Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7a will help students to clear their doubts quickly.
Inter 2nd Year Maths Integrals Solutions Exercise 7a
Integrals Exercise 7a Solutions
I. Find an antiderivative (or integral) of the following functions by the method of inspection.
Question 1.
sin 2x
Solution:
Consider sin 2x
We know that (cos 2x)’ = -sin 2x . 2
⇒ sin 2x = \(\frac {-1}{2}\)(cos 2x)’
∴ Anti derivative of sin 2x is \(\frac {-1}{2}\)(cos 2x)
Question 2.
cos 3x
Solution:
Consider cos 3x
We know that (sin 3x)’ = cos 3x . (3)
⇒ cos 3x = \(\frac {1}{3}\)(sin 3x)’
∴ Anti derivative of cos 3x is \(\frac {1}{3}\)(sin 3x)
Question 3.
e2x
Solution:
Consider sin e2x
We know that (e2x)’ = e2x . 2
⇒ e2x = \(\frac {1}{2}\)(e2x)’
∴ Anti derivative of e2x = \(\frac {1}{2}\) e2x
Question 4.
(ax + b)2
Solution:
Consider (ax + b)2
We know that [(ax + b)3]’ = 3(ax + b)2 . a
⇒ (ax + b)2 = \(\frac {1}{3a}\)[(ax + b)3]’
∴ Anti derivative of (ax + b)2 = \(\frac {1}{3a}\)(ax + b)3
Question 5.
sin 2x – 4e3x
Solution:
Consider sin 2x – 4e3x
We know that (cos 2x)’ = -sin 2x . 2
sin 2x = \(\frac {-1}{2}\)(cos 2x)’ ……..(1)
Also (4e3x)’ = 4e3x . 3
\(4 \mathrm{e}^{3 \mathrm{x}}=\frac{4}{3}\left(\mathrm{e}^{3 \mathrm{x}}\right)^{\prime}\) …….(2)
From (1) and (2)
∴ Anti derivative of sin 2x – 4e3x is \(\frac{-1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\)
II. Find the following integrals.
Question 6.
∫(4e3x + 1) dx
Solution:
∫(4e3x + 1) dx
= 4∫e3x dx + ∫1 dx [∵ \(\int e^{n x} d x=\frac{e^{n x}}{n}+C\)]
= \(\frac{4 e^{3 x}}{3}\) + x + C
Question 7.
\(\int \mathrm{x}^2\left(1-\frac{1}{\mathrm{x}^2}\right) \mathrm{dx}\)
Solution:
\(\int x^2\left(1-\frac{1}{x^2}\right) d x=\int\left(x^2-\frac{x^2}{x^2}\right) d x\)
= ∫(x2 – 1) dx [∵ \(\int x^n d x=\frac{x^n+1}{n+1}+C\)]
= ∫x2 dx – ∫1 dx
= \(\frac{x^3}{3}\) – x + C
Question 8.
∫(ax2 + bx + c) dx
Solution:
∫(ax2 + bx + c) dx = ∫ax2 dx + ∫bx dx + ∫c dx
= \(\frac{a x^3}{3}+\frac{b x^2}{2}\) + cx + k (where k is a constant of integration)
[∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\); ∫dx = x + C]
Question 9.
∫(2x2 + ex) dx
Solution:
∫(2x2 + ex) dx
= 2∫x2 dx + ∫ex dx [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C, \int e^x d x=e^x+C\)]
= \(\frac{2 x^3}{3}+e^x+C\)
Question 10.
\(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 d x\)
Solution:
\(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 d x\)
= \(\int\left[(\sqrt{x})^2+\left(\frac{1}{\sqrt{x}}\right)^2-2 \sqrt{x} \times \frac{1}{\sqrt{x}}\right] d x\)
= \(\int\left(x+\frac{1}{x}-2\right) d x\)
= ∫x dx + ∫\(\frac {1}{x}\) dx – 2 ∫1 dx [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C, \int \frac{1}{x} d x=\log |x|+C\)]
= \(\frac{x^2}{2}\) + log|x| – 2x + C
Question 11.
\(\int \frac{x^3+5 x^2-4}{x^2} d x\)
Solution:
\(\int \frac{x^3+5 x^2-4}{x^2} d x=\int \frac{x^3}{x^2} d x+5 \int \frac{x^2}{x^2} d x-4 \int \frac{1}{x^2} d x\)
= ∫x dx + 5 ∫1 dx – 4 ∫x-2 dx
= \(\frac{x^2}{2}+5 x-4\left(\frac{x^{-2+1}}{-2+1}\right)+C\)
= \(\frac{x^2}{2}+5 x-4\left(\frac{x^{-1}}{-1}\right)+C\) [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\)]
= \(\frac{x^2}{2}+5 x+\frac{4}{x}+C\)
Question 12.
\(\int \frac{x^3+3 x+4}{\sqrt{x}} d x\)
Solution:
\(\int \frac{\mathrm{x}^3+3 \mathrm{x}+4}{\sqrt{\mathrm{x}}} \mathrm{dx}=\int \frac{\mathrm{x}^3}{\sqrt{\mathrm{x}}} \mathrm{dx}+3 \int \frac{\mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}+4 \int \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}\)
= \(\int \mathrm{x}^3 \cdot \mathrm{x}^{-1 / 2} \mathrm{dx}+3 \int \mathrm{x}^{1 / 2} \mathrm{dx}+4 \int \mathrm{x}^{-1 / 2} \mathrm{dx}\)
= \(\int x^{(6-1) / 2} d x+3 \int x^{1 / 2} d x+4 \int x^{-1 / 2} d x\)
= \(\int x^{5 / 2} d x+3 \int x^{1 / 2} d x+4 \int x^{-1 / 2} d x\)
= \(\frac{x^{(5 / 2)+1}}{(5 / 2)+1}+\frac{3 x^{1 / 2+1}}{(1 / 2)+1}+\frac{3 x^{-1 / 2+1}}{(-1 / 2)+1}+C\)
= \(\frac{x^{7 / 2}}{7 / 2}+\frac{3 x^{3 / 2}}{3 / 2}+\frac{4 x^{1 / 2}}{1 / 2}+C\) [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\)]
= \(\frac{2}{7} x^{7 / 2}+2 \cdot x^{3 / 2}+8 x^{1 / 2}+C\)
Question 13.
\(\int \frac{x^3-x^2+x-1}{x-1} d x\)
Solution:
\(\int \frac{x^3-x^2+x-1}{x-1} d x=\int \frac{x^2(x-1)+1(x-1)}{(x-1)} d x\)
= \(\int \frac{\left(x^2+1\right)(x-1)}{(x-1)} d x\)
= ∫(x2 + 1) dx [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\)]
= \(\frac{x^3}{3}\) + x + C
Question 14.
∫(1 – x)√x dx
Solution:
\(\int(1-x) \sqrt{x} d x=\int\left(\sqrt{x}-x^{3 / 2}\right) d x\)
= \(\frac{x^{1 / 2+1}}{(1 / 2)+1}-\frac{x^{3 / 2+1}}{(3 / 2)+1}+C\)
= \(\frac{x^{(3 / 2)}}{(3 / 2)}-\frac{x^{(5 / 2)}}{(5 / 2)}+C\) [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\)]
= \(\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}+C\)
Question 15.
∫√x(3x2 + 2x + 3) dx
Solution:
∫√x(3x2 + 2x + 3) dx = \(3 \int x^{5 / 2} d x+2 \int x^{3 / 2} d x+3 \int x^{1 / 2} d x\)
= \(\frac{3 \cdot x^{5 / 2+1}}{(5 / 2)+1}+\frac{2 \cdot x^{3 / 2+1}}{(3 / 2)+1}+\frac{3 \cdot x^{1 / 2+1}}{(1 / 2)+1}+C\)
= \(\frac{3 x^{7 / 2}}{7 / 2}+\frac{2 x^{5 / 2}}{5 / 2}+\frac{3 x^{3 / 2}}{3 / 2}+C\) [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\)]
= \(\frac{6}{7} x^{7 / 2}+\frac{4}{5} x^{5 / 2}+2 x^{3 / 2}+C\)
Question 16.
∫(2x – 3 cos x + ex) dx
Solution:
∫(2x – 3 cos x + ex) dx = 2∫x dx – 3∫cos x dx + ∫ex dx
= \(\frac{2 x^2}{2}\) – 3 sin x + ex + C
= x2 – 3 sin x + ex + C
[∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\), ∫cos x dx = sin x + C, ∫ex dx = ex + C]
Question 17.
∫(2x2 – 3 sin x + 5√x) dx
Solution:
∫(2x2 – 3 sin x + 5√x) dx = \(\frac{2 x^3}{3}-3(-\cos x)+5 \frac{x^{1 / 2+1}}{(1 / 2)+1}+C\)
[∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\), ∫sin x dx = -cos x + C]
= \(\frac{2}{3} x^3+3 \cos x+\frac{5 x 2}{3} x^{3 / 2}+C\)
= \(\frac{2}{3} x^3+3 \cos x+\frac{10}{3} x^{3 / 2}+C\)
Question 18.
∫sec x (sec x + tan x) dx
Solution:
∫sec x (sec x + tan x) dx
= ∫sec2x dx + ∫sec x . tan x dx
= tan x + sec x + C [∵ ∫sec2x dx = tan x + C, ∫sec x tan x dx = sec x + C]
Question 19.
\(\int \frac{\sec ^2 x}{{cosec}^2 x} d x\)
Solution:
\(\int \frac{\sec ^2 x}{{cosec}^2 x} d x\)
= \(\int \frac{\sin ^2 x}{\cos ^2 x} d x\) [∵ \(\frac{1}{{cosec}^2 x}=\sin ^2 x\) & \(\sec ^2 x=\frac{1}{\cos ^2 x}\)]
= ∫tan2x dx
= ∫(sec2x – 1) dx [∵ ∫sec2x dx = tan x + C, ∫dx = x + C]
= ∫sec2x dx – ∫1 dx
= tan x – x + C
Question 20.
\(\int \frac{2-3 \sin x}{\cos ^2 x} d x\)
Solution:
\(\int \frac{2-3 \sin x}{\cos ^2 x} d x=\int \frac{2}{\cos ^2 x} d x-3 \int \frac{\sin x}{\cos ^2 x} d x\)
= 2∫sec2x dx – 3∫tan x . sec x dx
= 2 tan x – 3 sec x + C [∵ ∫sec2x dx = tan x + C, ∫sec x tan x dx = sec x + C]
Question 21.
\(\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x\)
Solution:
\(\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x\)
= \(\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}} d x\)
= \(\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} d x\) [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\)]
= \(\frac{1}{a-b} \int\left\{(x+a)^{1 / 2}-(x+b)^{1 / 2}\right\} d x\)
= \(\frac{1}{a-b}\left\{\frac{(x+a)^{1 / 2+1}}{1 / 2+1}-\frac{(x+b)^{1 / 2+1}}{1 / 2+1}\right\}+C\)
= \(\frac{1}{a-b}\left\{\frac{(x+a)^{3 / 2}}{3 / 2}-\frac{(x+b)^{3 / 2}}{3 / 2}\right\}+C\)
= \(\frac{2}{3(a-b)}\left[(x+a)^{3 / 2}-(x+b)^{3 / 2}\right]+C\)
Question 22.
\(\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x\)
Solution:
\(\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=\int \frac{e^{\log x^5}-e^{\log x^4}}{e^{\log x^3}-e^{\log x^2}} d x\)
= \(\int \frac{x^5-x^4}{x^3-x^2} d x\) [∵ elog x = x]
= \(\int \frac{x^4(x-1)}{x^2(x-1)} d x\) [∵ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C\)]
= ∫x2 dx
= \(\frac{1}{3} x^3+C\)