NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 1.
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
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 1
i. The given curves are y = x² , x = 1 and x = 2. The required area lies between the lines x = 1 and x = 2. Hence the limits of integration are 1 and 2.
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 2

ii. The given curves are y = x4, x = 1 and x = 5. The required area lies between the lines x = 1 and x = 5. Hence the limits of integration are 1 and 5.
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 3

Question 2.
Find the area between the curves y = x and y = x².
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 4
y = x … (1)
y = x² … (2)
(2) → x = x²
⇒ x² – x = 0 ⇒ x(x – 1) = 0
⇒ x = 0 and x = 1
Hence the limits of integration are 0 and 1. Required area = (Area under the line y = x) – (Area under the parabola y = x²)
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 5

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 3.
Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4.
Solution:
The given curves are y = 4x²
⇒ x² = \(\frac { y }{ 4 }\) ⇒ x = \(\frac{1}{2} \sqrt{y}\)
x = 0, y= 1, y = 4
The required area lies in the first quadrant between the lines y= 1 and y = 4
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 6
Hence the limits of integration are 1 and 4.
Required area = Shaded area
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 7

Question 4.
Sketch the graph of y = |x + 3| and evaluate \(\int_{-6}^{0}|x+3| d x\).
Solution:
y = |x + 3| is redefined as follows,
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 8

Question 5.
Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
Solution:
The area bounded by the curve y = sinx between x = 0 and x = 2π is shaded in the figure.1
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 9
The area bounded by the curve between x = 0 and x = π is same as the area bounded by it between x = π and x = 2π
Required area 2 x Area of the shaded
region above x – axis between x = 0
and π = 2\(2 \int_{0}^{\pi} y d x=2 \int_{0}^{\pi} \sin x d x\)
= 2\([-\cos x]_{0}^{\pi}=-2[\cos x]_{0}^{\pi}\)
= – 2(cos π – cos0) = – 2(- 1 – 1) = 4 sq.unit

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 6.
Find the area enclosed between the parabola y² = 4ax and the line y = mx.
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 10
The curves are y² = 4ax … (1)
and y = mx … (2)
Solving (1) and (2), we get
m²x² = 4ax ⇒ m²x² – 4ax = 0
i.e., x(m²x – 4a) = 0 ⇒ x = 0 or x = \(\frac{4 a}{m^{2}}\)
The x coordinate of the point of intersection
are x = 0 and x = \(\frac{4 a}{m^{2}}\)
Required area = (Area under the parabola y² = 4ax) – (Area under the line y = mx)
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 11

Question 7.
Find the area enclosed by the parabola 4y = 3x² and the line 2y = 3x+ 12.
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 12
The curves are 4y = 3x²
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 13
∴ The points of intersection are (4, 12) and (-2, 3)
Required area is shaded in the figure
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 14

Question 8.
Find the area of the smaller region bounded by the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1 and the line \(\frac{x}{3}+\frac{y}{2}=1\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 15
\(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1 is the ellipse with centre at origin intersecting the positive x-axis at A(3, 0) and positive y-axis at B(0, 2)
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 16
The required area is shaded in the figure.
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 17

Question 9.
Find the area of the smaller region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) and the line \(\frac{x}{a}+\frac{y}{b}\) = 1.
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 18
The ellipse intersects the positive x-axis at A (a, 0) and y-axis at B(0, b).
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 19
Required area is shaded in the figure.
Required area = Area under the ellipse in the first quadrant – Area under the line AB
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 20

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 10.
Find the area of the region enclosed by the parabola x² = y, the line y = x + 2 and the x-axis.
Solution:
The given curves are x² = y … (1)
and y = x + 2 … (2)
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 21
(1) → x² = x + 2 ⇒ x² – x – 2 = 0
⇒ (x – 2)(x + 1) = 0 ⇒ x = 2, x = – 1
When x = 2, y = 4
When x = – 1, y = 1
The points of intersection are (-1, 1) and (2,4)
Required area is shaded in the figure
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 22

Question 11.
Using the method of integration find the area bounded by the curve |x| + |y| = 1.
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 23
The given curve is |x| + |y| = 1
x + y = 1 … (1)
x – y = 1 … (2)
– x + y = 1 … (3)
– x – y = 1 … (4)
Required area is symmetric w.r.t. both the axes.
Required area = 4 x Area of shaded region
= \(4 \int_{0}^{1}(1-x) d x=4\left[x-\frac{x^{2}}{2}\right]_{0}^{1}\)
= \(4\left(1-\frac{1}{2}\right)=4\left(\frac{1}{2}\right)\)
= 2 sq.units

Question 12.
Find the area bounded by curves {(x, y) : y ≥ x² and y = |x|}
Solution:
The area bounded by the curves y ≥ x² and y = |x| is same as the area bounded by y = x², y = |x| are same
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 24
The point B is obtained by solving y = x
and y = x²
i.e., x = x² ⇒ x² – x = 0
x(x – 1) = 0 ⇒ x = 0 or x = 1
When x = 0, y = 0 and when x = 1, y = 1
∴ B is(1, 1)
Required area = 2 x (Area of the shaded portion in the first quadrant)
= 2[(Area under the line y = x) – (Area under the parabola y = x²)]
= \(2\left[\int_{0}^{1} x d x-\int_{0}^{1} x^{2} d x\right]=2\left[\left[\frac{x^{2}}{2}\right]_{0}^{1}-\left[\frac{x^{3}}{3}\right]_{0}^{1}\right]\)
= \(2\left(\frac{1}{2}-\frac{1}{3}\right)=2\left(\frac{1}{6}\right)=\frac{1}{3} \text { sq.unit }\)

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 13.
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6,3).
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 25
Equation of BC: y – 5 = \(\left(\frac{3-5}{6-4}\right)\)(x – 4)
y – 5 = – (x – 4) ⇒ y = 9 – x
Equation of AC : y – 0 = \(\left(\frac{3-0}{6-2}\right)\)(x – 2)
y = \(\frac{3}{4}\)(x – 2)
Area of ∆ABC = Area under AB + Area under BC – Area under AC
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 26
= 5 + 8 – 6 = 7 sq. unit

Question 14.
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
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 27
The given lines are 2x + y = 4 … (1)
3x – 2y = 6 … (2)
x – 3y + 5 = 0 … (3)
Solving (1) and (2), we get x = 2, y = 0
Let A be the point (2, 0)
Solving (2) and (3), we get x = 4, y = 3
Let B be the point (4, 3)
Solving (1) and (3), we get x = 1, y = 2
Let C be the point (1, 2)
Equation of AB is y = \(\frac{3x-6}{2}\)
Equation of BC is y = \(\frac{x+5}{3}\)
Equation of AC is y = 4 – 2x
Area of ∆ABC Area of trapezium DCBE – Area of ∆DCA – Area of ∆ABE
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 28

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 15.
Find the area of the region \(\left\{(x, y): y^{2} \leq 4 x, 4 x^{2}+4 y^{2} \leq 9\right\}\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 29
circle with centre at origin and intersecting positive x-axis at C(\(\frac{3}{2}\), 0)
Solving 4x² + 4y² = 9 and y² = 4x, we get 4x² + 16x = 9 ⇒ 4x² + 16x – 9 = 0
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 30
Area of the shaded region = Area under OA + Area under AC
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 31
The required region is symmetric with respect to jt-axis.
∴ Required area = 2 x shaded area
= \(2\left(\frac{\sqrt{2}}{12}+\frac{9 \pi}{16}-\frac{9}{8} \sin ^{-1}\left(\frac{1}{3}\right)\right)\)
= \(\frac{\sqrt{2}}{6}+\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right) \text { sq.unit }\)

Question 16.
Area bounded by the curve y = x³, the x- axis and the ordinates x = – 2 and x = 1 is
a. – 9
b. \(\frac{-15}{4}\)
c. \(\frac{15}{4}\)
d. \(\frac{17}{4}\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 32

Question 17.
The area bounded by the curve y = x|x|, x-axis and the ordinates x = – 1 and x = 1 is given by
a. 0
b. \(\frac{1}{3}\)
c. \(\frac{2}{3}\)
d. \(\frac{4}{3}\)
Solution:
y = x|x| is redefined as
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 33

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 18.
The area of the circle x² + y² = 16 exterior to the parabola y² = 6x is
a. \(\frac{4}{3}(4 \pi-\sqrt{3})\)
b. \(\frac{4}{3}(4 \pi+\sqrt{3})\)
c. \(\frac{4}{3}(8 \pi-\sqrt{3})\)
d. \(\frac{4}{3}(8 \pi+\sqrt{3})\)
Solution:
c. \(\frac{4}{3}(8 \pi-\sqrt{3})\)
 NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 34
x² + y² = 16 is a circle with centre at origin
and intersecting positive x-axis at C(4,0).
Solving x² + y² = 16 and y² = 6x, we get
x² + 6x = 16
x² + 6x – 16 = 0 ⇒ (x + 8)(x – 2) = 0
∴ x = – 8 and x = 2
But x = – 8 is not possible ∴ x = 2
When x = 2, y² = 12 ∴ y = ±2\(\sqrt{3}\)
The points of intersection are
A(2, \(\sqrt{3}\)) and B(2, -2\(\sqrt{3}\))
The required area is shaded in the figure.
The nonshaded region OACBO is symmetric w.r.t. x-axis and lies between x = 0 and x = 4.
∴ Area of the nonshaded region = 2 x Area of the nonshaded region in the first quadrant.
= 2 [Area under OA + Area under AC]
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 35

Question 19.
The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ \(\frac { π }{ 2 }\) is
a. 2(\(\sqrt{2}\) – 1)
b. \(\sqrt{2}\) – 1
c. \(\sqrt{2}\) + 1
d. \(\sqrt{2}\)
Solution:
b. \(\sqrt{2}\) – 1
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 36
The curves are y = sinx … (1)
and y = cos x … (2)
From (1) and (2), we get sin x = cos x
⇒ x = \(\frac { π }{ 4 }\)
∴ y = sin\(\frac { π }{ 4 }\) = \(\frac{1}{\sqrt{2}}\)
Point of intersection of curves is (\(\frac { π }{ 4 }\), \(\frac{1}{\sqrt{2}}\))
Required area = Area of the shaded region = (Area under the curve y = cos x) – (Area above the curve y = sinx)
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise 37

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