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Integrals

Multiple Choice Questions

51.

The value of $\int \frac{(x - 2)}{{\{{(x - 2)}^{2} {(x + 3)}^{7}\}}^{1 / 3}} dx$

$\frac{3}{20} {(\frac{x - 2}{x + 3})}^{4 / 3} + C$
$\frac{3}{20} {(\frac{x - 2}{x + 3})}^{3 / 4} + C$
$\frac{5}{12} {(\frac{x - 2}{x + 3})}^{4 / 3} + C$
$\frac{3}{20} {(\frac{x - 2}{x + 3})}^{5 / 3} + C$

52.

If f(x) = $\{2 x^{2} + 1, x \leq 1 4 x^{3} - 1, x > 1\}$ , then $\int_{0}^{2} f (x) d x$ is

47/3
50/3
1/3
47/2

53.

If $I = \int_{0}^{2} e^{x^{4}} (x - α) d x = 0$ , then $α$ lies in the interval

(0, 2)
(- 1, 0)
(2, 3)
(- 2, - 1)

54.

The value of $\lim_{x \to 0} \frac{\int_{0}^{x^{2}} \cos (t^{2}) d x}{xsin (x)}$

1
- 1
2
$\log_{e} (2)$

55.

Let f(x) = $\max \{x + |x|, x - [x]\}$ , where [x] denotes the greatest integer $\leq x$ . Then, the values of $\int_{- 3}^{3} f (x) d x$ is

0
51/2
21/2
1

56.

Suppose $M = \int_{0}^{π / 2} \frac{\cos (x)}{x + 2} d x, N = \int_{0}^{π / 4} \frac{\sin (x) \cos (x)}{{(x + 1)}^{2}} d x$ . Then, the values of (M - N) equals

$\frac{3}{π + 2}$
$\frac{2}{π - 4}$
$\frac{4}{π - 2}$
$\frac{2}{π + 4}$

57.

The value of the integral

$\int_{- 1}^{1} \{\frac{x^{2013}}{e^{|x|} (x^{2} + \cos (x))} + \frac{1}{e^{|x|}}\} d x$ is equal to

0
$1 - e^{- 1}$
$2 e^{- 1}$
$2 (1 - e^{- 1})$

58.
The value of I = $\int_{0}^{π / 4} (\tan^{n + 1} (x)) d x + \frac{1}{2} \int_{0}^{π / 2} (\tan^{n + 1} (\frac{x}{2})) d x$ is

$\frac{1}{n}$

$\frac{n + 2}{2 n + 1}$

$\frac{2 n - 1}{n}$

$\frac{2 n - 3}{3 n - 2}$

$\frac{1}{n}$

Given,

$I = \int_{0}^{π / 4} (\tan^{n + 1} (x)) d x + \frac{1}{2} \int_{0}^{π / 2} (\tan^{n - 1} (\frac{x}{2})) d x In second integral, put t = \frac{x}{2} \Rightarrow dx = 2 dt$

$\Rightarrow Also, when x = 0 then t = 0, When x = \frac{π}{2 then t = \frac{π}{4}} Then, I = \int_{0}^{\frac{π}{4}} \tan^{n + 1} (x) d x + \int_{0}^{π / 4} \tan^{n - 1} (t) d t I = \int_{0}^{\frac{π}{4}} \tan^{n + 1} (x) d x + \int_{0}^{\frac{π}{4}} \tan^{n - 1} (x) d x [∵ \int_{a}^{b} f (x) d x = \int_{0}^{\frac{π}{4}} f (y) dy] \Rightarrow I = \int_{0}^{\frac{π}{4}} (\tan^{n + 1} (x) + \tan^{n - 1} (x)) d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \tan^{n - 1} (x) . (\tan^{2} (x) + 1) d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \tan^{n - 1} \times (\sec^{2} (x)) d x Put t = \tan (x) \Rightarrow dt = \sec^{2} (x) dx Also, when x = 0, then t = 0 when x = \frac{π}{4,} then t = 1$

$I = \int_{0}^{1} t^{n - 1} d t = {[\frac{t^{n}}{n}]}_{0}^{1} = \frac{1}{n}$

59.

The value of the integral

$\int_{1}^{2} e^{x} (\log_{e} (x) + \frac{x + 1}{x}) d x$

$e^{2} (1 + \log_{e} (2))$
$e^{2} - e$
$e^{2} (1 + \log_{e} (2)) - e$
$e^{2} - e (1 + \log_{e} (2))$

60.

If [a] denote the greatest integer which is less than or equal to a. Then, the value of the integral $\int_{- \frac{π}{2}}^{\frac{π}{2}} [\sin (x) \cos (x)] d x$ is

$\frac{π}{2}$
$π$
$- π$
$- \frac{π}{2}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

58. The value of I = ∫0π/4tann + 1xdx + 12∫0π/2tann + 1x2dx is 1n n + 22n + 1 2n - 1n 2n - 33n - 2

58.
The value of I = $\int_{0}^{π / 4} (\tan^{n + 1} (x)) d x + \frac{1}{2} \int_{0}^{π / 2} (\tan^{n + 1} (\frac{x}{2})) d x$ is

$\frac{1}{n}$

$\frac{n + 2}{2 n + 1}$

$\frac{2 n - 1}{n}$

$\frac{2 n - 3}{3 n - 2}$