Students must practice this TS Intermediate Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(d) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(d)

I. Find the area of the region enclosed by the given curves.

Question 1.

y = cos x, y = 1 – \(\frac{2 x}{\pi}\)

Solution:

Consider the graphs of the functions

f(x) = cos x and g(x) = 1 – \(\frac{2 x}{\pi}\)

The two curves intersect at (\(\frac{\pi}{2}\), 0) and (π, -1)

∴ The area enclosed between the curves y = cos x, y = 1 – \(\frac{2 \mathrm{x}}{\pi}\) is (2 – \(\frac{\pi}{2}\)).

Question 2.

y = cos x, y = sin 2x, x = 0, x = \(\frac{\pi}{2}\).

Solution:

Consider the graphs of the functions f(x) = cos x and g(x) = sin 2x.

Taking f(x) = cos x and g(x) = sin 2x in [0, \(\frac{\pi}{6}\)]

we have f(x) ≥ g(x) and also in \(\left[\frac{\pi}{6}, \frac{\pi}{2}\right]\) have g(x) ≥ f(x).

Hence area enclosed between f(x) = cos x and g(x) = sin 2x is

∴ Area enclosed between the curves y = cos x, y = sin 2x, x = 0, x = \(\frac{\pi}{2}\) is \(\frac{1}{2}\) sq.units.

Question 3.

y = x^{3} + 3, y = 0, x = – 1, x = 2. (Mar. ’12)

Solution:

Required area enclosed between y = x^{3} + 3, y = 0, x = – 1, x = 2 is

Question 4.

y = e^{x}, y = x, x = 0, x = 1.

Solution:

Question 5.

y = sin x, y = cos x, x = 0, x = \(\frac{\pi}{2}\).

Solution:

Taking f(x) = sin x and g(x) = cos x

We have sin x – cos x < 0 for x ∈ \(\left[0, \frac{\pi}{4}\right]\) and sin x – cos x > 0 for x ∈ \(\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\)

Area enclosed between y = sin x, y = cos x, x = 0, x = \(\frac{\pi}{2}\) is

Question 6.

x = 4 – y^{2}, x = 0.

Solution:

The curve x = 4 – y^{2} cuts Y – axis at x = 0.

Hence 4 – y^{2} = 0

⇒ y = ± 2

∴ Points of intersection are given by (0, 2) and (0, – 2).

∴ Area subtended by Y-axis, the curve x = 4 – y^{2} is given by \(A=\int_{-2}^2 x d y=\int_{-2}^2\left(4-y^2\right) d y\)

Question 7.

Find the area enclosed with in the curve |x| + |y| = 1.

Solution:

The given equation of the curve is |x| + |y| = 1 which represents ± x ± y = 1 representing four different lines forming a square.

Consider the line x + y = 1

⇒ y = 1 – x

If the line touches the X-axis then y = 0 and one of the points of intersection with X-axis is (1, 0).

Since the curve is symmetric with respect to coordinate axes, area bounded by |x| + |y| = 1 is

II.

Question 1.

x = 2 – 5y – 3y^{2}, x = 0.

Solution:

Solving x = 2 – 5y – 3y^{2} and x = 0

We get – 3y^{2} – 5y + 2 = 0

⇒ 3y^{2} + 5y – 2 = 0

⇒ 3y^{2} + 6y – y – 2 = 0

⇒ 3y (y + 2) – 1 (y + 2) = 0

⇒ (y + 2)(3y – 1) = 0

⇒ y = -2 or y = \(\frac{1}{3}\)

Required area subtended by the curve x = 2 – 5y – 3y^{2}, Y-axis and y = \(\frac{1}{3}\) and y = – 2 is

Question 2.

x^{2} = 4y, x = 2, y = 0.

Solution:

The given equation x^{2} = 4y represents parabola which is symmetric with respect to Y-axis.

When x = 2 we have

4y = 4

⇒ y = 1

Hence P(2, 1) is the point of intersection.

∴ Area bounded by x^{2} = 4y, and x = 0, x = 2 is

A = \(\int_0^2 y d x=\int_0^2 \frac{x^2}{4} d x\)

= \(\frac{1}{4}\left[\frac{x^3}{3}\right]_0^2=\frac{1}{4}\left(\frac{8}{3}\right)=\frac{2}{3}\) sq. units.

Question 3.

y^{2} = 3x, x = 3.

Solution:

y^{2} = 3x is a parabola which is symmetrical to X-axis;

when x = 3 then y^{2} = 9

⇒ y = ± 3

Hence (3, 3) and (3, – 3) are the points of intersection.

Question 4.

y = x^{2}, y = 2x

Solution:

Given y = x^{2} and y = 2x

solving these we get x^{2} – 2x = 0

⇒ x (x – 2) = 0

⇒ x = 0 or x = 2

∴ y = 0 or y = 4

Hence O(0, 0) and P(2, 4) are the points of intersection of (1) and (2).

∴ Required area bounded by y = x^{2} and y = 2x is

= \(\int_0^2\left(2 x-x^2\right) d x=2\left[\frac{x^2}{2}\right]_0^2-\left[\frac{x^3}{3}\right]_0^2\)

= \(2\left[\frac{4}{2}\right]-\left[\frac{8}{3}\right]\)

= 4 – \(\frac{8}{3}\) = \(\frac{4}{3}\) sq. units.

Question 5.

y = sin 2x, y = √3 sin x, x = 0 and x = \(\frac{\pi}{6}\).

Solution:

Given curves are y = sin 2x …….(1)

and y = √3 sin x ……..(2)

When x = \(\frac{\pi}{6}\) then y = \(\frac{\sqrt{3}}{2}\)

∴ \(\left(\frac{\pi}{6}, \frac{\sqrt{3}}{2}\right)\) is a point of intersection (1) and (2).

∴ Area enclosed between the curves

Question 6.

y = x^{2}, y = x^{3}.

Solution:

Given curves are y = x^{2} and y = x^{3}

By solving we get x^{2} = x^{3}

⇒ x^{2} (x – 1) = 0

⇒ x = 0 or x = 1

∴ y = 0 or y = 1

Hence (0, 0) and (1, 1) are the points of intersection.

Also f(x) = x^{2} and g(x) = x^{3} and f(x) > g(x) in [0, 1].

∴ Area bounded by the curves

Question 7.

y = 4x – x^{2}, y = 5 – 2x.

Solution:

The given equations of curves are denoted by

y = f(x) = 4x – x^{2} ……(1)

y = g(x) = 5 – 2x ……..(2)

Solving (1) and (2)

4x – x^{2} = 5 – 2x

⇒ x^{2} – 2x – 4x + 5 = 0

⇒ x^{2} – 6x + 5 = 0

⇒ (x – 5) (x – 1) = 0

⇒ x = 1 or 5

When x = 1, y = 3 and x = 5, y = -5

Hence P(1, 3) and Q(5, – 5) are the two points of intersection.

Now f(x) > g(x) in [1, 5].

∴ Area bounded between the curves is

Question 8.

Find the area in sq.units bounded by the X-axis, part of the curve y = 1 + \(\frac{8}{x^2}\) and the ordinates x = 2 and x = 4.

Solution:

In [2, 4] we have the equation of the curve given by y = 1 + \(\frac{8}{x^2}\)

∴ Area bounded by the curve y = 1 + \(\frac{8}{x^2}\)

X-axis and the ordinates x = 2 and x = 4 is

Question 9.

Find the area of the region bounded by the parabolas y2 = 4x and x2 = 4y.

Solution:

Given equations of curves are

y^{2} = 4x ……..(1)

x^{2} = 4y ……..(2)

Solving (1) and (2) the points of intersection can be obtained.

y^{2} = 4x

⇒ y^{4} = 16x^{2}

⇒ y^{4} = 64y

⇒ y = 4

∴ 4x = y^{2}

⇒ 4x = 16

⇒ x = 4

Points of intersection are (0, 0) and (4, 4).

∴ Area bounded between the parabolas

Question 10.

Find the area bounded by the curve y = log x, the X-axis and the straight line x = e.

Solution:

Area bounded by the curve y = \(\log _e x\), X-axis

and the straight line x = e is \(\int_1^e \log _e x d x\)

= \([\mathrm{x} \log \mathrm{x}]_{\mathrm{1}}^{\mathrm{e}}-\int_1^{\mathrm{e}} \mathrm{dx}\) (∵ when x = e, y = loge e = 1)

= (e – 0) – (e – 1)

= 1 sq.units

III.

Question 1.

y = x^{2} + 1, y = 2x – 2, x = -1, x = 2.

Solution:

Equations of given curves are

y = x^{2} + 1 ……..(1)

y = 2x – 2 ………(2)

Take f(x) = x^{2} + 1 and g(x) = 2x – 1

we find f(x) > g(x) ∀ x ∈ [- 1, 2]

Hence area bounded by y = x^{2} + 1 and y = 2x – 2 is

Question 2.

y^{2} = 4x, y^{2} = 4(4 – x).

Solution:

Equations of the curves are

y^{2} = 4x ………(1)

y^{2} = 4(4 – x) ………(2)

From (1) and (2)

4x = 4(4 – x)

⇒ 8x = 16

⇒ x = 2

∴ y^{2} = 8

⇒ y = ± 2√2

∴ The points of intersection are (2, 2√2), (2, – 2√2)

Question 3.

y = 2 – x^{2}, y = x^{2}.

Solution:

The given curves are

y = 2 – x^{2} ………(1)

y = x^{2} …….(2)

Solving (1) and (2) we get

2 – x^{2} = x^{2}

⇒ 2x^{2} = 2

⇒ x^{2} = 1

⇒ x = ± 1

∴ y = 1

Hence points of intersection of curves are (1, 1), (- 1, 1)

∴ Area bounded between the curves

Question 4.

Show that the area enclosed between the curves y^{2} = 12 (x + 3) and y^{2} = 20 (5 – x) is 64 \(\sqrt{\frac{5}{3}}\).

Solution:

Given curves are y^{2} = 12 (x + 3) …….(1)

and y^{2} = 20 (5 – x) ………(2)

Solving (1) and (2) we get

12 (x + 3) = 20 (5 – x)

⇒ 12x + 36 = 100 – 20x

⇒ 32x = 64

⇒ x = 2

∴ y^{2} = 20(5 – 2) = 60

⇒ y = ± 2√15

∴ Points of intersection are B(2, 2√15) and B'(2, – 2√5).

The given equations represent these parabolas

(y – β)^{2} = 4a(x – α) and

(y – β)^{2} = – 4a(x – α)

∴ Vertex of (1) is A(- 3, 0) and vertex of (2) is A'(5, 0).

∴ The required area is symmetrical about X-axis.

∴ Area bounded by curves is ABA’B’

Question 5.

Find the area of the region {(x, y): x^{2} – x – 1 ≤ y ≤ -1}.

Solution:

Consider y = x^{2} – x – 1 ……(1)

and y = – 1 ……(2)

Which represents a parabola in the form (x – h)^{2} = 4a(y – k) where vertex \(\left(\frac{1}{2}, \frac{-5}{4}\right)\) with 4a = 1.

Also x^{2} – x – 1 = -1

⇒ x^{2} – x = 0

⇒ x (x – 1) = 0

⇒ x = 0 or 1

∴ Area subtended by the region {(x, y) : x^{2} – x – 1 ≤ y ≤ -1}

∴ Area of the region {(x, y); x^{2} – x – 1 ≤ y ≤ – 1} is \(\frac{1}{6}\).

Question 6.

The circle x^{2} + y^{2} = 8 is divided into two parts by the parabola 2y = x^{2}. Find the area of both the points.

Solution:

Given curves are x^{2} + y^{2} = 8 ……(1)

and 2y = x^{2} …….(2)

Solving (1) and (2)

x^{2} + \(\frac{x^4}{4}\) = 8

⇒ x^{4} + 4x^{2} – 32 = 0

⇒ x^{4} + 8x^{2} – 4x^{2} – 32 = 0

⇒ x^{2} (x^{2} + 8) – 4(x^{2} + 8) = 0

⇒ (x^{2} – 4) (x^{2} + 8) = 0

⇒ x = ±2 (∵ x^{2} + 8 = 0 is not admissible)

∴ y = 2

Hence the points of intersection of (1) and (2) are (2, 2) (- 2, 2).

Let the area bounded between (1) and (2) be sum of two areas A_{1} and A_{2} as shown in the figure. Then

Again

A_{2} = π + \(\frac{2}{3}\) (Since the bounded position is symmetric)

∴ Area bounded by the curves = 2π + \(\frac{4}{3}\) sq.units

Area bounded by the circle = πr^{2} = 8π when r = 2√2

∴ Area of the remaining portion = 8π – (2π + \(\frac{4}{3}\)) = 6π – \(\frac{4}{3}\) sq.units.

Question 7.

Show that the area of the region bounded by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (ellipse) is πab. Also deduce the area of the circle x^{2} + y^{2} = a^{2}. (June ’10)

Solution:

Given equation of ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1

Since the ellipse is symmetric with respect to coordinate axes we have

∴ Area of the region bounded by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 is πab sq.units

Substituting b = a in the above result we have the area of circle given by πa(a) = πa^{2} sq.units

Question 8.

Find the area of the region enclosed by the curves y = sin πx, y = x^{2} – x, x = 2.

Solution:

The graphs of the given equations y = sin πx and y = x^{2} – x, x = 2 are shown below.

Question 9.

Let AOB be the positive quadrant of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 with OA = a, OB = b. Then show that the area bounded between the chord AB and the arc AB of the ellipse is \(\frac{(\pi-2) a b}{4}\).

Solution:

Question 10.

Prove that the curves y^{2} = 4x and x^{2} = 4y divide the area of square bounded by the lines x = 0, x = 4, y = 4 and y = 0 into three equal parts.

Solution:

The given curves are y^{2} = 4x ……(1) and x^{2} = 4y ………(2)

Solving y^{4} = 16x^{2} = 64y

⇒ y(y^{3} – 64) = o

⇒ y = 0 or y = 4

When y = 4 we have 4x = 16 ⇒ x = 4

∴ Points of intersection of parabolas is P(4, 4).

∴ Area bounded by the parabolas

Area of the square formed = (OA)^{2} = (4)^{2} = 16

Since the area bounded by the parabolas x^{2} = 4y and y^{2} = 4x is \(\frac{1}{2}\) sq.units.

Which is one third of the area of square we conclude that the curves y^{2} = 4x and x^{2} = 4y divide the area of the square bounded by x = 0, x = 4, y = 0, y = 4 into three equal parts.