Integral Calculus Questions and Answers Part-17

1.If \[I=\int_{0}^{\pi/4} \frac{\sin 2\theta}{\sin^{2}\theta+\cos^{4}\theta}d\theta\]
then I equals
a) \[\pi/2\]
b) \[\pi/\sqrt{3}\]
c) \[\pi/2\sqrt{3}\]
d) \[\pi/3\sqrt{3}\]

Answer: d
Explanation:
q161
q161a

2. Let \[u\left(x\right)=\log\left(x+\sqrt{1+x^{2}}\right),x\epsilon R\]
and \[f\left(x\right)=\frac{\left(1-x+x^{2}\right)\cos u\left(x\right)-\sqrt{1+x^{2}}\sin u\left(x\right)}{\left(1+x^{2}\right)^{3/2}},x\epsilon R\]
then \[\int_{0}^{1} f\left(x\right)dx\]     is equal to
a) \[\sin \left(u\left(1\right)\right)+\cos \left(u\left(1\right)\right)-1\]
b) \[\sin \left(u\left(1\right)\right)+\frac{1}{\sqrt{2}}\cos \left(u\left(1\right)\right)-1\]
c) \[\frac{1}{\sqrt{2}}\sin \left(u\left(1\right)\right)+\cos \left(u\left(1\right)\right)-1\]
d) \[\frac{1}{\sqrt{2}}\left[\sin \left(u\left(1\right)\right)+\cos \left(u\left(1\right)\right)-1\right]\]

Answer: b
Explanation:
q162
q162a

3.If \[I=\int_{8}^{15}\frac{dx}{\left(x-3\right)\sqrt{x+1}}\]
then I equals
a) \[\frac{1}{2}\log\frac{5}{3}\]
b) \[2\log\frac{1}{3}\]
c) \[\frac{1}{2}\log\frac{1}{5}\]
d) \[2\log\frac{5}{3}\]

Answer: a
Explanation:
q163
q163a

4. If \[I=\int_{1}^{\infty}\frac{dx}{x^{2}\sqrt{1+x}}\]
then I equals
a) \[\sqrt{2}+\log(\sqrt{2}-1)\]
b) \[\sqrt{2}-\log(\sqrt{2}-1)\]
c) \[\log(\sqrt{2}+1)-\sqrt{2}\]
d) \[\sqrt{2}+\log(3-\sqrt{2})\]

Answer: a
Explanation:
q164

5. \[\lim_{x \rightarrow \pi/4}\frac{\int_{2}^{\sec^{2} x}f\left(t\right)dt }{x^{2}-\pi^{2}/16}\]     equals
a) \[\frac{8}{\pi}f\left(2\right)\]
b) \[\frac{2}{\pi}f\left(2\right)\]
c) \[\frac{2}{\pi}f\left(\frac{1}{2}\right)\]
d) 4 f (2)

Answer: a
Explanation:
q165
q165a
q165b

6. The area of the region between the curves \[y=\sqrt{\frac{1+\sin x}{\cos x}}\]     and \[y=\sqrt{\frac{1-\sin x}{\cos x}}\]
bounded by the lines x = 0 and \[x=\frac{\pi}{4}\]
is
a) \[\int_{0}^{\sqrt{2}-1} \frac{t}{\left(1+t^{2}\right)\sqrt{1-t^{2}}}dt\]
b) \[\int_{0}^{\sqrt{2}-1} \frac{4t}{\left(1+t^{2}\right)\sqrt{1-t^{2}}}dt\]
c) \[\int_{0}^{\sqrt{2}+1} \frac{4t}{\left(1+t^{2}\right)\sqrt{1-t^{2}}}dt\]
d) \[\int_{0}^{\sqrt{2}+1} \frac{t}{\left(1+t^{2}\right)\sqrt{1-t^{2}}}dt\]

Answer: b
Explanation:
q166
q166a

7. For \[k\epsilon N\] , Let \[a_{k}=\int_{0}^{1} \frac{x\sin \pi x}{x+\left(1-x\right)k^{1-2x}}dx\]
a) \[a_{k}\] is independent of k
b) \[\lim_{n \rightarrow \infty}\left(\frac{1}{n}\sum_{k=1}^na_{k}\right)=\pi\]
c) \[a_{1},a_{2},a_{3}....\]     an A.P
d) All of the Above

Answer: d
Explanation:
q167
q167a

8. For \[k\epsilon N\] ,Let \[b_{k}=\int_{1}^{e} \frac{1}{\left(k+\log x\right)\left(k+1+\log x\right)}\frac{dx}{x}\]
and \[a_{n}=\sum_{k=1}^n b_{k}\]
then
a) \[b_{k}=\log\left(\frac{\left(k+1\right)^{2}}{k\left(k+2\right)}\right)\]
b) \[a_{n}=\log\left(\frac{2n+2}{n+2}\right)\]
c) \[\lim_{n \rightarrow \infty}a_{n}=\log2\]
d) All of the Above

Answer: d
Explanation:
q168
q168a

9. a, \[b\epsilon R\] , \[\mid a\mid \leq 1,\mid b\mid \leq 1,\]
let \[I \left(a,b\right)=\int_{0}^{\pi} \left(a\sin x+b\cos x\right)^{3}dx\]
then
a) I (a, b) is independent of a
b) minimum value of I (–1, b) is \[\frac{-10}{3}\]
c) Maximum value of I (a, 1) is \[\frac{10}{3}\]
d) Both b and c

Answer: d
Explanation:
q169
q169a

10. For \[n\epsilon N\] , n > 2, let \[I_{n}=\int_{0}^{1} \frac{dx}{\left(1+x^{n}\right)^{1+1/n}}dx\]
then
a) \[\left(I_{n}\right)^{n}=\frac{1}{2}\]
b) \[\lim_{n \rightarrow \infty}I_{n}=1\]
c) \[\lim_{n \rightarrow \infty}I_{n}=ln 2\]
d) Both a and b

Answer: d
Explanation:
q170
q170a