GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 1.
Find the area under the following 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:
(i) Graphs of parabola y = x2, lines x = 1, x = 2 and x-axis are shown in the figure.
The area of region bounded by y = x2, x = 1, x = 2 and x-axis.
= Area of region PLMQP
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 1

(ii) The curve y = x4 passes through (0, 0) and it is symmetric about y-axis for all x ∈ R.
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 2
∴ \(\frac{dy}{dx}\) = 4x3.
\(\frac{dy}{dx}\) = 0 at x = 0.
\(\frac{dy}{dx}\) changes sign from – ve to + ve as x increases through x = 0
∴ y is minimum at x = 0.
When x > 0, y is the increasing function.
Graph is as shown in the figure.
Area of region bounded by
y = x4, x = 1,
x = 5 and x-axis
= Area of region ABQPA
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 3

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 2.
Find the area between the curves y = x and y = x2.
Solution:
Given curves are
y = x …………… (1)
y = x2 ……………… (2)
Putting y = x in eq.(2), we get
x = x2
∴ x = 0, 1
When x = 1, y = 1
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 4
These curves intersect at (0, 0) and (1, 1).
∴ Area between y = x
and y = x2
= Area of the region OCPQ
= Area of OAP – Area of region OAPCO
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 5

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 3.
Find the area of the region lying in first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4.
Solution:
The given curve and lines which bound the region are
y = 4x2, x = 0, y = 1 and y = 4.
Graph is as shown in the figure.
The area of the region lying in the first quadrant and bounded by
y = 4x2, x = 0, y = 1 and y = 4
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 6
= Area of the region = LPQML
= Area OQMO – Area OPLO
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 7

Question 4.
Sketch the graph of y = |x + 3| and evaluate \(\int_{6}^{0}\) |x + 3| dx.
Solution:
y = |x + 3|
At x = – 3, y = 0
AQ is the line y = x + 3.
When x + 3 < 0,
y = -(x + 3)
= – x – 3
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 8
Graph of the line is AP.
∴ Graph of y = |x + 3| is as shown in the figure.
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 9

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 5.
Find the area bounded by the curve y = sin x between x = 0 and x = 2Ï€.
Solution:
Some points on the sine graph are:
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 10
Plotting these points, we get the graph OPAQB.
Since sin(2Ï€ – x) = – sin x, therefore
graph between x = π and x = 2π has the same shape but it is below the x-axis.
Area of the region OPA = Area of the region AQB.
∴ Area bounded by the curve y = sin x between x = 0 and x = 2π.
= 2 × Area of the region OPA
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 11
= – 2[cos Ï€ – cos 0] = 2[1 + 1] = 4.

Question 6.
Find the area enclosed between the parabola y2 = 4ax and the line y = mx.
Solution:
The given curves are
y = mx …………….. (1)
y2 = 4ax …………….. (2)
Putting value of y from (1) in (2), we get
m2x2 = 4ax
or x(m2x – 4a) = 0
∴ x = 0, x = \(\frac{4 a}{m^{2}}\).
Putting x = \(\frac{4 a}{m^{2}}\) in (1),
y = m.\(\frac{4 a}{m^{2}}\) = \(\frac{4a}{m}\)
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 12
∴ The curve y2 = 4ax and OP intersect at O(0, 0) and
∴ Area enclosed between the parabola y2 = 4ax and the line y = mx.
= Area of the region OPOQ [As per figure]
= Area of the region OMPQO – Area of ∆ OMP
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 13

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 7.
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12.
Solution:
The parabola and the line are
4y = 3x2 …………………. (1)
2y = 3sx + 12 ………………. (2)
Multiplying (2) by 2 and
Subtracting from (1), we get
0 = 3x2 – 6x – 24
or x2 – 2x – 8 = 0
or (x – 4)(x + 2) = 0
∴ x = 4, – 2
From (2), y = 12, 3
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 14
The graph of parabola and lines are shown in the figure. They intersect at P(- 2, 3) and Q(4, 12).
∴ The area enclosed by the parabola 4y = 3x2
and the line 2y = 3x + 12
= Area of the region POPQP
= Area of trapezium PLMQP – Area of the region LMQROP
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 15
= 45 – 18 = 27 sq.units.

Question 8.
Find the area of the smaller region bounded by the ellipse
\(\frac{x^{2}}{9}\) + \(\frac{y^{2}}{4}\) = 1 and the straight line \(\frac{x}{3}\) + \(\frac{y}{2}\) = 1.
Solution:
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 16
It is an ellipse with vertices at A(3, 0) and B(0, 2) and length of the major axis = 2(3) = 6 and length of the minor axis 2(2) = 4.
Line \(\frac{x}{3}\) + \(\frac{y}{2}\) = 1 ⇒ y = (\(\frac{6-2x}{3}\))
It is a straight line passing through A(3, 0) and B(0, 2).
Smaller area common to both is shaded.
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 17
For I1 put x = 3 sin θ so that dx = 3 cos θ dθ.
When x = 0, θ = 0 and when x = 3, θ = \(\frac{π}{2}\)
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Ex 8.1 img 18

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 9.
Find the area of the smaller region bounded by the ellipse
\(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 and the staraight line \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.
Solution:
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 19

Question 10.
Find the area of the region enclosed by parabola x2 = y, the line y = x + 2 and the x-axis.
Solution:
We have x2 = y. It represents a parabola with vertex at (0, 0), axis along the positive direction of y-axis and it opens upwards.
Also, y = x + 2 represents a straight line cutting x-axis at (-2, 0).
Solving x2 = y and y = x + 2, we get
x2 = x + 2
⇒ x2 – x – 2 = 0
⇒ (x – 2)(x + 1) = 0
⇒ x = 2, x = 1.
⇒ When x = 2, y = (2)2 = 4
and when x = – 1, y = (- 1)2 = 1.
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 20
So the two curves x2 = y and y = x + 2 intersect at the points (2, 4) and (- 1, 1).
Required area = Shaded region shown in the figure
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 21

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 11.
Using method of integration, find the area bounded by |x| + |y| = 1.
Solution:
In I Quadrant, x > 0 and y > 0.
⇒ |x| = x, |y| = y.
The line is x + y = 1 …………….. (1)
In II Quadrant, x < 0 and y > 0.
⇒ |x| = – x. |y| = y.
The line is – x + y = 1
or x – y = -1 ………………… (2)
In III Quadrant, x < 0 and y < 0. ⇒ |x| = – x, |y| = – y. The line is – x – y = 1 or x + y = -1 ………….. (3)
In IV Quadrant, x > 0 and y < 0.
⇒ |x| = x, |y| = – y,
The line is x – y = 1 …………. (4)
Thus, |x| + |y| = 1 represent four lines forming a square ABCD.
Area of square ABCD
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 22
= 4 × Area of ∆ AOB
= 4 × \(\int_{0}^{1}\) (1 – x) dx
Since x + y = 1 is the equation of the line AB.
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 23
= 4 × \(\frac{1}{2}\) = 2 sq.units.

Question 12.
Find the area bounded by the curves {(x, y) y = ≥ x2 and y = |x|}.
Solution:
Clearly, x2 = y represents parabola with vertex at (0, 0), positive direction of y-axis as its axis and it opens upwards.
y= | x |, i.e., y = x and y = – x represent two lines passing through the origin and making an angle of 45° and 135° with the positive direction of the x-axis.
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 24
The required region is the shaded region as shown in the figure. Since both the curves are symmetrical about y-axis, therefore
required area = 2(shaded area in the first quadrant)
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 25

GSEB Solutions 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:
Equation of the line AB is
y – 0 = \(\frac{5-0}{4-2}\)(x – 2)
⇒ y = \(\frac{5}{2}\)(x – 2).
Equation of the line BC is
y – 5 = \(\frac{3-5}{6-4}\)(x – 4)
⇒ y = – x + 9.
Equation of the line CA is
y – 3 = \(\frac{0-3}{2-6}\)(x – 6)
⇒ y = \(\frac{3}{4}\)(x – 2).
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 26
Required area = area of the region bounded by ∆ ABC
= area of the region AMB + area of the region BMNC – area of the region ANC
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 27
= 5 + 8 – 6 = 7 sq. units.

Question 14.
Using the method of integration, find the area of the region bounded by the lines
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0.
Solution:
The given lines are 2x + y = 4 ………….. (1)
3x – 2y = 6 ……………….. (2)
x – 3y = – 5 ……………….. (3)
Multiplying (3) by 2 and subtracting from (1), we get
7y = 14 ⇒ y = 2.
So, from (3) x – 6 = 5 ⇒ x = 1.
∴ Lines (1) and (3) intersect at (1, 2).
Again multiplying (3) by 3 and subtract it from (2). We get
7y = 21 ∴ y – 3.
From (3), X – 9 = – 5 ⇒ x = 4.
∴ Lines (2) and (3) intersect at (4, 3).
Multiply eq. (1) by 2 and add it to (2).
We get: 7x = 14 ∴ x = 2
From (2), 4 + y – 4 ∴ y = 0.
Lines (1) and (2) intersect at (2, 0).
The points A(1, 2), B(4, 3) and C(2, 3) are plotted and joined obtaining triangle ABC.
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 28
Area of ∆ ABC = Area of trapezium ALMB – Area of ∆ ALC – Area of ∆ BCM
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 29

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 15.
Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}.
Solution:
y2 = 4x is a parabola whose vertex is the origin and 4x2 + 4y2 = 9
represents a circle whose centre is (0, 0) and radius = \(\frac{3}{2}\).
On solving y2 = 4x
and x2 + y2 = \(\frac{9}{4}\).
The points of intersection are P(\(\frac{1}{2}\), \(\sqrt{2}\)) and Q(\(\frac{1}{2}\), – \(\sqrt{2}\)).
Both the curves are symmetrical about x-axis
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 30
Required area = area of the shaded region
= 2(area of the region OAPO)
= 2[(area of the region OMPO) + (area of the region MAPM)]
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 31

Choose the correct answers in the following questions 16 to 19:

Question 16.
The area bounded by the curve y = x3, the x-axis and ordinates x = – 2 and x = 1 is
(A) – 9
(B) – \(\frac{15}{4}\)
(C) \(\frac{15}{4}\)
(D) \(\frac{17}{4}\)
Solution:
The curve is y = x3.
Differentiating, we get
\(\frac{dy}{dx}\) = 3x2 = +ve
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 32
∴ Curve is an increasing curve.
\(\frac{dy}{dx}\) = 0 ⇒ x = 0.
∴ x-axis is the tangent at x = 0.
f(- x) = – f(x) ∴ (- x)3 = – x3.
Curve is symmetrical in opposite quadrants,
Area bounded by the curve y = x3, the x-axis, x = – 2 and x = 1
= Area of the region AQOBPOA
= Area of the region AQOA + Area of the region ∆ BPO
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 33
∴ Part D is the correct answer.

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 17.
The area bounded by the curve y = x |x|, x-axis and the ordinates x = – 1 and x = 1 given by
(A) 0
(B) \(\frac{1}{3}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{4}{3}\)
Solution:
When x > 0, |x| = x.
∴ The equation of the curve is y = x2.
When x < 0, |x| = – x.
∴ Equation of the curve is
y = x2.
∴ Area bounded by the curve
y = x|x|,
x-axis and ordinates x = – 1, x = 1
= Area of region + Area of region ∆ BQO
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 34
= 2 × Area of region ∆ BQO
(∵ These areas are equal due to symmetry)
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 35
∴ Part (C) is the correct answer.

Question 18.
The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
(A) \(\frac{4}{3}\) (4Ï€ – \(\sqrt{3}\))
(B) \(\frac{4}{3}\) (4Ï€ + \(\sqrt{3}\))
(C) \(\frac{4}{3}\) (8Ï€ – \(\sqrt{3}\))
(D) \(\frac{4}{3}\) (8Ï€ + \(\sqrt{3}\))
Solution:
The given curves are
x2 + y2 = 16 ………….. (1)
y2 = 6x ……………….. (2)
Putting y2 = 6x in (1), we get
x2 + 6x = 16
or x2 + 6x – 16 = 0
(x + 8)(x – 2) = 0
∴ x = – 8, 2
But x ≠ – 8
So, x = 2.
From (2), y2 = 6x ⇒ y = ± 2\(\sqrt{3}\)
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 36
Area of the whole circle
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 37
Now circle and parabola intersect at P(2, 2\(\sqrt{3}\)) and Q (2, – 2\(\sqrt{3}\)).
Smaller area enclosed by circle and parabola
= Area of region OQAP
= 2 × Area of region OQAP
= 2 × [Area of region OMP + Area of region MAP]
= 2(\(\int_{0}^{2} y_{1} d x+\int_{2}^{4} y_{2} d x\))
When y1 is for parabola y2 = 6x. ∴ y = \(\sqrt{6x}\).
y2 is for circle x2 + y2 = 16 ∴ y = \(\sqrt{16-x^{2}}\)
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 38
Common area exterior to the parabola y2 = 6x is equal to
16Ï€ – (\(\frac{4 \sqrt{3}}{3}\) + \(\frac{16}{3}\)Ï€)
= \(\frac{32}{3}\)Ï€ – \(\frac{4 \sqrt{3}}{3}\) = \(\frac{4}{3}\)(8Ï€ – \(\sqrt{3}\)).
∴ Part (C) is the correct answer.

GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 19.
The area bounded by y-axis, y = cos x and y = sin x, 0 ≤ x ≤ \(\frac{π}{2}\) is
(A) 2(\(\sqrt{2}\) – 1)
(B) \(\sqrt{2}\) – 1
(C) \(\sqrt{2}\) + 1
(D) \(\sqrt{2}\)
Solution:
The curve are y = cos x, y = sin x, 0 ≤ x ≤ \(\frac{π}{2}\).
The curve meet
where sin x = cos x.
or tan x = 1
⇒ x = \(\frac{π}{4}\).
sin \(\frac{Ï€}{4}\) = cos \(\frac{Ï€}{4}\) = \(\frac{1}{\sqrt{2}}\).
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 39
Graphs of these curves are as shown in the figure.
They intersect at P(\(\frac{Ï€}{4}\), \(\frac{1}{\sqrt{2}}\)).
The area bounded by y-axis, y = cos x and y = sin x (0 ≤ x ≤ \(\frac{π}{2}\))
= shaded area
= Area of region ∆ OPBO
= Area of region ∆ PAO + Area of region ∆ PBA
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 40
where x1 is for y = sin x or x = sin-1y ,
and x2 is for y = cos x or x = cos-1 y.
GSEB Solutions Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise img 41
∴ Part (B) is the correct answer.

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