Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.11 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11
Direction (1 – 19):
By using the properties of definite integrals, evaluate the integrals.
Question 1.
\int_{0}^{\frac{\pi}{2}} cos x dx
Solution.
Let I = \int_{0}^{\frac{\pi}{2}} cos x dx ……….(i)
Question 2.
\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx
Solution.
Question 3.
\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x
Solution.
Adding equations (i) and (ii), we get
⇒ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} dx
⇒ 2I = \int_{0}^{\frac{\pi}{2}} 1 dx
⇒ 2I = [x]_{0}^{\frac{\pi}{2}}
⇒ 2I = \frac{\pi}{2}
⇒ I =\frac{\pi}{4}
Question 4.
\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} dx
Solution.
Question 5.
\int_{-5}^{5} |x + 2| dx
Solution.
Let I = \int_{-5}^{5} |x + 2| dx
It can be seen that (x + 2) ≤ 0 on [- 5, – 2]and (x + 2) ≥ 0 on [- 2, 5].
Question 6.
\int_{2}^{8} |x – 5| dx
Solution.
Let I = \int_{2}^{8} |x – 5| dx
It can be seen that (x – 5) ≤ 0 on [2, 5] and (x – 5) ≥ 0 on [5, 8].
Question 7.
\int_{0}^{1} x (1 – x)n dx
Solution.
Let I = \int_{0}^{1} x (1 – x)n dx
I = \int_{0}^{1} (1 – x) (1 – (1 – x))n dx
= \int_{0}^{1} (1 – x) (x)n dx
= \int_{0}^{1} (xn – xn+1) dx
= \left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_{0}^{1}
(∵ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a – x) dx
= \left[\frac{1}{n+1}-\frac{1}{n+2}\right]
= \frac{(n+2)-(n+1)}{(n+1)(n+2)}
= \frac{1}{(n+1)(n+2)}
Question 8.
\int_{0}^{\frac{\pi}{4}} log (1 + tan x) dx
Solution.
⇒ I = \int_{0}^{\frac{\pi}{4}} log 2 dx – \int_{0}^{\frac{\pi}{4}} log (1 + tan x) dx
⇒ I = \int_{0}^{\frac{\pi}{4}} log 2 dx – I [From Eq. (i)]
⇒ 2I = [x \log 2]_{0}^{\frac{\pi}{4}}
⇒ 2I = \frac{\pi}{4} log 2
⇒ I = \frac{\pi}{8} log 2
Question 9.
\int_{0}^{2} x \sqrt{2-x} dx
Solution.
Let I = \int_{0}^{2} x \sqrt{2-x} dx
(∵ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a – x) dx)
I = \int_{0}^{2} (2 – x) √x dx
Question 10.
\int_{0}^{\frac{\pi}{2}} (2 log sin x – log sin 2x) dx
Solution.
Let I = \int_{0}^{\frac{\pi}{2}} (2 log sin x – log sin 2x) dx
⇒ I = \int_{0}^{\frac{\pi}{2}} {2 log sin x – log (2 sin x cos x)} dx
⇒ I = \int_{0}^{\frac{\pi}{2}} {2 log sin x – log sin x – log cos x – log 2} dx
⇒ I = \int_{0}^{\frac{\pi}{2}} {log sin x – log cos x – log 2}dx ……………(i)
We know that,
∵ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a – x) dx
∴ I = \int_{0}^{\frac{\pi}{2}} {log cos x – log sin x – log 2} dx …………..(ii)
Adding equations (i) and (ii), we get
2I = \int_{0}^{\frac{\pi}{2}} (log 2 – log 2) dx
⇒ 2I = – 2 log 2 \int_{0}^{\frac{\pi}{2}} 1 dx
⇒ I = – log 2 [latex]\frac{\pi}{2}[/latex]
⇒ I = \frac{\pi}{2} (- log 2)
⇒ I = \frac{\pi}{2} [log \frac{1}{2}]
⇒ I = \frac{\pi}{2} log \frac{1}{2}
Question 11.
\int_{\frac{2}{2}}^{\frac{\pi}{2}} sin2 x dx
Solution.
Let I = \int_{\frac{2}{2}}^{\frac{\pi}{2}} sin2 x dx
As sin2 (- x) = (sin(- x))2
= (- sin x)2
= sin 2x,
therefore, sin2 x is an even function.
We know that if f(x)is an even function, then
\int_{-a}^{a} f(x)dx = 2 \int_{0}^{a} f(x) dx
I = 2 \int_{0}^{\frac{\pi}{2}} sin2 x dx
= 2 \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} dx
= \int_{0}^{\frac{\pi}{2}} (1 – cos 2x) dx
= \left[x-\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}
Question 12.
\int_{0}^{\pi} \frac{x d x}{1+\sin x}
Solution.
⇒ 2I = π \int_{0}^{\pi} {sec2 x – tan x sec x} dx
⇒ 2I = π [\tan x-\sec x]_{0}^{\pi}
⇒ 2I = π [(tan π – sec π) – (tan 0 – sec 0)
⇒ 2I = π [(0 – (- 1) – (0 – 1)]
⇒ 2I = π [2]
⇒ I = π
Question 13.
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} sin7 x dx
Solution.
Let I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} sin7 x dx
As sin7 (- x) = (sin(- x))7
= (- sin x)7
= – sin7 x,
therefore, sin2 x is an odd function.
We know that if f(x) is an odd function, then \int_{-a}^{a} f(x) dx = 0
∴ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} sin7 x dx = 0.
Question 14.
\int_{0}^{2 \pi} cos5 x dx
Solution.
\int_{0}^{2 \pi} cos5 x dx = 2 \int_{0}^{\pi} cos5 x dx
(∵ \int_{0}^{2 a} f(x) = 2 \int_{0}^{a} f(x), where f(2 – a))
= f(x); hence 2a = 2x
∴ cos5 (2x – x) = cos5 x
= 2 × 0 = 0
(∵ \int_{0}^{2 a} f(x) = 0, if f(2a – x) = – f(x); hence 2a = π
∴ cos5</sup (π – x) = – cos5 x)
Question 15.
\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} dx
Solution.
Let I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} dx …………(i)
Question 16.
\int_{0}^{\pi} log (1 + cos x) dx
Solution.
Let I = \int_{0}^{\pi} log (1 + cos x) dx …………….(i)
⇒ I = \int_{0}^{\pi} log (1 + cos (π – x) dx
(∵ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a – x) dx)
⇒ I = \int_{0}^{\pi} log (1 – cos x) dx ………….(ii)
⇒ 2I = \int_{0}^{\pi} {log (1 + cos x) + log (1 – cos x)} dx
⇒ 2I = \int_{0}^{\pi} log (1 + cos2 x) dx
⇒ 2I = \int_{0}^{\pi} log sin2 x dx
⇒ 2I = 2 \int_{0}^{\pi} log sin x
⇒ I = \int_{0}^{\pi} log sin x dx …………..(iii)
sin (π – x) = sin x
∴ I = 2 \int_{0}^{\frac{\pi}{2}} log sin x dx ………………(iv)
⇒ I = 2 \int_{0}^{\frac{\pi}{2}} log sin (\frac{\pi}{2} – x) dx
= 2 \int_{0}^{\frac{\pi}{2}} log cos x dx ………….(v)
Adding equations (iv) and (v), we get
2I = 2 \int_{0}^{\frac{\pi}{2} (log sin x + log cos x) dx
I = \int_{0}^{\frac{\pi}{2} (log sin x + log cos x + log 2 – log 2) dx
I = \int_{0}^{\frac{\pi}{2} (log 2 sin x cos x – log 2) dx
I = \int_{0}^{\frac{\pi}{2} log sin 2x dx – \int_{0}^{\frac{\pi}{2} log 2 dx
Let 2x = t
⇒ 2 dx = dt
When x = 0, t = o and when x = \frac{\pi}{2}, t = π
∴ I = log sin t dt – \frac{\pi}{2} log 2
I = \frac{1}{2} I – \frac{\pi}{2} log 2
\frac{I}{2} = – \frac{\pi}{2} log 2
⇒ I = – π log 2.
Question 17.
\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} dx
Solution.
Question 18.
\int_{0}^{4} |x – 1| dx
Solution.
Let I = \int_{0}^{4} |x – 1| dx
It can be seen that, (x – 1) ≤ 0 when 0 ≤ x ≤ 1 and (x – 1) ≥ 0 when 1 ≤ x ≤ 4.
∴ I = \int_{0}^{1} |x – 1| dx + \int_{1}^{4} |x – 1| dx
(∵ \int_{a}^{b} f(x) = \int_{a}^{c} f(x) + \int_{c}^{b} f(x)dx)
= \int_{0}^{1} – (x – 1) dx + \int_{1}^{4} (x – 1) dx
= \left[x-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4}
= 1-\frac{1}{2}+\frac{(4)^{2}}{2}-4-\frac{1}{2}+1
= 1-\frac{1}{2}+8-4-\frac{1}{2}+1 = 5.
Question 19.
Show that \int_{0}^{a} f(x) g(x) dx = 2 \int_{0}^{a} f(x)dx if f and g are defined as
f(x) = f(a – x) and g(x) = g(a – x) = 4
Solution.
Let I = \int_{0}^{a} f(x) g(x) dx ………(i)
I = \int_{0}^{a} (a – x) g(a – x) dx
(∵ \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a – x) dx)
= I = \int_{0}^{a} f(x)g (a – x) dx ……..(ii)
Adding equations (i) and (ii), we get
2I = \int_{0}^{a} {f(x) g(x) + f(x) g(a – x)} dx
⇒ 2I = \int_{0}^{a} f(x) {g(x) + g(a – x)}dx
⇒ 2I = \int_{0}^{a} f(x) × 4 dx
[∵ g(x) + g(a – x) = 4 given]
⇒ I = 2 \int_{0}^{a} f(x) dx.
Direction (20 – 21): Choose the correct answer.
Question 20.
The value of \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} (x3 + x cos x + tan5 x + 1) dx is
(A) 0
(B) 2
(C) π
(D) 1
Solution.
Question 21.
The value of \int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) dx is
(A) 2
(B) \frac{3}{4}
(C) 0
(D) – 1
Solution.