Here is a NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals. This solutions covers all questions of Maths Chapter 8 Application of Integrals Class 12 as per CBSE Board guidelines from the latest NCERT book for class 12 maths. Following topics and sub-topics in Class 12 Maths Chapter 8 Application of Integrals are covered.
8.1 Introduction
8.2 Area under Simple Curves
8.2.1 The area of the region bounded by a curve and a line
8.3 Area between Two Curves.

Application of Integrals NCERT Solutions – Class 12 Maths

NCERT Solutions for Class 12 Maths Chapter 8 – Free PDF Download

Exercise 8.1 : Solutions of Questions on Page Number : 365
Q1 : Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
Answer :

The area of the region bounded by the curve, y2 = x, the lines,x = 1 and x = 4, and the x-axis is the area ABCD.


Q2 : Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Answer :

The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.


Q3 : Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Answer :

The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.


Q4 : Find the area of the region bounded by the ellipse
Answer :
The given equation of the ellipse, , can be represented as

It can be observed that the ellipse is symmetrical about x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area of OAB

Therefore, area bounded by the ellipse = 4 × 3π = 12π units


Q5 : Find the area of the region bounded by the ellipse
Answer :
The given equation of the ellipse can be represented as


It can be observed that the ellipse is symmetrical about x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area OAB

Therefore, area bounded by the ellipse =


Q6 : Find the area of the region in the first quadrant enclosed by x-axis, line and the circle
Answer:
The area of the region bounded by the circle, , and the x-axis is the area OAB.


The point of intersection of the line and the circle in the first quadrant is .
Area OAB = Area ΔOCA + Area ACB
Area of OAC
Area of ABC


Q7 : Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line
Answer :
The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, , is the area ABCDA.

It can be observed that the area ABCD is symmetrical about x-axis.
∴ Area ABCD = 2 × Area ABC


Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line, is    units.


Q8 : The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Answer :
The line, x = a, divides the area bounded by parabola and x = 4 into two equal parts.
∴ Area OAD = Area ABCD
∴ Area OAD = Area ABCD

It can be observed that the given area is symmetrical about x-axis.
⇒ Area OED = Area EFCD

From (1) and (2), We obtain,
 

Therefore,the value of a is 


Q9 : Find the area of the region bounded by the parabola y = x2 and
Answer :
The area bounded by the parabola, x2 = y, and the line,, can be represented as

The given area is symmetrical about y-axis.
∴ Area OACO = Area ODBO
The point of intersection of parabola, x2 = y, and line, y = x, is A (1, 1).
Area of OACO = Area ΔOAM – Area OMACO
Area of ΔOAM
Area of OMACO
⇒ Area of OACO = Area of ΔOAM – Area of OMACO

Therefore, required area = units


Q10 : Find the area bounded by the curve x2 = 4y and the line x = 4y – 2
Answer :
The area bounded by the curve, x2 = 4y, and line, x = 4y – 2, is represented by the shaded area OBAO.

Let A and B be the points of intersection of the line and parabola.
Coordinates of point .
Coordinates of point B are (2, 1).
We draw AL and BM perpendicular to x-axis.
It can be observed that,
Area OBAO = Area OBCO + Area OACO … (1)
Then, Area OBCO = Area OMBC – Area OMBO

Similarly, Area OACO = Area OLAC – Area OLAO

Therefore, required area =


Q11 : Find the area of the region bounded by the curve y2 = 4x and the line x = 3
Answer :
The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.

The area OACO is symmetrical about x-axis.
∴ Area of OACO = 2 (Area of OAB)

Therefore, the required area is units.


Q12 : Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is
A. π
B.
C.
D.
Answer :
The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as


Thus, the correct answer is A.


Q13 : Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is
A. 2
B.
C.
D.
Answer :
The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as


Thus, the correct answer is B.


Exercise 8.2 : Solutions of Questions on Page Number : 371


Q1 : Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
Answer :
The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of intersection as.

It can be observed that the required area is symmetrical about y-axis.
∴ Area OBCDO = 2 × Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are.
Therefore, Area OBCO = Area OMBCO – Area OMBO




 


Q3 : Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3
Answer :
The area bounded by the curves, y = x2 + 2, y = x, x = 0, and x = 3, is represented by the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO – Area ODCO


Q4 : Using integration finds the area of the region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).
Answer :
BL and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment AC is

Therefore, from equation (1), we obtain
Area (ΔABC) = (3 + 5 – 4) = 4 units


Q5 : Using integration find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.
Answer :
The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and x = 4.
On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C (4, 9).

It can be observed that,
Area (ΔACB) = Area (OLBAO) – Area (OLCAO)


Q6 : 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 smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is represented by the shaded area ACBA as

It can be observed that,
Area ACBA = Area OACBO – Area (ΔOAB)

Thus, the correct answer is B.


Q7 : Area lying between the curve y2 = 4x and y = 2x is
A.
B.
C.
D.
Answer :
The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded area OBAO as

The points of intersection of these curves are O (0, 0) and A (1, 2).
We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0).
∴ Area OBAO = Area (OCABO) – Area (ΔOCA)


square units
Thus, the correct answer is B.


Exercise Miscellaneous : Solutions of Questions on Page Number : 375


Q1 : Find the area under the given curves and given lines:
(i) y = x2, x = 1, x = 2 and x-axis
(ii) y = x4, x = 1, x = 5 and x -axis
Answer :
The required area is represented by the shaded area ADCBA as


The required area is represented by the shaded area ADCBA as


Q2 : Find the area between the curves y = x and y = x2
Answer :
The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x2, is A (1, 1).
We draw AC perpendicular to x-axis.
∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)


Q3 : Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4
Answer :
The area in the first quadrant bounded by y = 4x2, x = 0, y = 1, and y = 4 is represented by the shaded area ABCDA as


Q4 : Sketch the graph of and evaluate
Answer :
The given equation isThe corresponding values of x and y are given in the following table.

x-6-5– 4-3-2– 10
y3210123

On plotting these points, we obtain the graph of
as follows.

It is known that,


Q5 : Find the area bounded by the curve y = sin x between x = 0 and x = 2π
Answer :
The graph of y = sin x can be drawn as

∴ Required area = Area OABO + Area BCDB


Q6 : Find the area enclosed between the parabola y2 = 4ax and the line y = mx
Answer :
The area enclosed between the parabola, y2 = 4ax, and the line, y = mx, is represented by the shaded area OABO as

The points of intersection of both the curves are (0, 0) and .
We draw AC perpendicular to x-axis.
∴ Area OABO = Area OCABO – Area (ΔOCA)


Q7 : Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Answer :
The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is represented by the shaded area OBAO as

The points of intersection of the given curves are A (-2, 3) and (4, 12).
We draw AC and BD perpendicular to x-axis.
∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)


Q8 : Find the area of the smaller region bounded by the ellipse and the line
Answer :
The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)


Q9 : Find the area of the smaller region bounded by the ellipse and the line
Answer :
The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)


Q10 : Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis
Answer :
The area of the region enclosed by the parabola, x2 = y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as

The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (-1, 1) and C(2, 4).


Q11 : Using the method of integration find the area bounded by the curve
[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]
Answer :
The area bounded by the curve, , is represented by the shaded region ADCB as

The curve intersects the axes at points A (0, 1), B (1, 0), C (0, -1), and D (-1, 0).
It can be observed that the given curve is symmetrical about x-axis and y-axis.
∴ Area ADCB = 4 × Area OBAO


Q12 : Find the area bounded by curves
Answer :
The area bounded by the curves, , is represented by the shaded region as

It can be observed that the required area is symmetrical about y-axis.



= 7 units


Q14 : Using the method of integration find the area of the region bounded by lines:
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Answer :
The given equations of lines are
2x + y = 4 … (1)
3x – 2y = 6 … (2)
And, x – 3y + 5 = 0 … (3)

The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x-axis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)


Q15 : Find the area of the regionAnswer :
The area bounded by the curves, , is represented as

The points of intersection of both the curves are.
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about x-axis.
∴ Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC


Therefore, the required area is units


Q16 : 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 :


Q17 : The area bounded by the curve, x-axis and the ordinates x = -1 and x = 1 is given by
[Hint: y = x2 if x > 0 and y = -x2 if x < 0]
A. 0
B.
C.
D.
Answer :



Thus, the correct answer is C.


Q18 : The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
A.
B.
C.
D.
Answer :
The given equations are
x2 + y2 = 16 … (1)
y2 = 6x … (2)

Area bounded by the circle and parabola

Area of circle = π (r)2
= π (4)2
= 16π units

Thus, the correct answer is C.


Q19 : The area bounded by the y-axis, y = cos x and y = sin x when
A.
B.
C.
D.
Answer :
The given equations are
y = cos x … (1)
And, y = sin x … (2)

Required area = Area (ABLA) + area (OBLO)

Integrating by parts, we obtain

Thus, the correct answer is B.