Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

I. Choose the correct option from the given alternatives:

Question 1.

The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by

(a) 12 sq units

(b) 8 sq units

(c) 25 sq units

(d) 32 sq units

Answer:

(a) 12 sq units

Question 2.

The area of the region enclosed by the curve y = \(\frac{1}{x}\), and the lines x = e, x = e^{2} is given by

(a) 1 sq unit

(b) \(\frac{1}{2}\) sq units

(c) \(\frac{3}{2}\) sq units

(d) \(\frac{5}{2}\) sq units

Answer:

(a) 1 sq unit

Question 3.

The area bounded by the curve y = x^{3}, the X-axis and the lines x = -2 and x = 1 is

(a) -9 sq units

(b) \(-\frac{15}{4}\) sq units

(c) \(\frac{15}{4}\) sq units

(d) \(\frac{17}{4}\) sq units

Answer:

(c) \(\frac{15}{4}\) sq units

Question 4.

The area enclosed between the parabola y^{2} = 4x and line y = 2x is

(a) \(\frac{2}{3}\) sq units

(b) \(\frac{1}{3}\) sq units

(c) \(\frac{1}{4}\) sq units

(d) \(\frac{3}{4}\) sq units

Answer:

(b) \(\frac{1}{3}\) sq units

Question 5.

The area of the region bounded between the line x = 4 and the parabola y^{2} = 16x is

(a) \(\frac{128}{3}\) sq units

(b) \(\frac{108}{3}\) sq units

(c) \(\frac{118}{3}\) sq units

(d) \(\frac{218}{3}\) sq units

Answer:

(a) \(\frac{128}{3}\) sq units

Question 6.

The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is

(a) 1 sq unit

(b) 2 sq units

(c) 3 sq units

(d) 4 sq units

Answer:

(d) 4 sq units

Question 7.

The area bounded by the parabola y^{2} = 8x, the X-axis and the latus rectum is

(a) \(\frac{31}{3}\) sq units

(b) \(\frac{32}{3}\) sq units

(c) \(\frac{32 \sqrt{2}}{3}\) sq units

(d) \(\frac{16}{3}\) sq units

Answer:

(b) \(\frac{32}{3}\) sq units

Question 8.

The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is

(a) 4 sq units

(b) \(\frac{3}{4}\) sq units

(c) \(\frac{2}{3}\) sq units

(d) \(\frac{4}{3}\) sq units

Answer:

(d) \(\frac{4}{3}\) sq units

Question 9.

The area of the circle x^{2} + y^{2} = 25 in first quadrant is

(a) \(\frac{25 \pi}{3}\) sq units

(b) 5π sq units

(c) 5 sq units

(d) 3 sq units

Answer:

(a) \(\frac{25 \pi}{3}\) sq units

Question 10.

The area of the region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is

(a) ab sq units

(b) πab sq units

(c) \(\frac{\pi}{a b}\) sq units ab

(d) πa^{2} sq units

Answer:

(b) πab sq units

Question 11.

The area bounded by the parabola y^{2} = x and the line 2y = x is

(a) \(\frac{4}{3}\) sq units

(b) 1 sq unit

(c) \(\frac{2}{3}\) sq unit

(d) \(\frac{1}{3}\) sq unit

Answer:

(a) \(\frac{4}{3}\) sq units

Question 12.

The area enclosed between the curve y = cos 3x, 0 ≤ x ≤ \(\frac{\pi}{6}\) and the X-axis is

(a) \(\frac{1}{2}\) sq unit

(b) 1 sq unit

(c) \(\frac{2}{3}\) sq unit

(d) \(\frac{1}{3}\) sq unit

Answer:

(d) \(\frac{1}{3}\) sq unit

Question 13.

The area bounded by y = √x and line x = 2y + 3, X-axis in first quadrant is

(a) 2√3 sq units

(b) 9 sq units

(c) \(\frac{34}{3}\) sq units

(d) 18 sq units

Answer:

(b) 9 sq units

Question 14.

The area bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and the line \(\frac{x}{a}+\frac{y}{b}=1\) is

(a) (πab – 2ab) sq units

(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units

(c) (πab – ab) sq units

(d) πab sq units

Answer:

(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units

Question 15.

The area bounded by the parabola y = x^{2} and the line y = x is

(a) \(\frac{1}{2}\) sq unit

(b) \(\frac{1}{3}\) sq unit

(c) \(\frac{1}{6}\) sq unit

(d) \(\frac{1}{12}\) sq unit

Answer:

(c) \(\frac{1}{6}\) sq unit

Question 16.

The area enclosed between the two parabolas y^{2} = 4x and y = x is

(a) \(\frac{8}{3}\) sq units

(b) \(\frac{32}{3}\) sq units

(c) \(\frac{16}{3}\) sq units

(d) \(\frac{4}{3}\) sq units

Answer:

(c) \(\frac{16}{3}\) sq units

Question 17.

The area bounded by the curve y = tan x, X-axis and the line x = \(\frac{\pi}{4}\) is

(a) \(\frac{1}{3}\) log 2 sq units

(b) log 2 sq units

(c) 2 log 2 sq units

(d) 3 log 2 sq units

Answer:

(a) \(\frac{1}{3}\) log 2 sq units

Question 18.

The area of the region bounded by x^{2} = 16y, y = 1, y = 4 and x = 0 in the first quadrant, is

(a) \(\frac{7}{3}\) sq units

(b) \(\frac{8}{3}\) sq units

(c) \(\frac{64}{3}\) sq units

(d) \(\frac{56}{3}\) sq units

Answer:

(d) \(\frac{56}{3}\) sq units

Question 19.

The area of the region included between the parabolas y^{2} = 4ax and x^{2} = 4ay, (a > 0) is given by

(a) \(\frac{16 a^{2}}{3}\) sq units

(b) \(\frac{8 a^{2}}{3}\) sq units

(c) \(\frac{4 a^{2}}{3}\) sq units

(d) \(\frac{32 a^{2}}{3}\) sq units

Answer:

(a) \(\frac{16 a^{2}}{3}\) sq units

Question 20.

The area of the region included between the line x + y = 1 and the circle x^{2} + y^{2} = 1 is

(a) \(\frac{\pi}{2}-1\) sq units

(b) π – 2 sq units

(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units

(d) π – \(\frac{1}{2}\) sq units

Answer:

(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units

(II) Solve the following:

Question 1.

Find the area of the region bounded by the following curve, the X-axis and the given lines:

(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2

(ii) y = sin x, x = 0, x = π

(iii) y = sin x, x = 0, x = \(\frac{\pi}{3}\)

Solution:

(i) Required area = \(\int_{0}^{5} y d x\), where y = 2

= \(\int_{0}^{5} 2 d x\)

= \([2 x]_{0}^{5}\)

= 2 × 5 – 0

= 10 sq units.

(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.

Two bounded regions A_{1} and A_{2} are obtained. Both the regions have equal areas.

∴ required area = A_{1} + A_{2} = 2A_{1}

(iii) Required area = \(\int_{0}^{\pi / 3} y d x\), where y = sin x

Question 2.

Find the area of the circle x^{2} + y^{2} = 9, using integration.

Solution:

By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.

Clearly, for this region, the limits of integration are 0 and 3.

From the equation of the circle, y^{2} = 9 – x^{2}.

In the first quadrant, y > 0

∴ y = \(\sqrt{9-x^{2}}\)

∴ area of the circle = 4 (area of the region OABO)

Question 3.

Find the area of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) using integration.

Solution:

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.

Clearly, for this region, the limits of integration are 0 and 5.

From the equation of the ellipse

\(\frac{y^{2}}{16}=1-\frac{x^{2}}{25}=\frac{25-x^{2}}{25}\)

∴ y^{2} = \(\frac{16}{25}\) (25 – x^{2})

In the first quadrant y > 0

∴ y = \(\frac{4}{5} \sqrt{25-x^{2}}\)

∴ area of the ellipse = 4(area of the region OABO)

Question 4.

Find the area of the region lying between the parabolas:

(i) y^{2} = 4x and x^{2} = 4y

(ii) 4y^{2} = 9x and 3x^{2} = 16y

(iii) y^{2} = x and x^{2} = y.

Solution:

(i)

For finding the points of intersection of the two parabolas, we equate the values of y^{2} from their equations.

From the equation x^{2} = 4y, y = \(\frac{x^{2}}{4}\)

y = \(\frac{x^{4}}{16}\)

\(\frac{x^{4}}{16}\) = 4x

∴ x^{4} – 64x = 0

∴ x(x^{3} – 64) = 0

∴ x = 0 or x^{3} = 64 i.e. x = 0 or x = 4

When x = 0, y = 0

When x = 4, y = \(\frac{4^{2}}{4}\) = 4

∴ the points of intersection are 0(0, 0) and A(4, 4).

Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]

Now, area of the region ODACO = area under the parabola y^{2} = 4x, i.e. y = 2√x between x = 0 and x = 4

(ii)

For finding the points of intersection of the two parabolas, we equate the values of 4y^{2} from their equations.

From the equation 3x^{2} = 16y, y = \(\frac{3 x^{2}}{16}\)

∴ y = \(\frac{3 x^{4}}{256}\)

∴ \(\frac{3 x^{4}}{256}\) = 9x

∴ 3x^{4} – 2304x = 0

∴ x(x^{3} – 2304) = 0

∴ x = 0 or x^{3} = 2304 i.e. x = 0 or x = 4

When x = 0, y = 0

When x = 4, y = \(\frac{4^{2}}{4}\)

∴ the points of intersection are O(0, 0) and A(4, 4).

Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]

Now, area of the region ODACO = area under the parabola y^{2} = 4x,

i.e. y = 2√x between x = 0 and x = 4

Area of the region ODABO = area under the rabola x^{2} = 4y,

i.e. y = \(\frac{x^{2}}{4}\) between x = 0 and x = 4

(iii)

For finding the points of intersection of the two parabolas, we equate the values of y^{2} from their equations.

From the equation x^{2} = y, y = \(\frac{x^{2}}{y}\)

∴ y = \(\frac{x^{2}}{y}\)

∴ \(\frac{x^{2}}{y}\) = x

∴ x^{2} – y = 0

∴ x(x^{3} – y) = 0

∴ x = 0 or x^{3} = y

i.e. x = 0 or x = 4

When x = 0, y = 0

When x = 4, y = \(\frac{4^{2}}{4}\) = 4

∴ the points of intersection are O(0, 0) and A(4, 4).

Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]

Now, area of the region ODACO = area under the parabola y^{2} = 4x,

i.e. y = 2√x between x = 0 and x = 4

Area ofthe region ODABO = area under the rabola x^{2} = 4y,

i.e. y = \(\frac{x^{2}}{3}\) between x = 0 and x = 4

Question 5.

Find the area of the region in the first quadrant bounded by the circle x^{2} + y^{2} = 4 and the X-axis and the line x = y√3.

Solution:

For finding the points of intersection of the circle and the line, we solve

x^{2} + y^{2} = 4 ………(1)

and x = y√3 ……..(2)

From (2), x^{2} = 3y^{2}

From (1), x^{2} = 4 – y^{2}

3y^{2} = 4 – y^{2}

4y^{2} = 4

y^{2} = 1

y = 1 in the first quadrant.

When y = 1, r = 1 × √3 = √3

∴ the circle and the line intersect at A(√3, 1) in the first quadrant

Required area = area of the region OCAEDO = area of the region OCADO + area of the region DAED

Now, area of the region OCADO = area under the line x = y√3, i.e. y = \(\frac{x}{\sqrt{3}}\) between x = 0

and x = √3

Question 6.

Find the area of the region bounded by the parabola y^{2} = x and the line y = x in the first quadrant.

Solution:

To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.

∴ y^{2} = y

∴ y^{2} – y = 0

∴ y(y – 1) = 0

∴ y = 0 or y = 1

When y = 0, x = 0

When y = 1, x = 1

∴ the points of intersection are O(0, 0) and A(1, 1).

Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO

Now, area of the region OCADO = area under the parabola y^{2} = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1

Area of the region OBADO = area under the line y = x between x = 0 and x = 1

Question 7.

Find the area enclosed between the circle x^{2} + y^{2} = 1 and the line x + y = 1, lying in the first quadrant.

Solution:

Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)

Now, area of the region OACBO = area under the circle x^{2} + y^{2} = 1 between x = 0 and x = 1

Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1

∴ required area = \(\left(\frac{\pi}{4}-\frac{1}{2}\right)\) sq units.

Question 8.

Find the area of the region bounded by the curve (y – 1)^{2} = 4(x + 1) and the line y = (x – 1).

Solution:

The equation of the curve is (y – 1)^{2} = 4(x + 1)

This is a parabola with vertex at A (-1, 1).

To find the points of intersection of the line y = x – 1 and the parabola.

Put y = x – 1 in the equation of the parabola, we get

(x – 1 – 1)^{2} = 4(x + 1)

∴ x^{2} – 4x + 4 = 4x + 4

∴ x^{2} – 8x = 0

∴ x(x – 8) = 0

∴ x = 0, x = 8

When x = 0, y = 0 – 1 = -1

When x = 8, y = 8 – 1 = 7

∴ the points of intersection are B (0, -1) and C (8, 7).

To find the points where the parabola (y – 1)^{2} = 4(x + 1) cuts the Y-axis.

Put x = 0 in the equation of the parabola, we get

(y – 1)^{2} = 4(0 + 1) = 4

∴ y – 1 = ±2

∴ y – 1 = 2 or y – 1 = -2

∴ y = 3 or y = -1

∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).

To find the point where the line y = x – 1 cuts the X-axis.

Put y = 0 in the equation of the line, we get

x – 1 = 0

∴ x = 1

∴ the line cuts the X-axis at the point G (1, 0).

Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO

Now, area of the region BFAB = area under the parabola (y – 1)^{2} = 4(x + 1), Y-axis from y = -1 to y = 3

Since, the area cannot be negative,

Area of the region BFAB = \(\left|-\frac{8}{3}\right|=\frac{8}{3}\) sq units.

Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG

Since, area cannot be negative,

area of the region = \(\left|-\frac{1}{2}\right|=\frac{1}{2}\) sq units.

∴ required area = \(\frac{8}{3}+\frac{109}{6}+\frac{1}{2}\)

= \(\frac{16+109+3}{6}\)

= \(\frac{128}{6}\)

= \(\frac{64}{3}\) sq units.

Question 9.

Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.

Solution:

The equation of the line is

2y = 5x + 7, i.e., y = \(\frac{5}{2} x+\frac{7}{2}\)

Required area = area of the region ABCDA = area under the line y = \(\frac{5}{2} x+\frac{7}{2}\) between x = 2 and x = 5

Question 10.

Find the area of the region bounded by the curve y = 4x^{2}, Y-axis and the lines y = 1, y = 4.

Solution:

By symmetry of the parabola, the required area is 2 times the area of the region ABCD.

From the equation of the parabola, x^{2} = \(\frac{y}{4}\)

In the first quadrant, x > 0

∴ x = \(\frac{1}{2} \sqrt{y}\)

∴ required area = \(\int_{1}^{4} x d y\)