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 ? = ?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 x4 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
= ∫0π sinx dx + | ∫π 2π sin x dx|
= [-cos x]0π + |[-cosx]x2π|
= [-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
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