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Integrals

Multiple Choice Questions

661.

$\int_{0}^{1} x^{\frac{3}{2}} \sqrt{1 - x} dx$ is equal to

$\frac{π}{6}$
$\frac{π}{9}$
$\frac{π}{12}$
$\frac{π}{16}$

662.

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \sin |x| dx$ is equal to

0
1
2
$π$

663.

$\int \frac{dx}{(x + 1) \sqrt{4 x + 3}} = ?$

${tan}^{- 1} (\sqrt{4 x + 3}) + c$
$3 \tan^{- 1} (\sqrt{4 x + 3}) + c$
$2 \tan^{- 1} (\sqrt{4 x + 3}) + c$
$4 \tan^{- 1} (\sqrt{4 x + 3}) + c$

664.

$\int (\frac{2 - \sin (2 x)}{1 - \cos (2 x)}) e^{x} dx = ?$

$- \cot (x) e^{x} + c$
$c o t (x) e^{x} + c$
$2 \cot (x) e^{x} + c$
$- 2 \cot (x) e^{x} + c$

665.

$If I_{n} = \int \sin^{n} xdx, then {nI}_{n} - (n - 1) I_{n - 2} = ?$

$\sin^{n - 1} (x) \cos (x)$
$\cos^{n - 1} (x) \sin (x)$
$- \sin^{n - 1} (x) \cos (x)$
$- \cos^{n - 1} (x) \sin (x)$

666.

$\int_{0}^{π} \frac{1}{1 + \sin (x)} d x = ?$

667.

$The line x = \frac{π}{4} divides the area of the region bounded by y = \sin (x), y = \cos (x) and x - axis (0 \leq x \frac{π}{2}) into two regions of areas A_{1} and A_{2} . Then A_{1}, A_{2} equals$

4 : 1
3 : 1
2 : 1
1 : 1

668.

The velocity of a particle which starts from rest is given by the following table

t (in sec) 0 2 4 6 8 10

v (in m/s) 0 12 16 20 35 60

The total distance travelled (in metre) by the particles in 10s, using trapezoidal rule is given by

669.

$If \int \frac{7 x^{8} + 8 x^{7}}{{(1 + x + x^{8})}^{2}} dx = f (x) + c, then f (x) = ?$

$\frac{x^{8}}{1 + x + x^{8}}$
$28 \log (1 + x + x^{8})$
$\frac{1}{1 + x + x^{8}}$
$\frac{- 1}{1 + x + x^{8}}$

670.
$If f_{n} (x) = \log \log \log . . . \log (x) (\log is repeated n - times), then \int {({xf}_{1} (x) f_{2} (x) . . . f_{n} (x))}^{- 1} dx is equal to$

$f_{n + 1} + c$

$\frac{f_{n + 1} (x)}{n + 1} + c$

${nf}_{n (x)} + c$

$\frac{f_{n} (x)}{n} + c$

$f_{n + 1} + c$

$f_{n} (x) = \log . \log . \log . . . \log (x) (upto n times) f_{1} (x) = \log (x) f_{2} (x) = \log \log (x) f_{3} (x) = \log \log \log (x) f_{n - 1} (x) = \log \log \log . . . . . \log (x) (up to (n - 1) times) Now, \int {((x) f_{1} (x) f_{2} (x) . . . f_{n} (x))}^{- 1} dx = \int \frac{dx}{[(x) f_{1} (x) f_{2} (x) . . . f_{n} (x)]} = \int \frac{[{xf}_{1} (x) f_{2} (x) . . . f_{n - 1} (x)] dt}{[{xf}_{1} (x) f_{2} (x) . . . f_{n - 1} (x)] . t} = \int \frac{dt}{t} = \log (t) + c$

$[Put f_{n} (x) = t \frac{dt}{dx} = \frac{1}{[f_{n - 1} (x) f_{n - 2} (x) . . . f_{1} (x) . x]} dx = [{xf}_{1} (x) . f_{2} (x) . . . f_{n - 1} (x)] dt] = \log (f_{n} (x)) + c = f_{n + 1} (x) + c Hence, ({xf}_{1} (x) f_{2} (x) . . . f_{n} {(x)}^{- 1}) dx = f_{n + 1} (x) + c$

Previous Year Papers

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Multiple Choice Questions

670. If fnx = log log log . . .logx log is repeated n-times, then∫xf1xf2x . . . fnx - 1dx is equal to fn + 1 + c fn + 1xn + 1 + c nfnx + c fnxn + c

670.
$If f_{n} (x) = \log \log \log . . . \log (x) (\log is repeated n - times), then \int {({xf}_{1} (x) f_{2} (x) . . . f_{n} (x))}^{- 1} dx is equal to$

$f_{n + 1} + c$

$\frac{f_{n + 1} (x)}{n + 1} + c$

${nf}_{n (x)} + c$

$\frac{f_{n} (x)}{n} + c$