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Integrals

Multiple Choice Questions

501.

The value of $\int_{0}^{4} |x - 1| dx$ is

$\frac{5}{2}$
5
4
1

502.

If I_n = $\int_{0}^{\frac{π}{4}} \tan^{n} (x) dx$ , where where n is apositive integer, then $I_{10} + I_{8}$ is

$\frac{1}{9}$
$\frac{1}{8}$
$\frac{1}{7}$
9

503.

If I_n = $\int e^{x} [\frac{\sin (x) + \cos (x)}{1 - \sin^{2} (x)}] dx$ is

$(e^{x} . \csc (x)) + C$
$(e^{x} . \cot (x)) + C$
$(e^{x} . \sec (x)) + C$
$(e^{x} . \tan (x)) + C$

504.

When x > 0, then $\int \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) dx$ is

$2 [{xtan}^{- 1} (x) - \log (1 + x^{2})] + C$
$2 [{xtan}^{- 1} (x) + \log (1 + x^{2})] + C$
$2 {xtan}^{- 1} (x) + \log (1 + x^{2}) + C$
$2 {xtan}^{- 1} (x) - \log (1 + x^{2}) + C$

505.

If the area between y = mx² and x = my² (m > 0) is 1/4 sq units, then the value of m is

$\pm 3 \sqrt{2}$
$\pm \frac{2}{\sqrt{3}}$
$\sqrt{2}$
$\sqrt{3}$

506.
$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin^{3} (x)}{\sin^{3} (x) + \cos^{3} (x)} dx$ is equal to

$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{12}$

$\frac{π}{6}$

$\frac{π}{6}$

$Let I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin^{3} (x)}{\sin^{3} (x) + \cos^{3} (x)} dx . . . (i) = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin^{3} (\frac{π}{3} + \frac{π}{6} - x)}{\sin^{3} (\frac{π}{3} + \frac{π}{6} - x) + \cos^{3} (\frac{π}{3} + \frac{π}{6} - x)} dx = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin^{3} (\frac{π}{2} - x)}{\sin^{3} (\frac{π}{2} - x) + \cos^{3} (\frac{π}{2} - x)} dx = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\cos^{3} (x)}{\sin^{3} (x) + \cos^{3} (x)} dx . . . (ii) On adding Eqs . (i) and (ii), we get = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin^{3} (x) + \cos^{3} (x)}{\sin^{3} (x) + \cos^{3} (x)} dx = {[x]}_{\frac{π}{6}}^{\frac{π}{3}} = \frac{π}{3} - \frac{π}{6} = \frac{π}{6}$