Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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

248

$Here, n = 4, h = \frac{9 - 1}{4} = 2 y_{0} = f (1) = {(1)}^{2} = 1 y_{1} = f (3) = {(3)}^{2} = 9 y_{2} = f (5) = {(5)}^{2} = 25 y_{3} = f (7) = {(7)}^{2} = 49 y_{4} = f (9) = {(9)}^{2} = 81 by trapezoidal rule \int_{1}^{9} x^{2} dx = \frac{h}{2} [y_{o} + 2 (y_{1} + y_{2} + y_{3}) + y_{4}] = \frac{1}{2} 2 [1 + 2 (9 + 25 + 49) + 81] = 248$