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Integrals

Multiple Choice Questions

951.

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

952.

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

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

953.

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

954.

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

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

955.
$If f_{n} (x) = \log \log \log . . . \log (x) (\log is repeated n - times), then \int {({xf}_{1} (x) f_{2} (x) . . . f_{n} (x))}^{- 1} dx is equal to$

$f_{n + 1} + c$

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

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

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

$f_{n + 1} + c$

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

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