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Integrals

Multiple Choice Questions

361.

Which of the following is true ?

$\int_{0}^{1} e^{x} dx = e$
$\int_{0}^{1} 2^{x} dx = \log (2)$
$\int_{0}^{1} \sqrt{x} dx = \frac{2}{3}$
$\int_{0}^{1} xdx = \frac{1}{3}$

362.

$\int [\sin (\log (x)) + \cos (\log (x))] dx$ is equal to

$xcos (\log (x)) + c$
$\cos (\log (x)) + c$
$xsin (\log (x)) + c$
$\sin (\log (x)) + c$

363.

$\int e^{x} \frac{(x - 1)}{x^{2}} dx$ is equal to

$\frac{e^{x}}{x^{2}} + c$
$\frac{- e^{x}}{x^{2}} + c$
$\frac{e^{x}}{x} + c$
$\frac{- e^{x}}{x} + c$

364.

$\int_{5}^{10} \frac{1}{(x - 1) (x - 2)} dx$

$\log (\frac{27}{32})$
$\log (\frac{32}{27})$
$\log (\frac{8}{9})$
$\log (\frac{3}{4})$

365.

$\int xlog (x) dx$ is equal to

$\frac{x^{2}}{4} (2 \log (x) - 1) + c$
$\frac{x^{2}}{2} (2 \log (x) - 1)$
$\frac{x^{2}}{4} (2 \log (x) + 1) + c$
$\frac{x^{2}}{2} (2 \log (x) + 1)$

366.
$\int_{0}^{\frac{π}{2}} \frac{\sin (x) - \cos (x)}{1 - \sin (x) . \cos (x)} dx$ is equal to

0

$\frac{π}{2}$

$\frac{π}{4}$

$π$

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin (x) - \cos (x)}{1 - \sin (x) . \cos (x)} dx . . . (i) On putting x = (\frac{π}{2} - x) in Eq . (i), we get I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - x) - \cos (\frac{π}{2} - x)}{1 - \sin (\frac{π}{2} - x) . \cos (\frac{π}{2} - x)} dx = \int_{0}^{\frac{π}{2}} \frac{\cos (x) - \sin (x)}{1 - \sin (x) . \cos (x)} dx = - \int_{0}^{\frac{π}{2}} (\frac{\sin (x) - \cos (x)}{1 - \sin (x) . \cos (x)}) dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} 0 dx = 0 \Rightarrow I = 0$