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

Chapter 8 of the Class 12 NCERT Mathematics Part II textbook, titled "tegrals - Miscellaneous Exercise

This section provides detailed solutions for the Miscellaneous Exercise from Chapter 8 of the Class 12 NCERT Mathematics Part II textbook. The exercise encompasses a variety of problems involving the application of integrals to compute areas under curves, volumes of solids of revolution, and other related applications. Solutions are presented step-by-step to aid students in understanding and applying integration techniques effectively.

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? = ?² between? = 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.

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 ?= ?³ ,the x-axis, and the ordinates ? = −2 and ?=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

Summary

Chapter 8 of the Class 12 NCERT Mathematics Part II textbook, "Application of Integrals," focuses on the practical uses of integration. The Miscellaneous Exercise covers various problems where integrals are used to compute areas under curves, volumes of solids of revolution, and other applications. The exercise helps students practice and apply integration techniques to solve real-world problems involving geometric shapes and physical quantities.

Related Articles:

• Area as Definite Integral
• Areas under Simple Curves
• Area Between Two curves
• Area defined by Polar Curves