Class 12 NCERT Solutions – Mathematics Part ii – Chapter 8 – Application of Integrals Miscellaneous Exercise

Question 1. Find the area under the given curves and given lines:

(i) y = x², x = 1, x = 2 and x-axis

(ii) y = x⁴, x = 1, x = 5 and x-axis

Solution:

(i) For the curve 𝑦 = 𝑥² between 𝑥 = 1 and x = 2 and the x-axis:

Area EFGH = ∫₁² x² dx

Area = [ x³/3 ]₁²

Area = (2³ /3) – (1³ / 3) = 8/3 – 1/3 = 7/3

Areas under the curves and lines are: 7/3 square units

(ii) For the curve y = x⁴ between x = 1 and x = 5 and the x-axis:

Area EFGH = ∫₁⁵ x^{4 dx}

Area = [ x⁵/ 5]₁⁵

Area = (5⁵/ 5) – (1⁵/ 5) = 3125/5 = 1/5 = 624.8

The areas under the curves and lines are: 624.8 square units

Question 2. Sketch the graph of y = |x + 3| and evaluate ∫_-6⁰ |x+3| dx.

Solution:

Given equation is y = |x + 3|

Corresponding values of x and y are given in the following table.

X	-6	-5	-4	-3	-2	-1	0
Y	3	2	1	0	1	2	3

On plotting these points, we obtain the graph of y = |x + 3| as follows

It is know that , (x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0

∴ ∫_-6⁰ |(x+3)|dx = – ∫_-6^-3 (x+3)dx + ∫_-3⁰ (x+3)dx

= -[x² / 2 + 3x]-₆^-3 + [x² / 2 + 3x]_-3⁰

= -[((-3)²/2 + 3(-3))-((-6)²/2 + 3(-6)] + [0-((-3)² + 3(-3))]

= -[-9/2]-[-9/2]

= 9 square units

Question 3. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.

Solution:

Graph of y = sin x can be drawn as:

Required area = Area OEFO + Area FGHF

= ∫₀^π sinx dx + | ∫_π ^2π sin x dx|

= [-cos x]₀^π + |[-cosx]_x^2π|

= [-cosπ+cos0] + |-cos 2π + cosπ]

= 1+1+|(-1-1)|

= 2 + |-2|

= 2+ 2

= 4 square units

Question 4. Area bounded by the curve y = x³ , the x-axis and the ordinates x = – 2 and x = 1 is

(A) – 9 (B) -15/4 (C) 15/4 (D) 17/4

Solution:

To find the area bounded by the curve 𝑦= 𝑥³ ,the x-axis, and the ordinates 𝑥 = −2 and 𝑥=1, we integrate x³ with respect to x over the interval [−2, 1].

Area= ∫¹_-2 x³dx

Using the antiderivative of x³ which is x⁴/4, we have:

Area = [x⁴/4]¹_-2

= (1⁴/4)- ((-2)⁴ /4)

= (1/4) – (16/4)

= 1/4- 4/1

= -15/4 square units

Option B is correct.

Question 5. The area bounded by the curve y = x | x | , x-axis and the ordinates x = – 1 and x = 1 is given by

(A) 0 (B) 1/ 3 (C) 2/3 (D) 4/3

Solution:

Curve y = x.∣x∣ has different definitions for positive and negative values of x:

For x ≥ 0, |x| = x, so y = x²

For x < 0, |x| = -x, so y = -x²

We can integrate each part separately over their respective intervals:

For x ≥ 0;

Area₁ = ∫₀¹ x² dx

For x < 0;

Area₂ = ∫_-1⁰ -x² dx

Then, the total area is sum of these two areas:

Total Area = Area₁ + Area₂

Let’s calculate:

For x ≥ 0:

Area₁ = ∫₀¹ x² dx = [x³ / 3]₀¹ = 1/3

For x < 0:

Area₂ = ∫_-1⁰ -x² dx = [-x³/3]₁⁰ = 1/3

such that total area is :

Total Area = Area₁ + Area₂ = 1/3 + 1/3 = 2/3

so, that correct option is 2/3 square units