RBSE Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 8 Application of Integrals Ex 8.2

Question 1.
Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.
Answer:
x² = 4y is upward parabola with its vertex at origin O(0, 0). 4x² + 4y² = 9 or x² + y² = \(\frac{9}{4}\) is a circle with its centre at O(0, 0) and radius \(\frac{3}{2}\).
Parabola x² = 4y is symmetrical to y-axis, wherever circle x² + y² = \(\frac{9}{4}\) is symmetrical to both axis. Shaded region is the required area common to the parabola and circle as shown in the figure below.

Equation of the parabola
x² = 4y ....... (1)
Equation.of the circle
x² + y² = \(\frac{9}{4}\) ........ (2)
Putting x² = 4y from equation (1) in equation (2), we get
4y + y² = \(\frac{9}{4}\)
or 4y² + 16y - 9 = 0
or 4y² + 18y - 2y - 9 = 0
or 2y(2y + 9) - 1(2y + 9) = 0
or (2y + 9) (2y - 1) = 0
or 2y + 9 = 0 or 2y - 1 = 0
y = - \(\frac{9}{2}\) or y = \(\frac{1}{2}\)
∴ When y = \(\frac{-9}{2}\), from x² = 4y]
x² = 4 × (- 9/2) = - 18
or x = ± √- 18 which is not real.
∴ When y = \(\frac{1}{2}\), then from x² = 4y
x² = 4 × \(\frac{1}{2}\) = 2
or x = ±√2
Thus, points of intersection of the curves are A(√2, \(\frac{1}{2}\)) and B (-√2, \(\frac{1}{2}\)). Co-ordinate of D is (0, \(\frac{3}{2}\)).
(∵ radius of circle is \(\frac{3}{2}\) and point D is on circle and y-axis both)
Required area = area of OPADBQQ
= 2 × Area of OPARBQO
2[Area of OPARO + area of RADR]
= 2[∫(x for parabola) di, + ∫(x for circle) dy]

Question 2.
Find the area bounded by curves (x - 1)² + y² = 1 and x² + y² = 1.
Answer:
Curve (x - 1)² + y² = 1 is a circle with its centre (1, 0) and radius I unit and curve x² + y² = 1 is also a circle with its centre origin (0, 0) and radius 1 unit. Shaded region is the required area shown in the given figure.
Equations of the circle are:
(x - 1)² + y² = 1 ....... (1)
and x² + y² = 1 ......... (2)

Putting y² = 1 - x² from (2) in equation (1), we get
(x - 1)² + 1 - x² = 1
or x² - 2x + 1 + 1 - x² = 1
or - 2x = 1 - 2
or - 2x = - 1
or x = \(\frac{1}{2}\)
∴ y² = 1 - x²
= 1 - \(\left(\frac{1}{2}\right)^{2}\) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)
or y = ± \(\frac{\sqrt{3}}{2}\)
Hence, both the circle intersect each other, at points P\(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) and Q\(\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\).
Required area = area of OQAPO
= 2 × area of OAPO

Question 3.
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.
Answer:
Curve y = x² + 2 is a parabola with its vertex (0, 2). y = x is a straight line which passes through origin (0, 0). Shaded region is the required area shown in the figure below. Point of intersection of curve y = x² + 2 and line x 3 is Q (3, 11).

Required area area of PQSOP
= area of OAPQSOA - area of ∆OAP
=∫ y(for parabola) dx + ∫ y(for line) dx

Question 4.
Using integration, find the area of region bounded by the triangle whose vertices are (- 1, 0), (1, 3) and (3, 2).
Answer:
The points A(- 1, 0), B(1, 3) and C(3, 2) are plotted and joined. Shaded region is the required area shown in the figure below.

Required area = area of ∆ABC
= area of ∆ABP + area of trapezium BPQC - area of ∆AQC ...... (1)
Equation of line AB


Now, putting the values of area of ∆ABP, area of trapezium BPAQ and area of ∆AQC in equation (1),
we get Required area of ∆ABC
= 3 + 5 - 4 = 4 sq. unit
∴ Area of ∆ABC = 4sq.unit

Question 5.
Using integration, find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.
Answer:
Let the sides of the triangle are AB, BC and AC.
Then, equation of AB y = 2x + 1 ...(1)
Equation of BC y = 3x + 1 ......... (2)
and equation of AC x = 4 ......... (3)
From equations (1) and (2), we get
3x + 1 = 2x + 1
∴ 3x - 2x = 0 ⇒ x = 0
Then y = 1
∴ Co-ordinate of point B are (0, 1).
From equations (2) and (3), we get
y = 3 × 4 + 1 = 13
Co-ordinate of point C are (4, 13).
From equations (1) and (3), we get
y = 2 × 4 + 1 = 9
∴ Co-ordinate of point A are (4, 9).
∆ABC is shown in the figure.

Area of ∆ABC = area of trapezium BOPC - area of trapezium BOPA
= ∫ (y for line BC) dx - ∫ (y for line AB) dx

= 24 + 14 - 16 - 4 = 8 sq.unit
Thus, area of ∆ABC = 8 sq.unit

Question 6.
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:
Shaded region enclosed by circle x² + y² = 4 and line x + y = 2, is the required area as shown in the figure below.

Required area = Area of OBPAO - Area of ∆OAB
= ∫ y(for circle) dx - ∫y (for line) dx

= 0 + 2 sin^-1 1 - 0 + 0 - 4 + 2
= 2 × \(\frac{\pi}{2}\) - 2 = (π - 2) sq.units
Hence, (B) is the correct answer.

Question 7.
Area lying between the curves y² = 4x and y = 2x is:
(A) \(\frac{2}{3}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{3}{4}\)
Answer:
Shaded region enclosed by the parabola y² = 4x and line y = 2x, is the required area shown in the figure below.
Both the curves pass through the origin O and the curve y² = 4x is symmetrical about x-axis.
Equations of the curves are:
y² = 4x ..... (1)
y = 2x ....... (2)
Solving the equations, we get,
(2x)² = 4x
or 4x² = 4x
or 4x(x - 1) = 0
x = 0 or x = 1
Co-ordinate of point A = (1, 2)

Required area = area of OPAO
= area of OBAPO - area of AOAB
= ∫y (for parabola)dx - ∫(for line) dx

Hence, (B) is the correct answer.

Question 8.
Find the area of the parabola y² = 4ax bounded by its latus rectum.
Answer:
From figure, the vertex of the parabola y² = 4ax is at origin (0, 0). The equation of the latus rectum LSL is x = a. Also, parabola is symmetrical about the x-axis.

The required area of the region OLL'O
= 2(area of the region OLSO)