The value of ∫- π2π2x3 + xcosx

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Integrals

Multiple Choice Questions

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391.
The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{3} + xcos (x) + \tan^{5} (x) + 1) d x$ is equal to

0

2

$π$

None of these

C.

$π$

Let

$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{3} + xcos (x) + \tan^{5} (x) + 1) d x$

Since, x³, x cos(x) and tan⁵(x) are odd functions,therefore

$\int_{- \frac{π}{2}}^{\frac{π}{2}} x^{3} d x = 0 \int_{- \frac{π}{2}}^{\frac{π}{2}} xcos (x) d x = 0 and \int_{- \frac{π}{2}}^{\frac{π}{2}} \tan^{5} (x) d x = 0 ∴ I = \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d x = {[x]}_{- \frac{π}{2}}^{\frac{π}{2}} = π$

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392.

The value of the integral $\int_{- \frac{π}{4}}^{\frac{π}{4}} \sin (- 4 x) d x$ is

$- \frac{8}{3}$
$\frac{3}{2}$
$\frac{8}{3}$
None of these

393.

$\int {(\frac{x + 2}{x + 4})}^{2} e^{x} dx$ is equal to

$e^{x} (\frac{x}{x + 4}) + C$
$e^{x} (\frac{x + 2}{x + 4}) + C$
$e^{x} (\frac{x - 2}{x + 4}) + C$
$e^{x} (\frac{2 {xe}^{x}}{x + 4}) + C$

394.

The value of $\int_{3}^{5} \frac{x^{2}}{x^{2} - 4} d x$

$2 - \log_{e} (\frac{15}{7})$
$2 + \log_{e} (\frac{15}{7})$
2 + 4 $\log_{e} (3) - 4 \log_{e} (7) + 4 \log_{e} (5)$
$2 - \tan^{- 1} (\frac{5}{7})$

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395.

$\int_{0}^{\infty} \frac{d x}{{(x + \sqrt{x^{2} + 1})}^{3}}$ is equal to

$\frac{3}{8}$
$\frac{1}{8}$
$\frac{- 3}{8}$
None of these

396.

If for every integer n, $\int_{n}^{n + 1} f (x) d x = n^{2}$ , then the value of $\int_{- 2}^{4} f (x) d x$ is

16
14
19
None of these

397.

The value of integral $\int_{0}^{π} xf (\sin (x)) d x$ is

0
$π \int_{0}^{\frac{π}{2}} f (\sin (x)) d x$
$\frac{π}{4} \int_{0}^{π} f (\sin (x)) d x$
None of these

398.

$\lim_{x \to \infty} \frac{\int_{0}^{2 x} {xe}^{x} dx}{e^{4 x^{2}}}$ equals

0
$\infty$
2
$\frac{1}{2}$

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399.

The value of $\int_{- 2}^{3} |1 - x^{2}| d x$ , is

$\frac{1}{3}$
14/3
7/3
28/3

400.

$\int \frac{1}{\tan (x) + \cot (x) + \sec (x) + \csc (x)} dx$ is equal to

$\frac{1}{2} (\sin (x) + \cos (x) + x)$ + C
$\frac{1}{2} (\sin (x) - \cos (x) - x)$ + C
$\frac{1}{2} (\cos (x) - x + \sin (x))$ + C
None of the above

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