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Integrals

Multiple Choice Questions

901.

Evaluate $\int_{- 2}^{1} f (x) dx$ , where f(x) = $\{1 - 2 x, x \leq 0 1 + 2 x, x \geq 0\}$

902.

$\int \frac{dx}{\sqrt{x} (x + 9)}$ is equal to

$\frac{2}{3} \tan^{- 1} (\sqrt{x}) + C$
$\frac{2}{3} \tan^{- 1} (\frac{\sqrt{x}}{3}) + C$
$\tan^{- 1} (\sqrt{x}) + C$
$\tan^{- 1} (\frac{\sqrt{x}}{3}) + C$

903.

$\int {(x + 1)}^{2} e^{x} dx$ is equal to

xe^x + C
x²x^x + C
(x + 1)e^x + C
(x² + 1)e^x + C

904.

$\int \frac{dx}{a^{2} \sin^{2} (x) + b^{2} \cos^{2} (x)}$ is equal to

$\frac{1}{ab} \tan^{- 1} (\frac{atan (x)}{b}) + C$
$\tan^{- 1} (\frac{atan (x)}{b}) + C$
$\frac{1}{ab} \tan^{- 1} (\frac{btan (x)}{a}) + C$
$\tan^{- 1} (\frac{btan (x)}{a}) + C$

905.

$\int_{0}^{\frac{π}{2}} \sin^{8} (x) \cos^{2} (x) dx$ is equal to

$\frac{π}{512}$
$\frac{3 π}{512}$
$\frac{5 π}{512}$
$\frac{7 π}{512}$

906.

$\int_{- 1}^{1} ({ax}^{3} + bx) dx = 0$ for

any value of a and b
a > 0, b > 0 only
a > 0, b > 0 only
a < 0, b > 0 only

907.

Using the Trapezoidal rule, the approximate value of $\int_{1}^{4} ydx$

x 1 2 3 4

y 0.7111 0.7222 0.7333 0.7444

0.1833
1.1833
2.1833
3.1833

908.
$\int \frac{dx}{1 - \cos (x) - \sin (x)} is equal to$

$\log (|\begin{matrix} 1 + \cot (\frac{x}{2}) \end{matrix}|) + c$

$\log (|\begin{matrix} 1 - \tan (\frac{x}{2}) \end{matrix}|) + c$

$\log (|\begin{matrix} 1 - \cot (\frac{x}{2}) \end{matrix}|) + c$

$\log (|\begin{matrix} 1 + \tan (\frac{x}{2}) \end{matrix}|) + c$

$\log (|\begin{matrix} 1 - \cot (\frac{x}{2}) \end{matrix}|) + c$

$We have, I = \int \frac{dx}{1 - \cos (x) - \sin (x)} = \int \frac{dx}{1 - (\frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}) - \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} = \int \frac{\sec^{2} (\frac{x}{2}) dx}{1 + \tan^{2} (\frac{x}{2}) + \tan^{2} (\frac{x}{2}) - 2 \tan (\frac{x}{2})} = \frac{1}{2} \int \frac{\sec^{2} (\frac{x}{2}) dx}{\tan (\frac{x}{2}) (\tan (\frac{x}{2}) - 1)} Let \tan (\frac{x}{2}) = z \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) dx = dz \Rightarrow \sec^{2} (\frac{x}{2}) dx = 2 dz ∴ I = \frac{1}{2} \int \frac{2 dz}{z (z - 1)} = \int (\frac{1}{z - 1} - \frac{1}{z}) = \log (z - 1) - \log (z) + c = \log (\frac{z - 1}{z}) + c = \log ([\frac{\tan \frac{x}{2} - 1}{\tan \frac{x}{2}}]) + C = \log (|\begin{matrix} 1 - \cot (\frac{x}{2}) \end{matrix}|) + c$

909.

$\int \frac{dx}{7 + 5 \cos (x)} is equal to$

$\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}} \tan \frac{x}{2}) + c$
$\frac{1}{\sqrt{6}} \tan^{- 1} (\frac{1}{\sqrt{6}} \tan \frac{x}{2}) + c$
$\frac{1}{7} \tan^{- 1} (\tan \frac{x}{2}) + c$
$\frac{1}{4} \tan^{- 1} (\tan \frac{x}{2}) + c$

910.

$\int \frac{3^{x} dx}{\sqrt{9^{x} - 1}} is equal to$

$\frac{1}{\log (3)} \log (|\begin{matrix} 3^{x} + \sqrt{9^{x} - 1} \end{matrix}|) + c$
$\frac{1}{\log (3)} \log (|\begin{matrix} 3^{x} - \sqrt{9^{x} - 1} \end{matrix}|) + c$
$\frac{1}{\log (9)} \log (|\begin{matrix} 3^{x} + \sqrt{9^{x} - 1} \end{matrix}|) + c$
$\frac{1}{\log (3)} \log (|\begin{matrix} 9^{x} + \sqrt{9^{x} - 1} \end{matrix}|) + c$

Previous Year Papers

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Multiple Choice Questions

908. ∫dx1 - cosx - sinx is equal to log1 + cotx2 + c log1 - tanx2 + c log1 - cotx2 + c log1 + tanx2 + c