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Integrals

Multiple Choice Questions

141.

The value of $\int_{1}^{4} |x - 3| dx$ is equal to :

1
$\frac{5}{2}$
$\frac{1}{2}$
$\frac{3}{2}$

142.

The value of $\int_{0}^{\infty} \frac{dx}{(a^{2} + x^{2})}$ is equal to

$\frac{π}{2}$
$\frac{π}{2 a}$
$\frac{π}{a}$
$\frac{1}{2 a}$

143.

The value of the integral $\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} dx$ is

$e^{x} (\frac{1 - x}{1 + x^{2}}) + c$
$e^{x} (\frac{1 + x}{1 + x^{2}})$
$\frac{e^{x}}{1 + x^{2}} + c$
$e^{x} (1 - x) + c$

144.

If g(x) = $\int_{0}^{x} \cos^{4} (t) d t$ , then $g (x + π)$ is equsl to

g(x) + g( $π$ )
g(x) - g( $π$ )
g(x) . g( $π$ )
$\frac{g (x)}{g (π)}$

145.

$\int (\sin^{6} (x) + \cos^{6} (x) + 3 \sin^{2} (x) \cos^{2} (x)) dx$ is equal to

x + C
$\frac{3}{2} \sin (2 x) + C$
$- \frac{3}{2} \cos (2 x) + C$
$\frac{1}{3} \sin (3 x) - \cos (3 x) + C$

146.

$\int \frac{4^{x + 1} - 7^{x - 1}}{28^{x}} dx$ is equal to

$\frac{1}{7 \log_{e} (4)} 4^{- x} - \frac{4}{\log_{e} (7)} 7^{- x} + C$
$\frac{1}{7 \log_{e} (4)} 4^{- x} + \frac{4}{\log_{e} (7)} 7^{- x} + C$
$\frac{4^{- x}}{\log_{e} (7)} - \frac{7^{- x}}{\log_{e} (4)} + C$
$\frac{4^{- x}}{\log_{e} (4)} - \frac{7^{- x}}{\log_{e} (7)} + C$

147.

The value of $\int e^{\tan^{- 1} (x)} \frac{(1 + x + x^{2})}{(1 + x^{2})} dx$

$\tan^{- 1} (x) + c$
$e^{\tan^{- 1} (x)} + 2 x + C$
$e^{\tan^{- 1} (x)} + C$
${xe}^{\tan^{- 1} (x)} + C$

148.

The value of $\int \frac{e^{5 \log_{e} (x)} - e^{4 \log_{e} (x)}}{e^{3 \log_{e} (x)} - e^{2 \log_{e} (x)}} dx$ is

x² + C
$\frac{x^{2}}{2} + C$
$\frac{x^{3}}{3} + C$
$\frac{x}{2} + C$

149.
If $f (x) = \frac{\sin^{- 1} (x)}{\sqrt{1 - x^{2}}} and g (x) = e^{\sin^{- 1} (x)}$ , then $\int f (x) g (x) dx$ is equal to

$e^{\sin^{- 1} (x)} (\sin^{- 1} (x) - 1) + C$

$e^{\sin^{- 1} (x)} + C$

$e^{{(\sin^{- 1} (x))}^{2}} + C$

$e^{2 \sin^{- 1} (x)} + C$

$e^{\sin^{- 1} (x)} (\sin^{- 1} (x) - 1) + C$

Given that

$f (x) = \frac{\sin^{- 1} (x)}{\sqrt{1 - x^{2}}} and g (x) = e^{\sin^{- 1} (x)} ∴ \int f (x) g (x) dx = \int \frac{\sin^{- 1} (x)}{\sqrt{1 - x^{2}}} e^{\sin^{- 1} (x)} dx Let \sin^{- 1} (x) = t and \frac{1}{\sqrt{1 - x^{2}}} dx = dt \Rightarrow \int f (x) g (x) dx = \int {te}^{t} dt = {te}^{t} - e^{t} + C = e^{\sin^{- 1} (x)} (\sin^{- 1} (x) - 1) + C$