Practice AP Inter 2nd Year Maths Study Material Chapter 7 Integrals MCQ to identify your strengths and weak areas.
AP Inter 2nd Year Maths Integrals MCQ
Indefinite Integrals
Question 1.
The anti derivative of \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\) =
1) \(\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+C\)
2) \(\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^2+C\)
3) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)
4) \(\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+C\)
Solution:
3) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)
Anti derivative of \(\sqrt{x}+\frac{1}{\sqrt{x}}=\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x\)
⇒ I = \(\int x^{\frac{1}{2}} d x+\int x^{\frac{1}{2}} d x=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c\)
Question 2.
The anti derivative of e2logcotx
1) cot x – 1
2) tan x – cot x
3) -cot x – x
4) – 1 – cot x
Solution:
3) -cot x – x
I = \(\int e^{2 \log \cot x} d x=\int e^{\log _e \cot ^2 x} d x=\int \cot ^2 x d x=\int\left({cosec}^2 x-1\right) d x\) = -cot x – x + c
Question 3.
If \(\frac{d}{d x}\)f(x) = 4x3 – \(\frac{3}{x^4}\) such that f(2) = 0. Then f(x) is
1) \(x^4+\frac{1}{x^3}-\frac{129}{8}\)
2) \(x^3+\frac{1}{x^4}+\frac{129}{8}\)
3) \(x^4+\frac{1}{x^3}+\frac{129}{8}\)
4) \(x^3+\frac{1}{x^4}-\frac{129}{8}\)
Solution:
1) \(x^4+\frac{1}{x^3}-\frac{129}{8}\)
\(\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4} \Rightarrow f(x)=\int\left(4 x^3-\frac{3}{x^4}\right) d x=\not A \cdot \frac{x^4}{\not A}-\not z\left(\frac{-1}{\not \partial x^3}\right)=x^4+\frac{1}{x^3}+c\) …….(1)
Given, f(x) = 0 ⇒ 0 = 16 + \(\frac{1}{8}+c \Rightarrow c=-\left(\frac{129}{8}\right)(1) \Rightarrow f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}\)
Question 4.
\(\int \frac{10 x^9+10^x \log _e 10}{x^{10}+10^x}\)dx =
1) 10x – 1010 + C
2) 10x + x10 + C
3) (10x – x10)-1 + C
4) log(10x + x10) + C
Solution:
4) log(10x + x10) + C
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log|f(x)| + c
I = \(\int \frac{10 x^9+10^x \log _e^{10}}{x^{10}+10^x} d x\) = log(x10 + xx) + c
Question 5.
\(\int \frac{x^2}{1+x^3}\) dx =
1) \(\frac{1}{3} \log \left|1+x^3\right|+c\)
2) \(\frac{2}{3} \log \left|1+x^3\right|+c\)
3) log|1 + x3| + c
4) tan-1(x3/2 + c
Solution:
1) \(\frac{1}{3} \log \left|1+x^3\right|+c\)
Put 1 + x3 = t ⇒ 0 + 3x2dx = dt ⇒ x2 dx = \(\frac{1}{3}\)dt
I = \(\int \frac{x^2}{1+x^3} d x=\frac{1}{3} \int \frac{1}{t} d t=\frac{1}{3} \log |t|+c=\frac{1}{3} \log \left|1+x^3\right|+c\)
Question 6.
\(\int \frac{d x}{\sin ^2 x \cos ^2 x}\) =
1) tan x + cot x + C
2) tan x – cot x + C
3)tan x cot x + C
4) tan x – cot 2x + C
Solution:
2) tan x – cot x + C
I = \(\int \frac{1}{\sin ^2 x \cos ^2 x} d x=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x} d x\)
= \(\int \frac{\sin ^2 x}{\sin ^2 x \cos ^2 x} d x+\int \frac{\cos ^2 x}{\sin ^2 x \cos ^2 x} d x\) = ∫sec2 dx + ∫cosec2 dx = tan x – cot x + c
Question 7.
\(\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x}\)dx =
1) tanx + cot x + C
2) tan x + cosecx + C
3) -tan x + cot x + C
4) tan x + sec x + C
Solution:
1) tanx + cot x + C
I = \(\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x=\int \frac{1}{\cos ^2 x} d x-\int \frac{1}{\sin ^2 x} d x\)
= ∫sec2 x dx – ∫cosec2x dx = tan x + cot x + c
Question 8.
\(\int \frac{\cos x+x \sin x}{x(x+\cos x)}\)dx = log|f(x)| + c then f(x) =
1) x(x + cos x)
2) \(\frac{x+\cos x}{x}\)
3) \(\frac{x}{x+\cos x}\)
4) \(\frac{1}{x(x+\cos x)}\)
Solution:
3) \(\frac{x}{x+\cos x}\)
Question 9.
\(\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)}\)dx =
1) -cot(exx) + C
2) tan(xex) + C
3) tan(ex) + C
4) cot(ex) + C
Solution:
2) tan(xex) + C
Put, x.ex = t ⇒ (xex + ex)dx = dt ⇒ ex(x + 1)dx = dt
I = \(\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)} d x \Rightarrow I=\int \frac{d x}{\cos ^2 t}\) = ∫sec2 dt = tan t + c = tan(xex) + c
Question 10.
\(\int \frac{d x}{x^2+2 x+2}\) =
1) x tan-1(x + 1) + C
2) tan-1 (x + 1) + C
3) (x + 1)tan-1x + C
4) tan-1x + C
Solution:
2) tan-1 (x + 1) + C
I = \(\int \frac{\mathrm{dx}}{\mathrm{x}^2+2 \mathrm{x}+2} \mathrm{dx}=\int \frac{1}{\mathrm{x}^2+2 \mathrm{x}+1+1} \mathrm{dx}=\int \frac{1}{(\mathrm{x}+1)^2+1^2} \mathrm{dx}\) = tan-1 (x + 1) + C
Question 11.
\(\int \frac{d x}{\sqrt{9 x-4 x^2}}\) =
1) \(\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C\)
2) \(\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C\)
3) \(\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C\)
4) \(\frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{9}\right)+C\)
Solution:
2) \(\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C\)
9x – 4x2 = \(-4\left[x^2-\frac{9}{4} x\right]=-4\left[x^2-2 \cdot x \cdot \frac{9}{8}+\left(\frac{9}{8}\right)^2-\left(\frac{9}{8}\right)^2\right]=4\left[\left(\frac{9}{8}\right)^2-\left(x-\frac{9}{8}\right)^2\right]\)
Question 12.
\(\int \frac{x d x}{(x-1)(x-2)}\) =
1) \(\log \left|\frac{(x-1)^2}{x-2}\right|+C\)
2) \(\log \left|\frac{(x-2)^2}{x-1}\right|+C\)
3) \(\log \left|\left(\frac{x-1}{x-2}\right)^2\right|+C\)
4) log|(x – 1)(x -2)| + C
Solution:
2) \(\log \left|\frac{(x-2)^2}{x-1}\right|+C\)
Using partial fractions \(\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\) ⇒ x = A(x – 2) + B(x – 1)
ar x = 1 we get A = -1; at x = 2 we get B = 2
I = \(\int \frac{x}{(x-1)(x-2)} d x=\int \frac{-1}{x-1} d x+\int \frac{2}{x-2} d x\) = -log|x – 1| + 2log|x – 2|
= -log|x – 1| + log|(x – 2)|2 = \(\log \left|\frac{(x-2)^2}{(x-1)}\right|+c\)
Question 13.
\(\int \frac{d x}{x\left(x^2+1\right)}\) =
1) \(\log |x|-\frac{1}{2} \log \left|x^2+1\right|+C\)
2) \(\log |x|+\frac{1}{2} \log \left|x^2+1\right|+C\)
3) \(-\log |x|+\frac{1}{2} \log \left|x^2+1\right|+C\)
4) \(\frac{1}{2} \log |x|+\log \left|x^2+1\right|+C\)
Solution:
1) \(\log |x|-\frac{1}{2} \log \left|x^2+1\right|+C\)
\(\frac{1}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{B x+C}{x^2+1}\) we get A = 1; B = -1; C = 0
I = \(\int \frac{1}{x\left(x^2+1\right)} d x=\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right) d x=\int \frac{1}{x} d x-\frac{1}{2} \int \frac{2 x}{x^2+1} d x=\log |x|-\frac{1}{2} \log \left|x^2+1\right|+c\)
Question 14.
\(\int \frac{x^2}{1-x^4} d x\) =
1) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|-\frac{1}{2} \tan ^{-1} x+C\)
2) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|+\frac{1}{2} \tan ^{-1} x+C\)
3) \(\frac{1}{4} \log \left|\frac{1+x^2}{1-x^2}\right|-\frac{1}{2} \tan ^{-1} x^2+C\)
4) \(\frac{1}{4} \log \left|\frac{1-x^2}{1+x^2}\right|-\frac{1}{2} \tan ^{-1} x^2+C\)
Solution:
1) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|-\frac{1}{2} \tan ^{-1} x+C\)
Question 15.
∫x2ex dx =
1) \(\frac{1}{3}\)ex3 + C
2) \(\frac{1}{3}\)ex2 + C
3) \(\frac{1}{2}\)ex3 + C
4) \(\frac{1}{2}\)ex2 + C
Solution:
1) \(\frac{1}{3}\)ex3 + C
Put, x3 = t ⇒ 3x2dx = dt ⇒ x2dx = \(\frac{1}{3}\)dt
I = ∫x2ex3 dx = \(\frac{1}{3}\)∫etdt = \(\frac{1}{3}\). et + c = \(\frac{1}{3}\) ex3 + c
Question 16.
∫ x sec2 x dx =
1) xtanx – log|sec x| + C
2) xtanx – log|cos x| + C
3) xtanx – log|cosec x| + C
4) xtanx – log|sin x| + C
Solution:
1) xtanx – log|sec x| + C
Integration by parts we have
I = x(tan x) – ∫tan x dx = ∫ x sec2x dx = x(tan x) – log|sec| + c
Question 17.
∫ex sec x( + tan x) dx =
1) excos x + C
2) ex sec x + C
3) ex sin x + C
4) ex tan x + C
Solution:
2) ex sec x + C
∫ ex(f(x) + f'(x)) dx = exf(x) + c
I = ∫exsecx(1 + tan x) dx = ∫ex[sec x + sec x tan x]dx = ex sec x + c
Question 18.
\(\int e^x\left(\frac{1+x \log x}{x}\right) d x\) =
1) xelog x + C
2) ex log x +C
3) ex log x2 + C
4) None
Solution:
2) ex log x +C
I = ∫ex[f(x) + f'(x)]dx = exf(x) + c
I = \(\int e^x\left(\frac{1+x \log x}{x}\right) d x=\int e^x\left(\frac{1}{x}+\log x\right) d x=\int e^x\left(\log x+\frac{1}{x}\right) d x\) = ex(log x) + c
Question 19.
\(\int \sqrt{1+x^2} d x\) =
1) \(\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^2}\right)\right|+C\)
2) \(\frac{2}{3}\left(1+x^2\right)^{\frac{3}{2}}+C\)
3) \(\frac{2}{3} x\left(1+x^2\right)^{\frac{3}{2}}+C\)
4) \(\frac{x^2}{2} \sqrt{1+x^2}+\frac{1}{2} x^2 \log \left|x+\sqrt{1+x^2}\right|+C\)
Solution:
1) \(\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^2}\right)\right|+C\)
\(\int \sqrt{a^2+x^2} d x=\frac{x}{2} \sqrt{a^2+x^2}+\frac{a^2}{2} \log \left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}\right|+c\)
I = \(\int \sqrt{1+\mathrm{x}^2} \mathrm{dx}=\frac{\mathrm{x}}{2} \sqrt{1+\mathrm{x}^2}+\frac{1}{2} \log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+1}\right|+\mathrm{c}\)
Question 20.
\(\int \sqrt{x^2-8 x+7} d x\) =
1) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}+9 \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
2) \(\frac{1}{2}(x+4) \sqrt{x^2-8 x+7}+9 \log \left|x+4+\sqrt{x^2-8 x+7}\right|+C\)
3) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
4) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
Solution:
4) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
Question 21.
\(\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\) =
1) tan-1(ex) + C
2) tan-1(e-x) + C
3) log(ex – e-x) + C
4) log(ex + e-x) + C
Solution:
1) tan-1(ex) + C
I = \(\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}=\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\frac{1}{\mathrm{e}^{\mathrm{x}}}}=\int \frac{\mathrm{e}^{\mathrm{x}}}{\left(\mathrm{e}^{\mathrm{x}}\right)^2+1^2} \mathrm{dx}\). Put ex = t ⇒ ex dx = dt
I = \(\int \frac{d t}{t^2+1}\) = tan-1(t) + c = tan-1(ex) + c
Question 22.
\(\int \frac{\cos 2 x}{(\sin x+\cos x)^2} d x\) =
1) \(\frac{-1}{\sin x+\cos x}+C\)
2) log|sin x – cos x| + C
3) log|sin x – cos x| + C
4) \(\frac{1}{(\sin x+\cos x)^2}\)
Solution:
2) log|sin x – cos x| + C
I = \(\int \frac{\cos 2 x}{(\sin x+\cos x)^2}=d x=\int \frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2} d x\)
= \(\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)(\cos x+\sin x)} d x\) = log|cos x + sin x| + c
Definite Intervals
Question 1.
\(\int_1^{\sqrt{3}} \frac{d x}{1+x^2}\) =
1) \(\frac{\pi}{3}\)
2) \(\frac{2 \pi}{3}\)
3) \(\frac{\pi}{6}\)
4) \(\frac{\pi}{12}\)
Solution:
4) \(\frac{\pi}{12}\)
Textual given key is 1.
I = \(\int \frac{1}{\left(1+x^2\right)} d x=\left(\tan ^{-1}(x)\right)_1^{\sqrt{3}}\) = tan-1(\(\sqrt{3}\)) – tan-1(1) = 60 – 45 = 15 = \(\frac{\pi}{12}\)
Question 2.
\(\int_0^{\frac{2}{3}} \frac{d x}{4+9 x^2}\) =
1) \(\frac{\pi}{6}\)
2) \(\frac{\pi}{12}\)
3) \(\frac{\pi}{24}\)
4) \(\frac{\pi}{4}\)
Solution:
3) \(\frac{\pi}{24}\)
Textual given key is 4.
Question 3.
The value of the integral \(\int_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^2} d x\) is
1) 6
2) 0
3) 3
4) 4
Solution:
1) 6
Question 4.
If f(x) = \(\int_0^x t\) sin t dt, then f'(x) is
1) cos x + x sin x
2) x sin x
3) x cos x
4) sinx + x cosx
Solution:
2) x sin x
f(x) = \(\int_0^x t \sin t d x\) Diff. w.r.t we get f'(x) = x sin x
Question 5.
The value \(\int_{-\pi / 2}^{\pi / 2}\left(x^3+x \cos x+\tan ^5 x+1\right) d x\) is
1) 0
2) 2
3) π
4) 1
Solution:
3) π
I = \(\int_{-\pi / 2}^{\pi / 2}\left(x^3+x \cos x+\tan ^5 x+1\right)=\int_{-\pi / 2}^{\pi / 2} 1 d x=(x)_{-\pi / 2}^{\pi / 2}=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\frac{\pi}{2}+\frac{\pi}{2}=\pi\)
Question 6.
The value of \(\int_0^\pi 2 \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\) is
1) 2
2) 3/4
3) 0
4) -2
Solution:
3) 0
\(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)
I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x \quad \ldots \ldots \ldots .(1) \quad I=\int_0^{\pi / 2} \log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] d x\) …..(2)
I + I = \(\int_0^{\pi / 2}\left[\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)+\log \left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right] d x \Rightarrow 2 I=\int_0^{\pi / 2} \log (1) d x \Rightarrow 2 I=0 \Rightarrow I=0\)
Question 7.
If f(a + b – x) = f(x), then \(\int_a^b f(x) d x\) =
1) \(\frac{(a+b)}{2} \int_a^b f(b-x) d x\)
2) \(\frac{(a+b)}{2} \int_a^b f(b+x) d x\)
3) \(\frac{b-a}{2} \int_a^b f(x) d x\)
4) \(\frac{a+b}{2} \int_a^b f(x) d x\)
Solution:
4) \(\frac{a+b}{2} \int_a^b f(x) d x\)
\(\int_a^b x f(x) d x=\int_a^b(a+b-x) f(a+b-x) d x=\int_a^b[(a+b)-x] f(x) d x=\int_a^b(a+b) f(x) d x-\int_a^b x f(x) d x\)
\(\int_a^b x f(x) d x=(a+b) \int_a^b f(x) d x-\int_a^b x f(x) d x\) ⇒ \(2 \int_a^b x f(x) d x=(a+b) \int_a^b f(x) d x\)
⇒ \(\int_a^b x f(x) d x=\left(\frac{a+b}{2}\right) \int_a^b f(x) d x\)
Question 8.
\(\int_0^{\pi / 2} \frac{3 \sin x+5 \cos x}{\sin x+\cos x} d x\) =
1) 2π
2) π
3) 4π
4) 8π
Solution:
1) 2π
\(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)
Question 9.
\(\int_0^4|2-x| d x\) =
1) 12
2) 4
3) 8
4) 2
Solution:
2) 4
I = \(\int_0^4|2-x| d x\) |2 – x| = 2x if 2 – x ≥ 0; 2 ≥ x; x ≤ 2
= \(\int_0^2|2-x| d x+\int_2^4|2-x| d x=\int_0^2(2-x) d x+\int_2^4-(2-x) d x=\left(2 x-\frac{x^2}{2}\right)_0^2-\left(2 x-\frac{x^2}{2}\right)_2^4\)
= (4 – 2) – 0 – [(8 – 8) – (4 – 2)] = 2 – [0 – 2] = 2 + 2 = 4
Question 10.
\(\int_{-2}^2\left(4-x^2\right)^{\frac{3}{2}} d x\) =
1) 2π
2) 4π
3) 6π
4) 8π
Solution:
3) 6π
