Application Of Integrals

y² = 9x is the equation of a parabola and y = 3x is a straight line passing through origin

let us first roughly draw those equations

In y² = 9x negative values of x are not allowed hence the graph is on right of X-axis that is the parabola opening to the right

To find point of intersection solve both the equations simultaneously

Put y = 3x in y² = 9x

⇒ (3x)² = 9x

⇒ 9x² = 9x

⇒ x² – x = 0

⇒ x(x – 1) = 0

⇒ x = 0 and x = 1

Put x = 1 and x = 0 in y = 3x we will get y = 3 and y = 0 respectively hence the parabola and straight line intersect at (1, 3) and (0, 0)

We have to find the area between parabola and straight line

That area will be the area under parabola minus the area under the straight line from x = 0 to x = 1 as shown in figure below

⇒ area between parabola and straight line = area under parabola – area under straight line …(i)

Let us calculate the area under parabola

y² = 9x

⇒ y = 3√x

Integrate from 0 to 1

Now let us calculate area under the line y = 3x that is area of triangle OAB

⇒ y = 3x

Integrate from 0 to 1

Using (i)

⇒ area between parabola and straight line = 2 – = 1/2 unit²

Hence area of the region bounded by the curves y² = 9x and y = 3x is 1/2 unit²

In y² = 2px parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

Similarly for x² = 2py parabola it is not defined for negative values of y hence parabola will be above X-axis opening upwards and passing through (0, 0)

Let us find the point of intersection by solving the equations y² = 2px and x² = 2py simultaneously

Put in y² = 2px

⇒ x⁴ = 8p³x

⇒ x³ = 8p³

⇒ x = 2p

Put x = 2p in y² = 2px

⇒ y² = 2p(2p)

⇒ y = 2p

Hence the intersection point of two parabola is (2p, 2p)

We require the area between the two parabolas

⇒ area bounded by two parabolas given = area under parabola y² = 2px – area under parabola x² = 2py …(i)

Let us find area under parabola y² = 2px

⇒ y = √2p√x

Integrate from 0 to 2p

Now let us find area under parabola x² = 2py

⇒ x² = 2py

Integrate from 0 to 2p

Using (i)

⇒ area bounded by two parabolas given =

⇒ area bounded by two parabolas given =

Hence area is unit²

Roughly plot the curve y = x³ and the lines y = x + 6 and x = 0

x = 0 means Y-axis

We have to find the area between the curve y = x³ and the line y = x + 6 and Y-axis as shown

Solve y = x + 6 and y = x³ to find the intersection point

Put y = x³ in y = x + 6

⇒ x³ = x + 6

⇒ x³ – x – 6 = 0

Mentally checking if 0,1,2 satisfy the cubic we get that one root is 2 hence x – 2 is a factor

Take that factor out

⇒ (x – 2)(x² + 2x + 3) = 0

Observe that x² + 2x + 3 don’t have real roots

Hence x = 2

Put x = 2 in y = x + 6 we get y = 8

Hence both curves y = x³ and y = x + 2 intersect at (2, 8)

As we have to find area on Y-axis we should integrate x = f(y) that is here we are taking a horizontal strip of length dy

So the area bounded will be

⇒ area bounded = area by y = x³ on Y-axis – area by y = x + 6 on Y-axis …(i)

Let us find the area under y = x³

As we need in terms of x = f(y)

Integrate from 0 to 8

Now let us find area under y = x + 6 on Y axis

Observe in figure that we have to find area from 6 to 8 because the line intersects Y-axis at 6 and upto 8 because that is the y-coordinate where the curve and line intersects

⇒ x = y – 6

Integrate from 6 to 8

Using (i)

⇒ area bounded = 12 – 2 = 10 unit²

Hence area bounded by given curves is 10 unit²

In y² = 4x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

Similarly for x² = 4y parabola it is not defined for negative values of y hence parabola will be above X-axis opening upwards and passing through (0, 0)

Let us find the point of intersection by solving the equations y² = 4x and x² = 4y simultaneously

Put in y² = 4x

⇒ x⁴ = 64x

⇒ x³ = 64

⇒ x = 4

Put x = 4 in y² = 4x

⇒ y² = 4(4)

⇒ y = 4

Hence the intersection point of two parabola is (4, 4)

We require the area between the two parabolas

⇒ area bounded by two parabolas given = area under parabola y² = 4x – area under parabola x² = 4y …(i)

Let us find area under parabola y² = 4x

⇒ y = 2√x

Integrate from 0 to 4

Now let us find area under parabola x² = 4y

⇒ x² = 4y

Integrate from 0 to 4

Using (i)

⇒ area bounded by two parabolas given =

⇒ area bounded by two parabolas given =

Hence area is unit²

In y² = 9x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

And y = x is a straight line passing through origin

We have to find area between y² = 9x and y = x shown below

To find intersection point of parabola and line solve parabola equation and line equation simultaneously

Put y = x in y² = 9x

⇒ x² = 9x

⇒ x = 9

Put x = 9 in y = x we get y = 9

Hence point of intersection is (9, 9)

⇒ area between parabola and line = area under parabola – area under line …(i)

Let us find area under parabola

⇒ y² = 9x

⇒ y = 3√x

Integrate from 0 to 9

Now let us find area under straight line y = x

y = x

Integrate from 0 to 9

Using (i)

⇒ area between parabola and line = 54 – 40.5 = 13.5 unit²

Hence area bounded is 13.5 unit²

x² = y parabola is not defined for negative values of y hence parabola will be above X-axis opening upwards and passing through (0, 0)

y = x + 2 is a straight line

plot these equations and we have to find the area enclosed between them

To find the intersection points solve the equations x² = y and y = x + 2 simultaneously

Put y = x + 2 in x² = y

⇒ x² = x + 2

⇒ x² – x – 2 = 0

⇒ x² – 2x + x – 2 = 0

⇒ x(x – 2) + 1(x – 2) = 0

⇒ (x + 1)(x – 2) = 0

⇒ x = -1 and x = 2

Put x = -1 and x = 2 in x² = y we get y = 1 and y = 4 respectively

Hence (-1, 1) and (2, 4) are the points at which the line intersects the parabola

⇒ area enclosed by line and parabola = area under line – area under parabola …(i)

Let us find area under line y = x + 2

⇒ y = x + 2

Integrate from -1 to 2

Now let us find area under the parabola

x² = y

⇒ y = x²

Integrate from -1 to 2

Using (i)

⇒ area enclosed by line and parabola = – 3 = unit²

Hence area enclosed is unit²

In y² = 8x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

And x = 2 is a straight line parallel to Y-axis

Plot the equation y² = 8x and x = 2 and the area as shown

So we have to integrate y² = 8x that is y = 2√2√x from 0 to 2

But observe that integrating the parabola equation from 0 to 2 will give the area OBC that is area under the parabola in 1^st quadrant

We have to find the whole shaded region ODBC

Parabola y² = 8x is symmetric about X-axis hence the area above X-axis that is in 1^st quadrant is equal to area below X-axis that is in 4^th quadrant hence areaOBC = areaOBD

Hence area bounded by parabola and line will be twice the area which we will get by integration parabola from 0 to 2

areaODBC = 2 × areaOBC …(i)

let us find area under parabola

⇒ y² = 8x

⇒ y = 2√2√x

Integrate from 0 to 2

Using (i)

The shaded areaOCBD = 2 ×

Hence area bounded = unit²

Square both sides

⇒ y² = 4 – x²

⇒ x² + y² = 4

⇒ x² + y² = 2²

This is equation of circle with center origin and radius 2

Now in -2 ≤ x ≤ 2 and y ≥ 0 which means x and y both positive or x negative and y positive hence the curve has to be above X-axis in 1^st and 2^nd quadrant

Hence the graph of will be graph of circle x² + y² = 2² lying only above X-axis

Now equation of X-axis is y = 0

To find point of intersection of circle with X-axis put y = 0 in circle equation

⇒ x² = 4

⇒ x = ±2

Hence the intersection points with X-axis are (-2, 0) and (2, 0)

Hence the area is shown as below

Now let us find the area

Integrate from -2 to 2

Using uv rule of integration where u and v are functions of x

Here and v = 1

Hence

But

We know that

Hence area is 2π unit²

y = 2√x

squaring both sides we get

⇒ y² = 4x

y² = 4x is a equation of parabola

In y² = 4x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

Now for y = 2√x x and y both has to be greater than 0 that is both positive hence both lie in 1^st quadrant

Hence y = 2√x will be parabolic curve of y² = 4x only in 1^st quadrant

x = 0 is equation of Y-axis and x = 1 is a line parallel to Y-axis passing through (1, 0)

Plot equations y = 2√x and x = 1

So we have to integrate y = 2√x from 0 to 1

let us find area under parabola

⇒ y = 2√x

Integrate from 0 to 1

Hence area bounded = unit²

Plot the line 2y = 5x + 7

We need two points to plot the line, we can get those two points by putting x = 0 and then putting y = 0

Put x = 0

⇒ 2y = 5(0) + 7

Put y = 0

⇒ 2(0) = 5x + 7

⇒ 5x = -7

Hence and are the required two points to draw the line 2y = 5x + 7

x = 2 and x = 8 are lines parallel to Y-axis passing through (2, 0) and (8, 0) respectively

Plot lines 2y = 5x + 7, x = 2 and x = 8

We have to find area under 2y = 5x + 7 that is y = 1/2(5x + 7) from 2 to 8

⇒ y = 1/2(5x + 7)

Integrate from 2 to 8

Hence the area bounded by given lines is 96 unit²

Squaring both sides

⇒ y² = x – 1

y² = x – 1 is equation of a parabola

In y² = x – 1 parabola it is not defined for values of x less than 1 hence the parabola will be to the right of x = 1 passing through (1, 0)

Now observe that in x ≥ 1 and y has to positive because of square root hence x and y both positive hence the parabola will be drawn only in 1^st quadrant

We have to plot the curve in [1, 5] so just draw the parabolic curve from x = 1 to x = 5 in 1^st quadrant

x = 1 and x = 5 are lines parallel to Y-axis

So we have to integrate from 1 to 5

let us find area under parabolic curve

Integrate from 1 to 5

Hence area bounded = unit²

Squaring both sides

⇒ y² = a² - x²

⇒ x² + y² = a²

This is equation of circle having center as (0, 0) and radius a

Now in -a ≤ x ≤ a and y ≥ 0 which means x and y both positive or x negative and y positive hence the curve has to be above X-axis in 1^st and 2^nd quadrant

x = 0 is equation of Y-axis and x = a is a line parallel to Y-axis passing through (a, 0)

So we have to integrate from 0 to a

let us find area under curve

Integrate from 0 to a

Using uv rule of integration where u and v are functions of x

Here and v = 1

Hence

But

We know that

Hence area bounded = unit²

y = √x

squaring both sides

⇒ y² = x

In y² = x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

Now y = √x means y and x both has to be positive hence both lie in 1^st quadrant hence y = √x will be part of y² = x which is lying only in 1^st quadrant

And y = x is a straight line passing through origin

We have to find area between y = √x and y = x shown below

To find intersection point of parabola and line solve parabola equation and line equation simultaneously

Put y = x in y² = x

⇒ x² = x

⇒ x = 1

Put x = 1 in y = x we get y = 1

Hence point of intersection is (1, 1)

⇒ area between parabolic curve and line = area under parabolic curve – area under line …(i)

Let us find area under parabolic curve

⇒ y = √x

Integrate from 0 to 1

Now let us find area under straight line y = x

y = x

Integrate from 0 to 1

Using (i)

⇒ area between parabolic curve and line = = unit²

Hence area bounded is unit²

y = -x²⇒ x² = -y

x² = -y parabola is not defined for positive values of y hence parabola will be below X-axis opening downwards and passing through (0, 0)

x + y + 2 = 0 is a straight line

To find point of intersection of parabola and straight line solve the parabola equation and the straight line equation simultaneously

Put y = -(x + 2) in x² = -y

⇒ x² = -(-(x + 2))

⇒ x² = x + 2

⇒ x² – x – 2 = 0

⇒ x² – 2x + x – 2 = 0

⇒ x(x – 2) + 1(x – 2) = 0

⇒ (x + 1)(x – 2) = 0

⇒ x = -1 and x = 2

Put x = -1 in x² = -y

⇒ (-1)² = -y

⇒ y = -1

Put x = 2 in x² = -y

⇒ 2² = -y

⇒ y = -4

Hence the parabola and line intersects at (-1, -1) and (2, -4)

Plot the parabola and straight line and the bounded area is as shown

⇒ area enclosed by line and parabola = area under line – area under parabola …(i)

Let us find area under the straight line

x + y + 2 = 0

⇒ y = -(x + 2)

Integrate from -1 to 2

Now let us find area under parabola

x² = -y

⇒ y = -x²

Integrate from -1 to 2

Using (i)

⇒ area enclosed by line and parabola = = unit²

We are getting negative sign because the area is below X-axis as seen in figure

Hence area enclosed is unit²

y = √x

Squaring both sides

⇒ y² = x

In y² = x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

But we have to plot y = √x which means x,y both can only be positive hence the graph has to be only in 1^st quadrant

Hence y = √x will be part of parabola y² = x only above X-axis in 1^st quadrant

(why can’t y be negative? Because the symbol square root itself denotes primary root which means positive root)

x = 2y + 3 is a straight line

To get the intersection points of the straight line and the parabola solve the parabola equation and straight line equation simultaneously

Put x = 2y + 3 in y² = x

⇒ y² = 2y + 3

⇒ y² – 2y – 3 = 0

⇒ y² – 3y + y – 3 = 0

⇒ y(y – 3) + 1(y – 3) = 0

⇒ (y + 1)(y – 3) = 0

⇒ y = -1 and y = 3

Put y = 3 in y² = x

⇒ x = 3²

⇒ x = 9

y = -1 doesn’t matter because we are in 1^st quadrant

Hence the parabola and straight line intersects at (9, 3)

Plot roughly the parabola and the line and the required area is the shaded region as shown

The point of intersection of straight line with X-axis can be calculated by putting y = 0 in line equation

⇒ x = 2(0) + 3 ⇒ x = 3

Observe that

⇒ area bounded = area under y = √x – area under straight line …(i)

Let us fling area under y = √x

Integrate from 0 to 9

Now area under the straight line x = 2y + 3

Integrate from 3 to 9

(why 3 to 9? Because the line cuts X-axis at 3 and the curve y = √x at 9)

Using (i)

⇒ area bounded = 18 – 9 = 9 unit²

Hence area bounded by curve and straight line is 9 unit²

y² = 2x is a parabola

In y² = 2x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

x² + y² = 4x is equation of circle

The general equation of circle is given by x² + y² + 2gx + 2fy + c = 0

Centre of circle is (-g, -f) and radius is

In x² + y² – 4x = 0, 2g = -4 ⇒ g = -2 and f = c = 0

Hence center is (-(-2), 0) that is (2, 0) and radius is which is 2

Hence plot the circle and parabola roughly and to mark the intersection points solve the parabola equation and circle equation simultaneously

Put y² = 2x in x² + y² = 4x

⇒ x² + 2x = 4x

⇒ x² – 2x = 0

⇒ x(x – 2) = 0

⇒ x = 0 and x = 2

Put x = 2 in y² = 2x

⇒ y² = 2(2)

⇒ y = ±2

Hence the circle and parabola intersects at (0, 0), (2, 2) and (2, -2)

By integrating we will get the area only in the 1^st quadrant but given parabola and circle are symmetric about X-axis hence area above and below X-axis will be equal

Hence area of shaded region will be twice the area we will get by integration in 1^st quadrant …(a)

Observe that

⇒ area of shaded in 1^st quadrant = area under circle – area under parabola …(i)

Let us find area under circle

x² + y² = 4x

⇒ y² = 4x – x²

Integrate from 0 to 2

Using

Now let us find area under parabola

⇒ y² = 2x

⇒ y = √2√x

Integrate from 0 to 2

Using (i)

⇒ area of shaded in 1^st quadrant = unit²

Using (a)

The area required of shaded region = unit²

Plot the graph of sinx from 0 to 2π and the required area is shaded

Now observe that the area from 0 to π is above X-axis and the area from π to 2π is below X-axis

The area below X-axis will be negative

Also the areas under sinx from 0 to π and π to 2π are equal in magnitude but they will have opposite sign

Hence if we integrate sinx from 0 to 2π the two areas will cancel out each other as they have opposite signs and we will end up on 0

So either find area under sinx from 0 to π and multiply it by 2 or split the limit 0 to 2π into 0 to π and π to 2π

Here we will split the limit

y = sinx

Integrating from 0 to 2π

Because where c ∈ (a, b)

Here π ∈ (0, 2π)

Also now observe that when x is from π to 2π sinx is negative

Hence for π to 2π sinx will become -sinx

Hence area bounded by sinx from 0 to 2π is 4 unit²

Given;

The triangle whose vertices are (–1, 1), (0, 5) and (3, 2).

Let P(−1,1), Q(0,5) and R(3,2)

⇒ Equation of PQ is

⇒ Equation of QR is

⇒ Equation of RP is

Area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by .

Required area

Given;

The region {(x, y) : y² ≤ 6ax and x² + y² ≤ 16a²}

By solving the equations: y² ≤ 6ax and x² + y² ≤ 16a²

Through substituting for y²

⇒ x² + 6ax = 16a²

⇒ (x − 2a) (x + 8a) = 0

∴ x = 2a.

[as x = -8a is not possible]

Area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by .

[By the symmetry of the image w.r.t x axis]

Required area =

Given;

The lines x + 2y = 2

x = 2 – 2y …. (1)

y – x = 1

⇒ x = y - 1 …. (2)

and 2x + y = 7 …. (3)

Equate the values of x from 1 and 2 to get,

2 – 2y = y – 1

2+1 = y + 2y

3 = 3y

y = 1

put the value of y in (2) to get,

x = 1 -1

= 0

So, intersection point is (0,1).

By solving these equations, we get the points of intersection as (0,1), (2,3) and (4,-1).

Area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by .

Required area

=3 + (24 − 12) − (12 − 3) = 6 sq.units

Given;

The lines y = 4x + 5, y = 5 – x and 4y = x + 5.

By solving these equations,

y = 4x + 5 …… (1)

y = 5 − x …… (2)

4y = x + 5 …… (3)

From (1) and (2);

4x + 5 = 5 − x

⇒ x = 0; ∴ y = 5 − x = 5

From (2) and (3);

4 (5 − x) = x + 5

⇒ x = 3; ∴ y = 5 − x = 2

From (1) and (3);

4 (4x + 5) = x + 5

⇒ x = −1; ∴ y = 4x + 5 = 1

we get the points of intersection as (0,5), (3,2) and (-1,1).

Area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by .

Required area

Given;

The curve y = 2cos x and the x-axis from x = 0 to x = 2π.

Area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by .

From the figure;

Required area

= 8 sq.units

Given;

Curve y = 1 + |x +1|, x = –3, x = 3, y = 0.

Area of the region bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by .

Required area

= 4 + 12 = 16 sq.units

Given;

y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2

⇒ sin x = cos x

Required area

= (√2 − 1) sq.units

Given;

The curve x² = 4y and the straight line x = 4y – 2

By substituting for 4y;

⇒ x = x² − 2

⇒ x² − x − 2 = 0

⇒ (x + 1) (x − 2) = 0

∴ x = −1, 2.

Required area

Given;

The curve and x-axis; y=0

∴ x = ±4

Required area

= 8π sq.units

Given;

The x-axis, the line y = x and the circle x² + y² = 32.

By substitution;

⇒ x² + x² = 32

∴ x = 4

Radius of the circle

Required area

= (8 + 8π − 8 − 4π)

= 4π sq.units

Given;

The curve y = cos x and x = 0 and x = π

Required area

= 2 (1 − 0)

= 2 sq.units

Given;

Parabola y² = x and the straight line 2y = x

By substituting for x;

⇒ y² = 2y

⇒ y (y − 2) = 0

∴ y = 0, 2.

∴ x = 2y = 0, 4.

Required area

Given;

The curve y = sin x between the ordinates x = 0, x = π/2 and the x-axis

Required area

= 0 − (−1)

= 1 sq.units

Given;

The ellipse

By the symmetry of the ellipse with x-axis and y-axis.

Required area

= 20π sq.units

Given;

The circle x² + y² = 1

By the symmetry of the circle with x-axis and y-axis.

Required area

= π sq.units

Given;

The curve y = x + 1 and the lines x = 2 and x = 3

Required area

Given;

The curve x = 2y + 3 and the y lines; y = 1 and y = –1

Required area

= (1 + 3 − 1 +3)

= 6 sq.units