Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d)

I.

Question 1.

Fine the area of the region enclosed by the given curves.

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

Solution:

Equations of the given curves are

y = cos x ………….. (1)

y = 1 – \(\frac{2x}{\pi}\) ………….. (2)

Eliminating y from eq’s (1) and (2)

cos x = 1 – \(\frac{2x}{\pi}\)

When x = \(\frac{\pi}{2}\), cos x = cos\(\frac{\pi}{2}\) = 0

1 – \(\frac{2}{\pi}\), x = \(\frac{2}{\pi}\) . \(\frac{\pi}{2}\) = 1 – 1 = 0

When x = 0, cos x = cos 0 = 1

1 – \(\frac{2x}{\pi}\) = 1 – 0 = 1

∴ Point of intersection are A = (\(\frac{\pi}{2}\), 0) B = [π – 1]

Question 2.

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

Solution:

Question 3.

y = x³ + 3, y = 0, x = -1, x = 2.

Solution:

Required area PABQ

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:

Between 0 and \(\frac{\pi}{4}\).

cos x > sin x

Between \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\)

cos x < sin x

Required area

Question 6.

x = 4 – y², x = 0.

Solution:

The given parabola x = 4 – y² meets, the x – axis at A(4, 0) and Y – axis at P(0, 2) and Q(0, -2).

The parabola is symmetrical about X – axis

Required area = 2 Area of OAP

Question 7.

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

Solution:

II.

Question 1.

x = 2 – 5y – 3y², x = 0.

Solution:

Solving the equation of given curves

2 – 5y – 3y² = 0

3y² + 5y – 2 = 0

(y + 2)(3y – 1) = 0 y = -2 or \(\frac{1}{3}\)

Question 2.

x² = 4y, x = 2, y = 0.

Solution:

Question 3.

y² = 3x, x = 3.

Solution:

The parabola is symmetrical about X – axis

Required area = 2\(\int_0^3\)√3. √x dx

Question 4.

y = x², y = 2x.

Solution:

Given equation are y = x² ………….. (1)

y = 2x …………. (2)

Eliminating y, we get x² = 2x

x² – 2x = 0

x(x – 2) = 0

x = 0 or x = 2

y = 0, or y = 4

Point of intersection are O(0, 0), A(2, 4)

Question 5.

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

Solution:

Given equation are y = sin 2x ………… (1)

y = √3 sin x …………. (2)

sin 2x = √3 sin x

2 sin x. cos x = √3 sin x

sin x = 0 or 2 cos x = \(\frac{\sqrt{3}}{2}\)

x = 0, cos x = \(\frac{\sqrt{3}}{2}\) ⇒ x = \(\frac{\pi}{6}\)

Question 6.

y = x², y = x³.

Solution:

Given equations are y = x² ………….. (1)

y = x³ ………. (2)

From equation (1) and (2) x² = x³

x³ – x² = 0

x²(x – 1) = 0

x = 0 or 1

Question 7.

y = 4x – x², y = 5 – 2x

Solution:

y = 4x – x² ………… (i)

y = 5 – 2x …………..(ii)

y = -([x – 2]²) = 4

y – 4 = -(x – 2)²

Solving equations (i) and (ii) we get

4x – x² = 5 – 2x

x² – 6x + 5 = 0

(x – 5)(x – 1 ) = 0

x = 1, 5

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:

Given equations y = 1 + \(\frac{8}{x^2}\)

Question 9.

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

Solution:

Equations of the given curve are

y² = 4x

x² = 4y

(\(\frac{x^2}{4}\))² = 4x

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

x^{4} = 64x ⇒ x^{4} = 0 or x³ = 64, x = 4

Question 10.

Find the area bounded by the curve y = *l*nx the X – axis and the straight line x = e.

Solution:

Equation of the curve is y = *l*nx

x = 1 ⇒ y = 0

The curve y = *l*nx meets

X – axis at C(1, 0)

Required area = \(\int_1^e\)lnx dx

= (x.lnx)^{e}_{1} – \(\int_1^e\)x.\(\frac{1}{x}\) dx

= (e.*l*n e – 1.*l*n 1) – (x)^{e}_{1}

= e – (e – 1)

= e – e + 1 = 1 Sq.unit.

III.

Question 1.

y = x² + 1, y = 2x – 2, x = -1, x = 2.

Solution:

Equation of the curve are

y = x² + 1 …………. (1)

y = 2x – 2 ………….. (2)

Area between the given curves

Question 2.

y² = 4x, y² = 4(4 – x).

Solution:

Equations of the curve are y² = 4x ………… (1)

y² = 4(4 – x) …………. (2)

Eliminating y, we get

4x = 4(4 – x)

2x = 4 ⇒ x = 2

Substituting in equation (1), y² = 8

y = ± 2√2

Points of intersection are

A(2, 2√2), B(2, -2√2)

Required area is symmetrical about X – axis

Area OACB

Question 3.

y = 2 – x², y = x².

Solution:

y = 2 – x² …………. (1)

y = x² …………. (2)

x² = -(y – 2)

From equation (2)

2 – x² = x²

2 = 2x² or x² = 1

x = ±1

Area bounded by two curve be

Question 4.

Show that the area enclosed between the curve y² = 12(x + 3) and y² = 20(5 – x) is 64\(\sqrt{\frac{5}{3}}\).

Solution:

Equation of the curve are

y² = 12(x + 3) ……….. (1)

y² = 20(5 – x) ……….. (2)

Eliminating y

12(x + 3) = 20(5 – x)

3x + 9 = 25 – 5x

8x = 16

x = 2

y² = 12(2 + 3) = 60

y = √60 = ±2√15

Points of intersection are B’ (2, 2√15)

B’ (+2, -2√15)

The required area is symmetrical about X – axis

Area ABCB’

Question 5.

Find the area of the region {(x, y)/x² – x – 1 ≤ y ≤ -1}

Solution:

Question 6.

The circle x² + y² = 8 is divided into two parts by the parabola 2y = x². Find the area of both the parts.

Solution:

Equations of the curves are

x² + y² = 8 ………… (1)

2y = x² ………… (2)

Eliminating y between equations (1) and (2)

Let x² = t

4t + t² = 32

t² + 4t – 32 = 0

(t + 8)(t – 4) = 0

t = -8 (not possible) x² = 4 ⇒ x = ±2

As curve is symmetric about Y – axis, total area be

Question 7.

Show that the area of the region bounded \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (ellipse) is π ab. also deduce the area of the cricle x² + y² = a².

Solution:

The ellipse is symmetrical about X and Y axis

Area of the ellipse = 4 Area of CAB

= 4.\(\frac{\pi}{4}\) ab

Equation of elliple = \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1

Substituting b = a, we get the circle

x² + y² = a²

Area of the circle = πa(a) = πa² sq. units.

Question 8.

Find the area of region enclosed by the curves y = sin πx, y = x² – x, x = 2.

Solution:

Required area

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 the chord AB and the arc AB of the elliple is \(\frac{(\pi-2) a b}{4}\).

Solution:

Let OA = a, OB = b

Equation of AB is \(\frac{x}{a}+\frac{y}{b}\) = 1

\(\frac{y}{b}\) = 1 – \(\frac{x}{a}\)

y = b(1 – \(\frac{x}{a}\))

Question 10.

Prove that curves y² = 4x and x² = 4y divide the area of the square bounded by the lines x = 0, x = 4, y = 4 and x = 0 into three equal parts.

Solution:

The given equations are y² = 4x …………. (1)

x² = 4y …………. (2)

The points of intersection are O(0, 0) A(4, 4)