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Integrals

Multiple Choice Questions

701.

$If \int x^{3} e^{5 x} dx = \frac{e^{5 x}}{5^{4}} [f (x)] + C, then f (x) = ?$

$\frac{x^{3}}{5} - \frac{3 x^{2}}{5^{2}} + \frac{6 x}{5^{3}} - \frac{6}{5^{4}}$
$5 x^{3} - 5^{2} x^{2} + 5^{3} x - 6$
$5^{3} x^{3} - 15 x^{2} + 30 x - 6$
$5^{3} x^{3} - 75 x^{2} + 30 x - 6$

702.

$\int \frac{x}{{(x^{2} + 2 x + 2)}^{2}} dx = ?$

$\frac{x^{2} + 2}{x^{2} + 2 x + 2} - \frac{1}{2} \tan^{- 1} (x - 1) + C$
$\frac{x^{2} - 2}{4 (x^{2} + 2 x + 2)} - \frac{1}{2} \tan^{- 1} (x + 1) + C$
$\frac{x^{2} + 2}{2 (x^{2} + 2 x + 2)} - \frac{1}{2} \tan^{- 1} (x + 1) + C$
$\frac{2 (x - 1)}{x^{2} + 2 x + 2} + \frac{1}{2} \tan^{- 1} (x + 1) + C$

703.

$If \int \log (a^{2} + x^{2}) dx = h (x) + C, then h (x) = ?$

$xlog (a^{2} + x^{2}) + 2 \tan^{- 1} (\frac{x}{a})$
$x^{2} \log (a^{2} + x^{2}) + x + {atan}^{- 1} (\frac{x}{a})$
$xlog (a^{2} + x^{2}) - 2 x + 2 {atan}^{- 1} (\frac{x}{a})$
$x^{2} \log (a^{2} + x^{2}) + 2 x - a^{2} \tan^{- 1} (\frac{x}{a})$

704.

$For x > 0, if \int (\log {(x)}^{5}) dx = ? x [A (\log {(x)}^{5}) + B {(\log (x))}^{4} + C {(\log (x))}^{3} + D {(\log (x))}^{2} + E (\log (x)) + F] + Constant, then A + B + C + D + E + F = ?$

- 44
- 42
- 40
- 36

705.

By the definition of the definite integral, the value of $\lim_{n \to \infty} (\frac{1}{\sqrt{n^{2} - 1}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + . . . \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}})$ is equal to

$π$
$\frac{π}{2}$
$\frac{π}{4}$
$\frac{π}{6}$

706.

$\int_{\frac{π}{4}}^{\frac{π}{4}} (\frac{x + \frac{π}{4}}{2 - \cos (2 x)}) dx$ is equal to

$\frac{8 π \sqrt{3}}{5}$
$\frac{2 π \sqrt{3}}{9}$
$\frac{4 π^{2} \sqrt{3}}{9}$
$\frac{π^{2}}{6 \sqrt{3}}$

707.

$\int \frac{5 x + 3}{x^{2} (x^{2} - 2)} dx = ?$

$\frac{3}{2 x} + \frac{\sqrt{13}}{2 \sqrt{2}} \log (|\frac{\sqrt{2} - x}{\sqrt{2} + x}|) + C$
$\frac{3}{2 x} + \frac{13}{4 \sqrt{2}} \log (|\frac{x + \sqrt{2}}{x - \sqrt{2}}|) + C$
$\frac{3}{2 x} + \frac{13}{4 \sqrt{2}} \log (|\frac{x - \sqrt{2}}{x + \sqrt{2}}|) + C$
$\frac{3}{5} x + \frac{5}{3 \sqrt{2}} \log (|\frac{x + \sqrt{2}}{x - \sqrt{2}}|) + C$

708.

$If y = \tan^{- 1} (\{\frac{x}{1 + \sqrt{1 - x^{2}}}\}) + \sin (2 \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}})), then \frac{dy}{dx} = ?$

$\frac{1 - 2 x}{2 \sqrt{1 - x^{2}}}$
$\frac{1 - 2 x}{x \sqrt{1 - x^{2}}}$
$\frac{2 x + 1}{x \sqrt{1 - x}}$
$\frac{2 - x}{2 \sqrt{1 - x^{2}}}$

709.
If $\int_{0}^{10} f (x) dx = 5, then \sum_{k = 1}^{10} \int_{0}^{1} f (k - 1 + x) dx$ = ?

50

10

5

20

$Let I = \int_{0}^{1} f (k - 1 + x) dx Now, put k - 1 + x = t \Rightarrow dx = dt Also, when x = 0 then t = k - 1 and when x = 1, then t = k ∴ I = \int_{k - 1}^{k} f (t) dt = \int_{k - 1}^{k} f (x) dx [∵ t is dummy variable] Now, \sum_{k = 1}^{10} \int_{0}^{1} f (k - 1 + x) dx = \sum_{k = 1}^{10} \int_{k - 1}^{k} f (x) dx = \int_{0}^{1} f (x) dx + \int_{1}^{2} f (x) dx + . . . + \int_{9}^{10} f (x) dx = \int_{0}^{10} f (x) dx = 5$