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Integrals

Multiple Choice Questions

621.

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

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

622.

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

623.

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

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

624.

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

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

625.

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

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

626.

$\int_{2}^{3} \frac{dx}{x^{2} - x is equal to}$

$\log (\frac{2}{3})$
$\log (\frac{4}{3})$
$\log (\frac{8}{3})$
$\log (\frac{1}{4})$

627.

$\int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{4} (x) \cos^{6} (x) dx is equal to$

$\frac{3 π}{128}$
$\frac{3 π}{256}$
$\frac{3 π}{572}$
$\frac{3 π}{64}$

628.

The approximate value of $\int_{2}^{9} x^{2} dx$ by using trapezoidal rule with 4 equal intervals, is

248
242.5
242.8
243

629.
$A minimum value of \int_{0}^{x} {te}^{t^{2}} dt is$

0

1

2

3

$We have, \int_{0}^{x} {te}^{t^{2}} d t Let t^{2} = z \Rightarrow 2 tdt = dz f (x) = \int_{0}^{x^{2}} \frac{e^{z}}{2} d z = \frac{1}{2} {[e^{z}]}_{0}^{x^{2}} On differentiating both sides w . r . t . x, f' (x) = \frac{1}{2} [2 {xe}^{x^{2}}] For maxima or minima, put f' (x) = 0 \Rightarrow x = 0 \Rightarrow f'' (x) = e^{x^{2}} + 2 {xe}^{x^{2}} \Rightarrow f'' (0) = 1 > 0 ∴ f (x) is minirmum at x = 0 ∴ f (0) = \frac{1}{2} [e^{0} - 1] = 0$