NCERT Exemplar: Application of Integrals

# NCERT Exemplar: Application of Integrals | Mathematics (Maths) for JEE Main & Advanced PDF Download

Q.1. Find the area of the region bounded by the curves y2 = 9x, y = 3x.
Ans.
We have, y2 = 9x, y = 3x
Solving the two equations, we have

(3x)2 = 9x
⇒ 9x2 – 9x = 0 ⇒ 9x (x - 1) = 0
∴ x = 0, 1 Area of the shaded region
= ar (region OAB) – ar (DOAB)

Hence, the required area =sq. units.

Q.2. Find the area of the region bounded by the parabola y2 = 2px, and x2 = 2py.
Ans.
We are given that: x2 = 2py ...(i)
and y2 = 2px ...(ii)
From eqn. (i) we get y =

Putting the value of y in eqn. (ii) we have

⇒ x4 = 8p3x ⇒ x4 – 8p3x = 0
⇒ x (x3 – 8p3) = 0 ∴ x = 0, 2p
Required area = Area of the region (OCBA – ODBA)

Hence, the required area =sq. units.

Q.3. Find the area of the region bounded by the curve y = x3 and y = x + 6 and x = 0.
Ans.
We are given that: y = x3, y = x + 6 and x = 0
Solving y = x3 and y = x + 6, we get
x + 6 = x3
⇒ x3 – x – 6 = 0
⇒ x2 (x – 2) + 2x (x – 2) + 3 (x – 2) = 0
⇒ (x – 2) (x2 + 2x + 3) = 0
x2 + 2x + 3 = 0 has no real roots. ∴ x = 2
∴ Required area of the shaded region

10 sq. units.

Q.4. Find the area of the region bounded by the curve y2 = 4x, x2 = 4y.
Ans.
We have y2 = 4x and x2 = 4y.

⇒ x4 = 64x ⇒ x4 – 64x = 0
⇒ x(x3 – 64) = 0
∴ x = 0, x = 4
Required area

Hence, the required area =

Q.5. Find the area of the region included between y2 = 9x and y = x
Ans.
Given that: y2 = 9x ...(i)
and y = x...(ii)
Solving eqns. (i) and (ii) we have
x2 = 9x ⇒ x2 – 9x = 0
x (x – 9) = 0 ∴ x = 0, 9
Required area

Hence, the required area =

Q.6. Find the area of the region enclosed by the parabola x2 = y and the line y = x + 2
Ans.
Here, x2 = y and y = x + 2
∴ x2 = x + 2
⇒ x2 – x – 2 = 0
⇒ x2 – 2x + x – 2 = 0
⇒ x(x –2) + 1 (x – 2) = 0
⇒ (x – 2) (x + 1) = 0
∴ x = –1, 2
Graph of y = x + 2

Area of the required region

Hence, the required area =

Q.7. Find the area of region bounded by the line x = 2 and the parabola y2 = 8x
Ans.
Here, y2 = 8x and x = 2
y2 = 8(2) = 16
∴ y = ±4
Required area

Hence, the area of the region =

Q.8. Sketch the region {(x, 0) : y =and x-axis. Find the area of the region using integration.
Ans.
Given that {(x, 0) : y =
⇒ y2 = 4 – x2
⇒ x2 + y2 = 4 which is a circle.

Required area

[Since circle is symmetrical about y-axis]

Hence, the required area = 2π sq. units.

Q.9. Calculate the area under the curve y = 2 √x included between the lines x = 0 and x = 1.
Ans.
Given the curves y = 2√x , x = 0 and x = 1.
y = 2√x ⇒ y2 = 4x (Parabola)

Required area

Hence, required area =

Q.10. Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8.
Ans.
Given that: 2y = 5x + 7, x-axis, x = 2 and x = 8.
Let us draw the graph of 2y = 5x + 7 ⇒ y =

Area of the required shaded region

Hence, the required area = 96 sq. units.

Q.11. Draw a rough sketch of the curve y =in the interval [1, 5]. Find the area under the curve and between the lines x = 1 and x = 5.
Ans.
Here, we have y =

⇒ y2 = x – 1 (Parabola)
Area of the required region

Hence, the required area =

Q.12. Determine the area under the curve y = included between the lines x = 0 and x = a.
Ans.
Here, we are given y =

⇒ y2 = a2 – x2

⇒ x2 + y2 = a2
Area of the shaded region

Hence, the required area =

Q.13. Find the area of the region bounded by y = √x and y = x.
Ans.
We are given the equations of curve y = √x and line y = x.
Solving y = √x ⇒ y2 = x and y = x, we get
x2 = x ⇒ x2 – x = 0
⇒ x (x – 1) = 0 ∴ x = 0, 1
Required area of the shaded region

Hence, the required area =

Q.14. Find the area enclosed by the curve y = –x2 and the straight lilne x + y + 2 = 0.
Ans.
We are given that y = –x2 or x2 = –y
and the line x + y + 2 = 0
Solving the two equations, we get
x  – x2 + 2 = 0
⇒ x2  – x – 2 = 0
⇒ x2 – 2x + x – 2 = 0
⇒ x (x – 2) + 1 (x – 2) = 0
⇒ (x – 2) (x + 1) = 0
∴ x = –1, 2
Area of the required shaded region

Q.15. Find the area bounded by the curve y = √x , x = 2y + 3 in the first quadrant and x-axis.
Ans.
Given that: y = √x , x = 2y + 3, first quadrant and x-axis.
Solving y = √x and x = 2y + 3, we get
y =  ⇒ 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, 3
Area of shaded region

= 18 - 9 = 9 sq. units
Hence, the required area = 9 sq. units.

Long Answer (L.A.)
Q.16. Find the area of the region bounded by the curve y2 = 2x and x2 + y2 = 4x.
Ans.
Equations of the curves are given by
x2 + y2 = 4x ...(i)
and y2 = 2x ...(ii)
⇒ x2 – 4x + y2 = 0
⇒ x2 – 4x + 4 – 4 + y2 = 0
⇒  (x – 2)2 + y2 = 4
Clearly it is the equation of a circle having its centre (2, 0) and radius 2.
Solving x2 + y2 = 4x and y2 = 2x
x2 + 2x = 4x
⇒ x2 + 2x – 4x = 0
⇒  x2 – 2x = 0
⇒ x (x – 2) = 0
∴ x = 0, 2

Area of the required region

[∴ Parabola and circle both are symmetrical about x-axis.]

sq. units
Hence, the required area =

Q.17. Find the area bounded by the curve y = sinx between x = 0 and x = 2π.
Ans.
Required area =

Q.18. Find the area of region bounded by the triangle whose vertices are (–1, 1), (0, 5) and (3, 2), using integration.
Ans.
The coordinates of the vertices of ΔABC are given by A(–1, 1), B (0, 5) and C (3, 2).

Equation of AB is y – 1 =
⇒ y – 1 = 4x + 4
∴ y = 4x + 4 + 1 ⇒ y = 4x + 5 ...(i)
Equation of BC is y – 5 =
⇒ y – 5 = –x
∴ y = 5 – x ...(ii)
Equation of CA is

Area of ΔABC

Q.19. Draw a rough sketch of the region {(x, y) : y2 ≤ 6ax and x2 + y2 ≤ 16a2}. Also find the area of the region sketched using method of integration.
Ans.
Given that:
{(x, y) : y2 ≤ 6ax and x2 + y2 ≤ 16a2}
Equation of Parabola is
y2 = 6ax ...(i)
and equation of circle is
x2 + y2 ≤ 16a2 ...(ii)
Solving eqns. (i) and (ii) we get
x2 + 6ax = 16a2
⇒ x2 + 6ax – 16a2 = 0
⇒ x2 + 8ax – 2ax – 16a2 = 0
⇒ x(x + 8a) – 2a (x + 8a) = 0
⇒(x + 8a) (x – 2a) = 0
∴ x = 2a and x = – 8a. (Rejected as it is out of region)

Area of the required shaded region

Hence, required area =

Q.20. Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.
Ans.
Given that: x + 2y = 2 ...(i)
y – x = 1 ...(ii)
and 2x + y = 7 ...(iii)

Solving eqns. (ii) and (iii) we get
y = 1 + x
∴ 2x + 1 + x = 7
3x = 6
⇒ x = 2
∴ y = 1 + 2  = 3
Coordinates of B = (2, 3)

Solving eqns. (i) and (iii)
we get
x + 2y = 2
∴ x = 2 – 2y
2x + y = 7
2(2 – 2y) + y = 7
⇒ 4 – 4y + y = 7 ⇒ - 3y = 3
∴ y = –1 and x = 4
∴ Coordinates of C = (4, – 1) and coordinates of A = (0, 1).
Taking the limits on y-axis, we get

Hence, the required area = 6 sq. units.

Q.21. Find the area bounded by the lines y = 4x + 5, y = 5 – x and 4y = x + 5.
Ans.
Given that y = 4x + 5 ...(i)
y = 5 – x ...(ii)
and 4y = x + 5 ...(iii)

Solving eq. (i) and (ii) we get
4x + 5 = 5 – x
⇒ x = 0 and y = 5
∴ Coordinates of A = (0, 5)
Solving eq. (ii) and (iii)
y = 5 – x
4y = x + 5
5y = 10
∴ y = 2 and x = 3
∴ Coordinates of B = (3, 2)
Solving eq. (i) and (iii)
y = 4x + 5
4y = x + 5
⇒ 4 (4x + 5) = x + 5
⇒ 16x + 20 = x + 5 ⇒ 15x = - 15
∴ x = –1 and y = 1
∴ Coordinates of C = (–1, 1).
∴ Area of required regions

Hence, the required area =

Q.22. Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2π.
Ans.
Given equation of the curve is y = 2 cos x

∴ Area of the shaded region

Q.23. Draw a rough sketch of the given curve y = 1 + |x +1|, x = –3, x = 3, y = 0 and find the area of the region bounded by them, using integration.
Ans.
Given equations are y = 1 + |x + 1|, x = –3
and x  = 3, y = 0
Taking y = 1 + |x + 1|
⇒ y = 1 + x + 1
⇒ y = x + 2 and y = 1 – x – 1
⇒ y = –x
On solving we get x = –1

Area of the required regions

Hence, the required area = 16 sq. units.

Objective Type Questions
Q.24. The area of the region bounded by the y-axis, y = cosx and y = sinx, 0 ≤ x ≤
(a) √2 sq. units
(b) ( √2 + 1) sq. units
(c) ( √2 - 1) sq. units
(d) (2√2 - 1) sq. units
Ans. (c)
Solution .
Given that y-axis, y = cos x, y = sin x,

Required area

Hence, the correct option is (c).

Q.25. The area of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is
(a) 3/8 sq. units
(b) 5/8 sq. units
(c) 7/8 sq. units
(d) 9/8 sq. units
Ans. (d)
Solution .
Given that: The equation of parabola is x2 = 4y ...(i)
and equation of straight line is x = 4y – 2 ...(ii)

Solving eqn. (i) and (ii) we get

⇒ x = x2 – 2
⇒ x2 – x – 2 = 0 ⇒ x2 – 2x + x – 2 = 0
⇒ x (x – 2) + 1 (x – 2) = 0 ⇒ (x – 2) (x + 1) = 0 ∴ x = –1, x = 2
Required area =

Hence, the correct option is (d).

Q.26. The area of the region bounded by the curve y =and x-axis is
(a) 8 sq units
(b) 20π sq units
(c) 16π sq units
(d) 256π sq units
Ans. (a)
Solution .
Here, equation of curve is y =
Required area

Hence, the correct option is (a).

Q.27. Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32 is
(a) 16π sq units
(b) 4π sq units
(c) 32π sq units
(d) 24 sq units
Ans. (b)
Solution .

Given equation of circle is  x2 + y2 = 32 ⇒ x2 + y2 = (4√2 )2
and the line is y = x and the x-axis.
Solving the two equations we have

x2 + x2 = 32
⇒ 2x2 = 32
⇒ x2 = 16
∴ x = ± 4
Required area

Hence, the correct option is (b).

Q.28. Area of the region bounded by the curve y = cosx between x = 0 and x = π is
(a) 2 sq. units
(b) 4 sq. units
(c) 3 sq. units
(d) 1 sq. units
Ans. (a)
Solution .

Given that: y = cos x, x = 0, x = π
Required area

Hence, the correct option is (a).

Q.29. The area of the region bounded by parabola y2 = x and the straight line 2y = x is
(a) 4/3 sq. units
(b) 1 sq. unit
(c) 2/3 sq. units
(d) 1/3 sq. units
Ans. (a)
Solution .

Given equation of parabola is y2 = x ...(i)
and equation of straight line is 2y = x ...(ii)
Solving eqns. (i) and (ii) we get

⇒ x(x – 4) = 0 ∴ x = 0, 4
Required area

Hence, the correct answer is (a).

Q.30. The area of the region bounded by the curve y = sinx between the ordinates x = 0,and the x-axis is
(a) 2 sq. units
(b) 4 sq. units
(c) 3 sq. units
(d) 1 sq. units
Ans. (d)
Solution .

Given equation of curve is y = sin x between x = 0 and x =

Area of required region

Hence, the correct answer is (d).

Q.31. The area of the region bounded by the ellipse is
(a) 20π sq units
(b) 20π2 sq units
(c) 16π2 sq units
(d) 25 π sq units
Ans. (a)
Solution .

Given equation of ellipse is

∴ Since the ellipse is symmetrical about the axes.
∴ Required area

Hence, the correct answer is (a).

Q.32. The area of the region bounded by the circle x2 + y2 = 1 is
(a) 2π sq units
(b) π sq units
(c) 3π sq units
(d) 4π sq units
Ans. (b)
Solution .

Given equation of circle is
x2 + y2 = 1  y =

Since the circle is symmetrical about the axes.

Hence, the correct answer is (b).

Q.33. The area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is
(a) 7/2 sq. units
(b) 9/2 sq. units
(c) 11/2 sq. units
(d) 13/2 sq. units
Ans. (a)
Solution .

Given equation of lines are
y = x + 1, x = 2 and x = 3
Required area

Hence, the correct option is (a).

Q.34. The area of the region bounded by the curve x = 2y + 3 and the y lines. y = 1 and y = –1 is

(a) 4 sq. units
(b) 3/2 sq. units

(c) 6 sq. units
(d) 8 sq. units

Ans. (c)
Solution .

Given equations of lines are x = 2y + 3,  y = 1 and y = –1

Required area

Hence, the correct answer is (c).

The document NCERT Exemplar: Application of Integrals | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

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## Mathematics (Maths) for JEE Main & Advanced

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