enclosed by

Y = 2 – x2 ...(i)


And x + y = 0 ...(ii)


On solving the equation (i) and (ii),


x – x2 + 2 = 0


Or, x2 – x + 2 = 0


Or, x = 2 or – 1


∴ y = – 2 or 1


Equation (i) represents a parabola with vertex (0, 2) and downward, meets axes at (±√2, 0).


Equation (2) represents a line passing through (0, 0) and (2, – 2).


The points of intersection are A (2, – 2) and C ( – 1,1).


These are shown in the graph below: -



Required area = Region ABPCOA



The area enclosed by the curve Y = 2 – x2 and the straight line y + x = 0 is



Question 30.

Using the method of integration, find the area of the region bounded by the following line 3x – y – 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0.


Answer:

To find region enclosed by

3x – y – 3 = 0 ...(i)


2x + y – 12 = 0 ...(ii)


x – 2y – 1 = 0 ...(iii)


Solving (i) and (ii), we get,


5x – 15 = 0


Or x = 3


∴y = 6


The points of intersection of (i) and (ii) is B (3,6)


Solving (i) and (iii), we get,


5x = 5


Or x = 1


∴y = 0


The points of intersection of (i) and (iii) is A (1,0)


Solving (ii) and (iii), we get,


5x = 25


Or x = 5


∴y = 2


The points of intersection of (ii) and (iii) is C (5,2) .


These are shown in the graph below: -



Area of the bounded region


=




= 11 sq. units


The area of the region bounded by the following line 3x – y 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0 is



Question 31.

Sketch the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 1. Also, find the area of the region.


Answer:

To find area bounded x = 0, x = 1

And y = x ...(i)


y = x2 + 2 ...(ii)


Putting x = 1 in equation (ii) we get,


Y = 1 + 2 = 3


Putting x = 1 in equation (i) we get,


Y = 1


So the point of intersection B (1,3), A(1,1)


Equation (i) is a line passing through (1,1) and (0, 0)


Equation (2) is a parabola upward with vertex at (0, 2).


These are shown in the graph below: -



Required area = Region OABCO








The area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 1is



Question 32.

Find the area bounded by the curves x = y2 and x = 3 – 2 y2.


Answer:

To find area bounded by

X = y2 …(i)


And


X = 3 – 2y2 …(ii)


On solving the equation (i) and (ii),


y2 = 3 – 2y2


Or 3y2 = 3


Or y = 1


When y = 1 then x = 1 and when y = – 1 then x = 1


Equation (i) represents an upward parabola with vertex (0, 0) and axis – y.


Equation (ii) represents a parabola with vertex (3, 0) and axis as x – axis.


They intersect at A (1, – 1) and C (1, 1)


These are shown in the graph below: -



Required area = Region OABCO


= 2 Region OBCO


= 2[Region ODCO + Region BDCB]








The area bounded by the curves x = y2 and x = 3 – 2 y2 is



Question 33.

Using integration, find the area of the triangle ABC coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4).


Answer:

To find area of ΔABC with A (4,1), B(6,6) and C(8,4)

Equation of AB,





...(i)


Equation of BC,



y – 6 = – 1(x – 6)


y = – x + 12 ...(ii)


Equation of AC,



These are shown in the graph below: -



Clearly, Area of ΔABC = Area ADB + Area BDC


Area of ADB,





Similarly,








Thus, Area ABC = Area ADB + Area BDC


=


= 64 – 56


= 7 sq. units


The area of the triangle ABC coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4) is 7 sq. Units



Question 34.

Using integration find the area of the region {(x,y)|x – 1| ≤ y ≤ √5 – x2}.


Answer:

To find area of region

{(x,y)|x – 1|≤y≤√5 – x2}


|x – 1| = y



And x2 + y2 = 5 …(iii)


|x – 1|≤y≤√5 – x2


|x – 1| = √5 – x2


X = 2, – 1


Equation (i) and (ii) represent straight lines and equation (iii) is a circle with centre (0,0), meets axes at (±√5,0) and (0,±√5).


These are shown in the graph below: -



Required area = Region BCDB + Region CADC









The area of the region {(x,y)|x – 1|≤y≤√s – x2} is



Question 35.

Find the area of the region bounded by y – |x – 1| and y = 1.


Answer:

To find are bounded by

Y = |x – 1|


y = 1



y = x – 1


or, 1 = x – 1


or, x – 2 = 0


or, x = 2


C(2,1) is point of intersection of y = x – 1 and y = 1.


y = 1 – x


1 = 1 – x


X = 0


A(0,1) is point of intersection of y = 1 – x and y = 1.


Points of intersection are A (0, 1) and C(2,1)


These are shown in the graph below: -



Required area = Region ABCA


= Region ABDA + Region BCDB








= 1 sq. units


The area of the region bounded by y = |x – 1| and y = 1 is



Question 36.

Find the area of the region bounded by y = x and circle x2 + y2 = 32 in the 1st quadrant.


Answer:

To find area of in first quadrant enclose by the circle

X2 + y2 = 32 …(i)


And y = x …(ii)


Solving these two equations, we get


Or 2X2 = 32


Or X2 = 16


Or x = 4


∴y = 4


Equation (i) is a circle with centre (0, 0) and meets axes at A (±4√2, 0), (0,±4√2). And y = x is a line passes through (0, 0) and intersect circle at B (4, 4).


These are shown in the graph below:



Region OABO = Region OCBO + Region CABC








The area of the region bounded by y = x and circle X2 + y2 = 32 is



Question 37.

Find the area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x.


Answer:

The given equations are

x2 + y2 = 16 …(i)


And y2 = 6x …(ii)


On solving the equation (i) and (ii),


Or x2 + 6x = 16


Or x2 + 6x – 16 = 0


Or (x + 8)(x – 2) = 0


Or x = 2 or – 8 is not possible solution


∴ When x = 2, y = ± = ± = ±


Equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)


Equation (ii) represents a parabola with axis as x - axis.


Points of intersection are A () and C (2,)


These are shown in the graph below:



Area bounded by the circle and parabola


= 2[Area (OADO) + Area (ADBA)]










Area of circle = π(r) 2


= π(4)2


= 16π sq. Units


Thus, required area





The area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is




Question 38.

Find the area of the region enclosed by the parabola x2 = y and the line y = x + 2.


Answer:

To Find the area of the region enclosed by

x2 = y …(i)


y = x + 2 …(ii)


On solving the equation (i) and (ii),


Or x2 – x – 2 = 0


Or (x – 2) (x + 1) = 0


Or x = 2 or x = – 1


∴y = 4 or y = 1


Points of intersection are A () and C


These are shown in the graph below: -



Area of the bounded region






The area of the region enclosed by the parabola x2 = y and the line y = x + 2 is



Question 39.

Make a sketch of the region{(x,y): 0 ≤ y ≤ x2 + 3; 0 ≤ y ≤ 2x + 3; 0 ≤ x ≤ 3} and find its area using integration.


Answer:

To find area given equations are

Y = x2 + 3 …(i)


Y = 2x + 3 …(ii)


And x = 3 …(iii)


Solving the above three equations to get the intersection points,


x2 + 3 = 2x + 3


Or x2 – 2x = 0


Or x(x – 2) = 0


And x = 0 or x = 2


∴ y = 3 or y = 7


Equation (1) represents a parabola with vertex (3, 0) and axis as y – axis.


Equation (2) represents a line a passing through (0, 3) and ( – 3/2, 0)


The points of intersection are A (0,3) and B(2,7).


These are shown in the graph below:



Required area =









The area of the region{(x,y): 0≤y≤x2 + 3; 0 ≤ y ≤ 2x + 3; 0 ≤ x ≤ 3} is



Question 40.

Find the area of the region bounded by the curve y = √1 – x2, line y = x and the positive x - axis.


Answer:

To find the area of the region bounded by

y = √1 – x2 …(i)


X2 + y2 = 1


X = y …(ii)


On solving the equation (i) and (ii),


Or X2 + x2 = 1


Or 2X2 = 1


Or x =


Equation (i) represents a circle (0,0) and meets axes at (±1,0), (0,±1).


Equation (ii) represents a line passing through B and and they are also points of intersection.


These are shown in the graph below:



Required area = Region OABO


= Region OCBO + Region CABC








The area of the region bounded by the curve y = √1 – x2, line y = x is



Question 41.

Find the area bounded by the lines y = 4x + 5, y = 5 – x and 4y = x + 5.


Answer:

To find the area bounded by

Y = 4x + 5 (Say AB) …(i)


Y = 5 – x (Say BC) (ii)


4y = x + 5 (Say AC) ...(iii)


By solving equation (i) and (ii), points of intersection is B (0, 5)


By solving equation (ii) and (iii), points of intersection is C (3, 2)


By solving equation (i) and (iii), we get points of intersection is A ( – 1, 1)


These are shown in the graph below:



Required area = area of (ΔABD) + area of (ΔBDC) …(1)


Area of (ΔABD) =







…(2)


Area of (ΔBDC) =







Area of (ΔBDC) = …(3)


Using equation (1), (2) and (3)


Required area = Area (ΔABD) + Area (ΔBDC)




Required bounded area of = Area of (ΔABD) + Area of (ΔBDC) =



Question 42.

Find the area of the region enclosed between the two curves x2 + y2 = 9 and (x – 3)2 + y2 = 9.


Answer:

To find area enclosed by

X2 + y2 = 9 (i)


(x – 3)2 + y2 = 9 (ii)


On solving equation (i) and (ii) we get,


X = and y = ±


Equation (i) represents a circle with centre (0,0) and meets axes at (±3,0),(0,±3).


Equation (ii) is a circle with centre (3,0) and meets axe at (0,0), (6,0).


They intersect each other at and .


These are shown in the graph below:



Required area = Region OABCO


= 2(Region OBCO)


= 2(Region ODCO + Region DBCD)









The area of the region enclosed between the two curves curves x2 + y2 = 9 and (x – 3)2 + y2 = 9 is



Question 43.

Find the area of the region {(x,y): x2 + y2 ≤ 4, x + y ≥ 2}


Answer:

The equation of the given curves are

X2 + y2 = 4 (i)


X + y = 2 (ii)


Clearly X2 + y2 = 4 represents a circle X + y = 2 is the equation of a straight line cutting x and y axes at (0, 2) and (2, 0) respectively.


These are shown in the graph below:



The required area is given by



We have y1 = 2 – x and y2 =









the area of the region {(x,y): x2 + y2≤ 4, x + y ≥ 2}is



Question 44.

Using integration, find the area of the following region.


Answer:

To find area of region

(i)


(ii)


Equation (1) represents an ellipse with centre at origin and meets axes at (±3, 0), (0,±2).


Equation (2) is a line that meets axes at (3, 0), (0, 2).


The sketch of the two curves are shown below:



Required area =








The area of the region: is



Question 45.

Using integration find the area of the region bounded by the curve , x2 + y2 – 4x = 0 and the x-axis.


Answer:


First, let us find the intersection points of the curve,


Given Equations are x2 + y2 = 4 and x2 + y2 – 4 x = 0.


From both of the equations,


4 x = 4


x = 1


Putting this value in x2 + y2 = 4, we get,


1 + y2 = 4


y2 = 3


y = ±√3


Thus the curves intersect at A(1, √3) and B(1, - √3)


The area to be found is shaded in the figure above.


Area of Shaded region =





Area of Shaded region = square units.



Question 46.

Find the area enclosed by the curves y = |x – 1| and y = – |x – 1| + 1.


Answer:

To find the area enclosed by

y = |x – 1|




And y = – |x – 1| + 1




Solving both the equation for x<1


Y = 1 – x and y = x,


We get x = and y =


And solving both the equations for x≥1


Y = x – 1 and y = 2 – x,


We get x = and y =


These are shown in the graph below:



Required area = Region ABCDA


Required area = Region BDCB + Region ABDA …(1)








The area enclosed by the curves y = |x – 1| and y = – |x – 1| + 1 is



Question 47.

Find the area enclosed by the curves 3x2 + 5y = 32 and y = |x – 2|.


Answer:

To find area enclosed by

3x2 + 5y = 32


…(i)


And y = |x – 2|




Equation (i) represents a parabola with vertex (0, 32/5) and equation (ii) represents lines.


These are shown in the graph below:



Required area = Region ABECDA


= Region ABEA + Region AECDA











The area enclosed by the curves 3x2 + 5y = 32 and y = |x – 2| is



Question 48.

Find the area enclosed by the parabolas y = 4x – x2 and y = x2 – x.


Answer:

To area enclosed by

Y = 4x – x2 (1)


4x – x2 = x2 – x


2x2 – 5x = 0


x = 0 or x =


y = 0 or y =


And y = x2 – x


…(2)


Equation (1) represents a parabola downward with vertex at (2,4) and meets axes at (4,0), (0,0).


Equation (2) represents a parabola upward whose vertex is and meets axes at Q(1,0), (0,0).Points of intersection of parabolas are O (0,0) and A .


These are shown in the graph below:



Required area = Region OQAP








The area enclosed by the parabolas y = 4x – x2 and y = x2 – x is .



Question 49.

In what ratio does the x-axis divide the area of the region bounded by the parabolas y = 4x – x2 and y = x2 – x?


Answer:


Let us find the intersection points first,


We have the equations of curves,


y = 4 x – x2 …….(1)


y = x2 – x ………(2)


From (1) and (2) we can get,


x2 – x = 4 x – x2


2 x2 – 5x = 0


x(2 x – 5) = 0


x = 0 or x = 5/2


Putting these values of x in equation (2) we get,


At x = 0,


Y = 02 – 0 = 0


At x = 5/2



Hence intersection points are (0, 0) and


Area bounded by the curves is shown by the shaded region of the figure shown above.


Area of shaded region =


Area of shaded region =


Area of Shaded Region =


Area of Shaded Region = Square units.



Question 50.

Find the area of the figure bounded by the curves y = |x – 1| and y = 3 – |x|.


Answer:

To find the area of the figure bounded by

y = |x – 1|



Y = x – 1 is a straight line passing through A(1,0)


Y = 1 – x is a straight line passing through A(1,0) and cutting y - axis at B(0,1)


y = 3 – |x|



Y = 3 – x is a straight line passing through C(0,3)and O(3,0)


Y = 3 + x is a straight line passing through C(0,3)and D( – 3,0)


Point of intersection for


Y = x – 1


And y = 3 – x


We get


X – 1 = 3 – x


or, 2x – 4 = 0


or, x = 2


or, y = 2 – 1 = 1


Thus, point of intersection for y = x – 1 and y = 3 + x is B(2,1)


Point of intersection for


y = 1 – x


y = 3 + x


or,1 – x = 3 + x


or, 2x = – 2


or, x = – 1


or, y = 1 – ( – 1) = 2


Thus, point of intersection for y = 1 – x and y = 3 + x is D( – 1,2)


These are shown in the graph below:



Required area = Region ABCDA


= Region ABFA + Region AFCEA + Region CDEC









The area of the figure bounded by the curves y = |x – 1| and y = 3 – |x| is 4 sq. units



Question 51.

If the area bounded by the parabola y2 = 4ax and the line y = mx is a2/12 sq. Units, then using integration, find the value of m.


Answer:

Area of the bounded region = a2/12


Mathematically the area in integral form will be,,


So, on equating..







Question 52.

If the area enclosed by the parabolas y2 = 16ax and x2 = 16ay, a>0 is 1024/3 square units, find the value of a.


Answer:


Area of the bounded region - 1024/3





Exercise 21.4
Question 1.

Find the area of the region between the parabola x = 4y – y2 and the line x = 2y – 3.


Answer:

Given: - Two equation; Parabola x = 4y – y2 and Line x = 2y – 3


Now to find an area between these two curves, we have to find a common area or the shaded part.


From figure, we can see that,



Area of shaded portion = Area under the parabolic curve – Area under line


Now, Intersection points;


From parabola and line equation equate x, we get


⇒ 4y – y2 =2y – 3


⇒ y2 – 2y – 3 = 0


⇒ y2 – 3y + y – 3 = 0


⇒ y(y – 3) + 1(y – 3)


⇒ (y + 1)(y – 3)


⇒ y = – 1,3


So, by putting the value of x in any curve equation, we get,


⇒ x = 2y – 3


For y = – 1


⇒ x = 2( – 1) – 3


⇒ x = – 5


For


y = 3


⇒ x = 2(3) – 3


⇒ x = 3


Therefore two intersection points coordinates are ( – 5, – 1) and (3, 3)


Area of the bounded region


= (Area under the parabola curve from – 1 to 3) – (Area under line from – 1 to 3)


Tip: -Take limits as per strips. If strip is horizontal than take y limits or if integrating with respect to y then limits are of y.


Here, limits are for y i.e from - 1 to 3.







Now putting limits, we get









Question 2.

Find the area bounded by the parabola x = 8 + 2y – y2; the y - axis and the lines y = – 1 and y = 3.


Answer:

Given: - Two equation;

Parabola x = 8 + 2y – y2 ,


y - axis,


Line1 y = – 1, and Line2 y = 3



Now to find the area between these four curves, we have to find a common area (ABDC) or the shaded part.


The 1st intersection of a parabola with line y = – 1, we get,


Putting the value of y = 1 in parabolic equation


⇒ x = 8 + 2y – y2


⇒ x = 8 + 2( – 1) – 1


⇒ x = 5


Hence intersection point is D(5, – 1)


The 2nd intersection of parabola with y = 3


Putting the value of y in parabola equation


⇒ x = 8 + 2y – y2


⇒ x = 8 + 2(3) – 32


⇒ x = 8 + 6 – 9


⇒ x = 5


Hence, intersection point is C(5,3)


and other points are A(0,3), B(0, – 1)


From the figure, we can see that, By taking a horizontal strip


The areaunder shaded portion = Area under parabola from y =– 1 to y = 3.


Tip: -Take limits as per strips. If strip is horizontal than take y limits or if integrating concerning y then limits are of y.


Here, limits are for y i.e. from – 1 to 3







Now putting limits, we get,









Question 3.

Find the area bounded by the parabola y2 = 4x and the line y = 2x – 4.

(i) By using horizontal strips

(ii) By using vertical strips.


Answer:

Given: - Two curves are y2 = 4x and y = 2x – 4



Now to find the area between these two curves, we have to find common area i.e. Shaded portion


Intersection of parabola y2 = 4x with line y = 2x – 4


Putting the value of y from the equation of a line in parabola equation, we get,


y2 = 4x


⇒ (2x – 4)2 = 4x


⇒ 4x2 – 16x + 16 = 4x


⇒ 4x2 – 20x + 16 = 0


⇒ 4x2 – 16x – 4x + 16 = 0


⇒ 4x(x – 4) – 4(x – 4) = 0


⇒ 4(x – 1)(x – 4) = 0


⇒ (x – 1)(x – 4) = 0


⇒ x = 1,4


When x = 1, y = √4x


⇒ y = + 2, – 2; we take – 2 as the intersection is in the 4th quadrant and when x = 4, y = √4x


⇒ y = + 4, – 4; we take + 4 as the intersection is in 1st quadrant


Therefore intersection points are B(4,4) and C(1, – 2)


Area of the bounded region, taking strips


i) By using horizontal strips


Therefore, limits are for y and integrating with respect to y


Area bounded by region = {Area under line from – 2 to 4} –{Area under parabola from – 2 to 4}






Putting limits, we get







ii) By using vertical strips.


Therefore, limits are for x, and integrating with respect to x


Area bounded by region = {2(Area under parabola from 0 to 1) + (Area under parabola from 1 to 4)} – {Area under line from 1 to 4}


Tip: - Parabola is symmetrical about x - axis therefore its area is twice the area above x - axis. So, till its latus rectum i.e here a = 1, area is twice the area above x - axis.






Putting limits, we get,









Hence from both methods we get same answer



Question 4.

Find the area of the region bounded by the parabola y2 = 2x and the straight-line x – y = 4.


Answer:

Given: -

Two equation;


Parabola y2 = 2x and


Line x – y = 4



Now to find an area between these two curves, we have to find a common area or the shaded part.


From figure we can see that,


Area of shaded portion = Area under line curve – Area under parabola; horizontal strip


Now, Intersection points;


From parabola and line equation equate y, x – 4 = y we get


⇒ y2 = 2x


⇒ (x – 4)2 = 2x


⇒ x2 – 8x + 16 = 2x


⇒ x2 – 10x + 16 = 0


⇒ x2 – 8x – 2x + 16 = 0


⇒ x(x – 8) – 2(x – 8) = 0


⇒ (x – 8)(x – 2) = 0


⇒ x = 8,2


So, by putting the value of x in any curve equation, we get,


⇒ y = x – 4


For x = 8


⇒ y = 8 – 4


⇒ y = 4


For x = 2


⇒ y = 2 – 4


⇒ y = – 2


Therefore, two intersection points coordinates are (8, 4) and (2, – 2)


Area of the bounded region


= Area under the line curve from – 2 to 4 – Area under parabola from – 2 to 4


Tip: -Take limits as per strips. If the strip is horizontal than take y limits or if integrating with respect to y then limits are of y.


Area bounded by region = {Area under line from – 2 to 4} – {Area under parabola from – 2 to 4}






Putting limits, we get





= 6 + 24 – 12


= 18 sq units




Mcq
Question 1.

If the area above the x-axis, bounded by the curves y = 2kx and x = 0, and x = 2 is then the value of k is
A. 1/2

B. 1

C. –1

D. 2


Answer:

The area can be computed as –






Comparing with



⇒ 4k = 3k + 1


Using Binomial Theorem,




This equality holds only when k = 1, because only then you get two terms in the expansion.


Ans: k = 1


Question 2.

The area of the region bounded by the parabola (y – 2)2 =x – 1, the tangent to it at the point with the ordinate 3 and the x-axis is
A. 3

B. 6

C. 7

D. none of these


Answer:

The question describes something like –



To solve the problem, we need to find the points 1, 2, and 3 first. They are the key to setting up the bounds of our integration and understanding what function to integrate.


For point 1:-


We know it is a point on the parabola with ordinate 3, at which the tangent to the parabola is taken.


Plugging in y = 3 into the equation for the parabola (y – 2)2 = x – 1,


(3 – 2)2 = x – 1


⇒ 1 = x – 1


⇒ x = 2


So, Point 1 is (2, 3)


For point 2:-


We need to find the point of intersection of the parabola and the x – axis.


We know that ordinate at this point is 0.


Plugging in y = 0 into the equation for the parabola (y – 2)2 = x – 1,


(0 – 2)2 = x – 1


⇒ 4 = x – 1


⇒ x = 5


So, Point 2 is (5, 0)


For point 3:-


Tangent at any point (x1, y1) for a curve is –


y – y1 = (x – x1)


For parabola (y – 2)2 = x – 1, differentiating both sides of the equation –


2(y – 2) = 1



So, slope at point 1, i.e., (2, 3) is –



So, equation of tangent at (2, 3) is –


y – 3 = � (x – 2)


⇒ 2y – 6 = x – 2


⇒ x – 2y + 4 = 0


The tangent intersects the x – axis at point 3. At this point, ordinate is 0.


So, plugging y = 0 in the equation of the tangent x – 2y + 4 = 0, -


x – 2 �0 + 4 = 0


⇒ x = -4


So, point 3 is (-4, 0).


Now, that we have the 3 points, let’s figure out how to compute the area required.


The area we need can be divided into 3 sections –


i. x = -4 to x = 1


Here, area required = area enclosed by the tangent and the x – axis


ii. x = 1 to x = 2


Here, area required = (area enclosed by the tangent and the x – axis) – (area enclosed within the parabola)


iii. x = 2 to x = 5


Here, area required = area enclosed by the parabola and the x – axis


The parabola is (y – 2)2 = x – 1


Solving for y,


y = 2 ± √(x – 1)


So, area A we need is –









(Ans)


Question 3.

The area included between the parabolas y2 = 4x and x2 = 4y is (in square units)
A. 4/3

B. 1/3

C. 16/3

D. 8/3


Answer:


The blue area is what we need to compute. To do that we need the bounds, i.e., where the area starts and where it ends. At the points of intersection, both the equations are satisfied.


This means that at points of intersection


y2 = 4x


(from the other equation)


Let us solve this.



x4 = 64x


⇒ x(x3 – 64) = 0


⇒ x = 0 or x3 = 64, i.e, x = 4


So the points of intersection are (0,0) and (4,4)


Now, let’s compute the area.


If we integrate w.r.t x, we’ll have to integrate the space between the two curves from x = 0 to x =4.


i.e.,







(Ans)


Question 4.

The area bounded by the curve y = loge x and x-axis and the straight line x = e is
A. e sq. units

B. 1 sq. units

C.

D.


Answer:


We need to find the area of the blue shaded region.


At, x = 1, y = loge(1) = 0


And, at x = e, y = loge(e) = 1


These are our bounds.


So, this will be computed as –



Using Integration by parts,




=[e – e – 0 + 1]


= 1 (Ans)


Question 5.

The area bounded by y = 2 – x2 and x + y = 0 is
A.

B.

C. 9 sq. units

D. none of these


Answer:


- the blue shaded region above


To define the bounds, we need to find the points of intersection. We know that at the points of intersection, both the equations are satisfied.


⇒ x + y = 0


⇒ x + (2 – x2) = 0 (from the other equation)


So, x2 – x – 2 = 0 i.e., (x - 2)(x + 1) = 0 or x = -1,2


So, bounds are x = -1 to x = 2


Therefore, area shall be evaluated as –




(Ans)


Question 6.

The area bounded by the parabola x = 4 – y2 and y-axis, in square units, is
A. 3/32

B. 32/3

C. 33/2

D. 16/3


Answer:


Is the blue shaded region.


At points of intersection on y-axis, x = 0


So, equation becomes 0 = 4 – y2 or y = -2, 2


This becomes our bounds, and we integrate w.r.t. the y-axis –





(Ans)


Question 7.

If An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x = π/4, then for x > 2
A.

B.

C.

D. none of these


Answer:




Now, let u = tan x du = sec2 x dx


When x = 0, u = 0 and when x= π/4, u = 1


So,





Ans:


Question 8.

The area of the region formed by x2 + y2 – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is
A.

B.

C.

D. none of these


Answer:

x2 + y2 – 6x – 4y + 12 ≤ 0 can be written as –


x2 – 6x + 9 – 9 +y2 – 4y + 4 – 4 + 12 ≤ 0


i.e., (x - 3)2 + (y – 2)2 ≤ 1


So, this indicates the area enclosed by a circle centred at (3,2) with radius 1.


Now, the area of the region formed by x2 + y2 – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is



So, our bounds are x = 2 to x = 2.5


The equation for the ordinate of point on the circle from x = 2 to x = 2.5 –


(x – 3)2 + (y – 2)2 = 1


(y – 2)2 = 1 – (x – 3)2


y – 2 = y = 2


Since we are considering the lower value of y in x = 2 to x = 2.5 (the one in y ≤ x),


y = 2 -


So, the area is –




(Ans)


Question 9.

The area enclosed between the curves y =loge (x + e), x = logeand the x-axis is
A. 2

B. 1

C. 4

D. none of these


Answer:

y =loge (x + e) and x = loge look like –


The curves intersect at (0, 1)


(Putting x = 0 in the 2 curves, y = loge(e) = 1 and 0 = loge(1/y),


i.e., y = 1/e0 = 1)


So, bounds are x = 1 – e to x = 0 for the first curve and then x = 0 to apparently x = ∞ for the second curve.


Therefore,




(Ans)


Question 10.

The area bounded by the curves y = sin x between the ordinates x = 0, x = π and the x-axis is
A. 2 sq. units

B. 4 sq. units

C. 3 sq. units

D. 1 sq. units


Answer:

This is as simple as –


A =


=


= [-(-1) + 1]


= 2 (Ans)


Question 11.

The area bounded by the parabola y2 = 4ax and x2 = 4ay is
A.

B.

C.

D.


Answer:

This problem is the generalized form of question 2.


Let’s proceed to solve it similarly –


We need the bounds, i.e., where the area starts and where it ends. At the points of intersection, both the equations are satisfied.


This means that at points of intersection


y2 = 4ax


(from the other equation)


Let us solve this.



x4 = 64a3x


⇒ x(x3 – 64a3) = 0


⇒ x = 0 or x3 = 64a3, i.e, x = 4a


So the points of intersection are (0,0) and (4a,4a)


Now, let’s compute the area.


If we integrate w.r.t x, we’ll have to integrate the space between the two curves from x = 0 to x =4a.


i.e.,






(Ans)


Question 12.

The area bounded by the curve y = x4 – 2x3 + x2 + 3 with x-axis and ordinates corresponding to the minima of y is
A. 1

B. 91/30

C. 30/9

D. 4


Answer:

y = x4 – 2x3 + x2 + 3


⇒ y’ = 4x3 – 6x2 + 2x


At extrema of y, y’ = 0


i.e., 4x3 – 6x2 + 2x = 0


or, 2x3 – 3x2 + x = 0


⇒ x(2x – 1)(x – 1) = 0


i.e., x = 0, x = 1/2 and x = 1 correspond to extrema of y


Now, y’’ = 12x2 – 12x + 2


y’’0 = 2 > 0 ⇒ x = 0 corresponds to a minima


y’’1/2 = -1 < 0 ⇒ x = 1/2 corresponds to a maxima


y’’1 = 2 > 0 ⇒ x = 1 corresponds to a minima


So, x = 0 and x = 1 are the ordinates we need.


So, the area A is –


A =


=




(Ans)


Question 13.

The area bounded by the parabola y2 = 4ax, latus rectum and x-axis is
A. 0

B. 4/3 a2

C. 2/3 a2

D. a2/3


Answer:

The area A is –


A =



(Ans)


Question 14.

The area of the region is
A. π/5

B. π/4

C. π/2 – 1/2

D. π2/2


Answer:

We need to determine what region is being talked about.


x2 + y2 = 1 is a circle with centre at origin and unit radius.


So, x2 + y2 ≤ 1 represents the region inside that circle.


x + y = 1 is a line that intersects the 2 axes at unit distances from the origin, in the positive direction.


So x + y ≥ 1 represents all the points, i.e., the region above the line x + y = 1.




So, the region {(x, y): x2 + y2 ≤ 1 ≤ x + y} is –



For each point (x, y) on circle in first quadrant, y = √(1 – x2)


For each point (x, y) on line, y = 1 – x


So, area A of the region described is –


A =


=


= π/4 – 1/2 (Ans)


Question 15.

The area common to the parabola y = 2x2 and y = x2 + 4 is
A. 2/3 sq. units

B. 3/2 sq. units

C. 32/3 sq. units

D. 3/32 sq. units


Answer:

The area we need looks like –



We need to find the points of intersection to set the bounds of integration.


At points of intersection, y = x2 + 4 = 2x2


or, x2 – 4 = 0


⇒ (x + 2)(x – 2) = 0


⇒ x = -2, 2


⇒ The points of intersection are (-2, 8) and (2, 8), which is also evident from the graph.


So, the area A of the shaded region is –


A =


=


Since 4 – x2 is an even function,


A =


=


=


= 32/3 (Ans)


Question 16.

The area of the region bounded by the parabola y = x2 + 1 and the straight line x + y = 3 is given by
A. 45/7

B. 25/4

C. π/18

D. 9/2


Answer:

The situation looks like this –



At the intersection points, y = 3 – x = x2 + 1


Or, x2 + x – 2 = 0


⇒ (x + 2)(x – 1) = 0


⇒ x = -2, 1 → these are our bounds


So, area A enclosed is –


A =


=


=


= 2 – 1/2 - 1/3 + 4 + 2 – 8/3


= 8 – 9/3 – 1/2


= 5 – 1/2


= 9/2 (Ans)


Question 17.

The ratio of the areas between the curves y = cosx and y = cos 2x and x-axis from x = 0 to x = π/3 is
A. 1 : 2

B. 2 : 1

C.

D. none of these


Answer:

Let us call the corresponding areas A1 and A2.


A1 =


=


= √3/2


A2 =


=


= √3/4


∴ A1 : A2 = 2 : 1 (Ans)


Question 18.

The area between x-axis and curve y = cos x when 0 ≤ x ≤ 2π is
A. 0

B. 2

C. 3

D. 4


Answer:

Let area be A.


So, area A is –


A =


Now, cos x is positive from x = 0 to x = π/2 and from x = 3π/2 to x = 2π and negative from x = π/2 to x = 3π/2.


So, A =


=


= [1 – 0] + [-(-1) + 1] + [0 – (-1)]


= 1 + 1 + 1 + 1


= 4 (Ans)


Question 19.

Area bounded by parabola y2 = x and straight line 2y = x is
A. 43

B. 1

C. 23

D. 13


Answer:

Construction:


The situation looks like this –



At the intersection points, x = 2y = y2


or, y2 – 2y = 0


⇒ y(y – 2) = 0


⇒ y = 0 or y = 2


So, intersection points are (0, 0) and (4, 2)


Hence, the area A enclosed is –


A =


=


= [4 – 8/3]


= 4/3 (Ans)


Question 20.

The area bounded by the curve y = 4x – x2 and the x-axis is
A.

B.

C.

D.


Answer:

y = 4x – x2


This is a parabola with negative co-efficient of x2, i.e., it’s a downward parabola.


So, area A enclosed is the area of the peak of the parabola above the x – axis.


We need to find the bounds of this peak.


Now, at the point where the peak starts/ends, y = 0,


i.e., 4x – x2 = 0


⇒ x(4 – x) = 0


⇒ x = 0 or x = 4


∴ A =


=


= [32 – 64/3]


= 32/3 sq. units (Ans)


Question 21.

Area enclosed between the curve y2 (2a – x) = x3 and the line x = 2a above x-axis is
A. πa2

B.

C. 2π a2

D. 3π a2


Answer:

y2(2a – x) = x3




Since we are concerned with the area above the x – axis, we’ll be considering the positive root.


We can see that at x = 0, y = 0.


So,


A =


Putting √x = u


We get du = (1/2√x)dx


or, dx = 2u du


So,


A =


Putting u = √2√a sin t


We get du = √2√a cos t dt


A =


=


=


=


=


=


=


=


=


= 4a2(π/2 – 0 – 0 + 0) – a2(π/2 – 0 – 0 + 0)


= πa2 (Ans)


Question 22.

The area of the region (in square units) bounded by the curve x2 = 4y, line x = 2 and x-axis is
A. 1

B. 2/3

C. 4/3

D. 8/3


Answer:

x2 = 4y, at y = 0, x = 0


So, our bounds are x = 0 and x = 2


Area A enclosed is –


A =


=


=


= 8/12, i.e., 2/3 (Ans)


Question 23.

The area bounded by the curve y = f(x), x-axis, and the ordinates x = 1 and x = b is (b – 1) sin (3b + 4). Then, f (x) is
A. (x – 1) cos (3x + 4)

B. sin (3x + 4)

C. sin (3x + 4) +3 (x – 1) cos (3x + 4)

D. none of these


Answer:

So, the area enclosed from x = 1 to x = x (say) is (x – 1) sin (3x + 4)


⇒ A = ∫f(x) dx = F(x) = (x – 1) sin (3x + 4)


⇒ f(x) = F’(x) = sin (3x + 4) + 3(x – 1) cos (3x + 4) (Using u-v rule of differentiation)


(Ans)


Question 24.

The area bounded by the curve y2 = 8x and x2 = 8y is
A.

B.

C.

D.


Answer:

This problem is similar to Problem 2. So we’ll solve it in a similar way.


At points of intersection,


y2 = 8x


(from the other equation)


Let us solve this.



x4 = 512x


⇒ x(x3 – 512) = 0


⇒ x = 0 or x3 = 512, i.e, x = 8


So the points of intersection are (0,0) and (8,8)


Now, let’s compute the area.


If we integrate w.r.t x, we’ll have to integrate the space between the two curves from x = 0 to x = 8.


i.e.,






(Ans)


Question 25.

The area bounded by the parabola y2 = 8x, the x-axis and the latusrectum is
A. 16/3

B. 23/3

C. 32/3

D.


Answer:

Latus rectum of parabola y2 = 4ax is x = a


So, latus rectum of parabola y2 = 8x is x = 2


Therefore, area A enclosed is –


A =


=


=


= 16/3 (Ans)


Question 26.

Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is
A. –9

B.

C.

D.


Answer:

The area A enclosed is –


A =


=


=


= 16/4 + 1/4


= 17/4 (Ans)


Question 27.

The area bounded by the curve y = x |x| and the ordinates x = –1 and x = 1 is given by
A. 0

B. 1/3

C. 2/3

D. 4/3


Answer:

The area A enclosed is –


A =


Now, y = x|x| =


So, |y| = |x|x|| = x2


∴ A =


=


= 1/3 – (-1/3)


= 2/3 (Ans)


Question 28.

The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ π/2 is
A.

B.

C.

D.


Answer:

The area we want is –



We’ll integrate w.r.t y, since area is enclosed by curves and y – axis.


Intersection point is at x = π/4, i.e., y = 1/√2


So, area A enclosed is –


A =


Using Integration by parts –


A =


Putting u = 1 – y2


We get du = -2y dy


A =


=


=


= π/4√2 + 1/√2 – 0 – 1 + 0 – 0 - π/4√2 + 1/√2


= √2 - 1 (Ans)


Question 29.

The area of the circle x2 + y2 = 16 enterior to the parabola y2 = 6x is
A.

B.

C.

D.


Answer:

The area we want is –



At intersection points, y2 = 6x = 16 – x2


Or, x2 + 6x – 16 = 0


i.e., (x + 8)(x – 2) = 0


i.e., x = -8, 2


Now x = -8 ⇒ y2 = 16 – (-8)2 = 16 – 64 = -48 < 0, which is not possible for y ∈ ℝ


So, x = 2


and y2 = 16 – 22


= 16 – 4


= 12


Or, y = √12 = ± 2√3


So, our bounds are y = -2√3 to y = 2√3


The area A enclosed is –


A =


Since both the functions are symmetrical about the x – axis,


A =


=





(Ans)


Question 30.

Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is
A. 2(π –2)

B. π –2

C. 2π –1

D. 2(π + 2)


Answer:

The area we need is –



This area A = area of quadrant OAB – area of triangle OAB


= πr2/4 – � x base x height


= π x 22/4 – � x 2 x 2


= π – 2 (Ans)


Question 31.

Area lying between the curves y2 = 4x and y = 2x is
A. 2/3

B. 1/3

C. 1/4

D. 3/4


Answer:

At the points of intersection,


x = y2/4 = y/2


Or, y2 = 2y


i.e., y2 – 2y = 0


i.e., y(y – 2) = 0


i.e., y = 0, 2


So, area A enclosed is –


A =


=




(Ans)


Question 32.

Area lying in first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2, is
A. π

B. π/2

C. π/3

D. π/4


Answer:

The part of the circle x2 + y2 = 4 in between x = 0 and x = 2 is the semicircle to the right of the y – axis.


And the part of this semicircle in the first quadrant is a quadrant of the circle.


So, area A of the portion is basically the area of a quadrant of the circle.


∴ A = πr2/4


= π × 22/4


= π (Ans)


Question 33.

Area of the region bounded by the cure y2 = 4x, y-axis and the line y = 3, is
A. 2

B. 9/4

C. 9/3

D. 9/2


Answer:

The area A is –


A =


=


= 9/4 (Ans)