Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 8 Application of Integrals Ex 8.2

Question 1.

Find the area of the circle 4x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y.

Solution:

Area is bounded by the circle 4x^{2} + 4y^{2} = 9 and interior of the parabola x^{2} = 4y.

Putting x^{2} = 4 in x^{2} + y^{2} = \(\frac{9}{4}\),

we get 4y + y^{2} = \(\frac{9}{4}\).

or y^{2} + 16y – 9 = 0

⇒ (2y + 9)(2y – 1) = 0

y = \(\frac{1}{2}\), – \(\frac{9}{2}\); y ≠ – \(\frac{9}{2}\) as y is positive.

∴ y = \(\frac{1}{2}\).

Radius of the circle = \(\frac{3}{2}\)

Area of the region required = Area of the region OPRQ

= (area of the region OPQ) + (area of the region PQR)

= 2 × area of the region OPT + 2 × area of the region TPQ

Question 2.

Find the area bounded by the curve (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1.

Solution:

Given circles are

x^{2} + y^{2} = 1 ………….. (1)

and (x – 1)^{2} + y^{2} = 1 ……………….. (2)

Centre of (1) is O (0, 0) and radius = 1.

Both these circles are symmetrical about x-axis.

Solving (1) and (2) we get,

– 2x + 1 = 0 ⇒ x = \(\frac{1}{2}\).

Then, y^{2} = 1 – (\(\frac{1}{2}\))^{2} = \(\frac{3}{4}\) ⇒ y = ±\(\frac{\sqrt{3}}{2}\).

∴ The points of intersection are P(\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\)) and Q(\(\frac{1}{2}\), – \(\frac{\sqrt{3}}{2}\)).

It is clear from the figure that the shaded portion is the region whose area is required.

Required area = area OQAPO

= 2 × area of the region CLAP

= 2 × (area of the region OLPO + area of the region LAPL)

Question 3.

Find the area of the region bounded by the curve y = x^{2} + 2 and the line y = x, x = 0 and x = 3.

Solution:

Equation of the parabola is

y = x^{2} + 2

or x^{2} = (y – 2)

Its vertex is (0, 2). Axis is y-axis.

Boundary lines are

y = x, x = 0 and x = 3.

Graphs of the curve and lines have been shown in the figure.

Area of the region PQRO

= Area of the region OAQR – Area of region OAP

Question 4.

Using integration, find the area of the region bounded by a traingle whose vertices are (- 1, 0), (1, 3) and (3, 2).

Solution:

The points A(- 1, 0), B(1, 3) and C(3, 2) are plotted and joined.

Area of ∆ ABC = Area of ∆ ABL + Area of trapezium BLMC – Area of ∆ ACM

The equation of the line joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is

Area of trapezium BCML.

= Area bounded by BC, x-axis x = 1, x = 3.

Area of ∆ ACM.

(Area bounded by AC, y = \(\frac{x+1}{2}\) x-axis and x = 3)

From (1),

∴ Area of ∆ ABC = 3 + 5 – 4 = 4 sq.units.

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.

Solution:

The given lines are

y = 2x + 1 ………… (1)

y = 3x + 1 …………. (2)

x = 4 ………….. (3)

Subtracting (1) from eq. (2), we get 0 = x,

i.e., x = 0.

Putting x = 0 in eq. (1), we get y = 1.

∴ Lines (2) and (1) intersect at A (0, 1).

Putting x = 4 in eq. (2), we get

y = 12 + 1 = 13.

∴ The lines (2) and (3) intersect at B(4, 13).

Putting x = 4 in eq. (1), we get

y = 8 + 1 = 9.

∴ Lines (1) and (2) intersect at C(4, 9).

Area of ABC

= Area of trapezium OLBA – Area of trapezium OLCA ………… (4)

Area of trapezium OLCA

= Area of region bounded of

OC : y = 2x + 1, x = 0 and x = 4 and y = 0

= (16 – 0) + (4 – 0) = 20.

Putting these values in (4),

area of ∆ ABC = 28 – 20 = 8 sq.units.

Choose the correct answers in the following questions 7 and 8:

Question 6.

Smaller area bounded by the circle x^{2} + y^{2} = 4 and the line x + y = 2

(A) 2(π – 2)

(B) π – 2

(C) 2π – 1

(D) 2(π + 2)

Solution:

A circle of radius 2 and centre at O is drawn. The line AB : x + y = 2 we passes through (2, 0) and (0, 2).

Area of the region ACB

= Area of quadrant OAB – Area of ∆ OAB …………. (1)

Now, x^{2} + y^{2} = 4

∴ y^{2} = 4 – x^{2} or y = \(\sqrt{4-x^{2}}\)

∴ Area of quadrant OAB

Area of ∆ ABC = Area of the region bounded by

AB : x + y = 2 or y = 2 – x

and x = 0, y = 0

Putting these values in (1), we get area of region ACB = π – 2

∴ Part (B) is the correct answer.

Question 7.

Area lying between the curves y^{2} = 4x and y = 2x is

(A) \(\frac{2}{3}\)

(B) \(\frac{1}{3}\)

(C) \(\frac{1}{4}\)

(D) \(\frac{3}{4}\)

Solution:

The curve is y^{2} = 4x …………… (1)

and the line is y = 2x …………….. (2)

From (1) and (2),

(4x)^{2} = 4

∴ x(x – 1) = 0

x = 0, x = 1.

Curves (1) and (2) intersect at O(0, 0) and A(1, 2).

∴ Area of region OACO

= Area of region OMACO – Area of OMA ………………. (3)

Now, area of region OMACO

= Area of region bounded by curve OCA

y = \(\sqrt{4x}\), x-axis and x = 1.

Area of ∆ OMA = Area of region bounded by

OA : y = 2x, x = 1 and x-axis

Putting these values in (3), we get

are of region OMACO

= \(\frac{4}{3}\) – 1 = \(\frac{1}{3}\) sq.units.

∴ Part(B) the correct answer.