Question

${\int }_0^2{x}^{5/2}\sqrt{2-x}dx=$

• A. $\frac{\pi }{8}$
• B. $\frac{3\pi }{8}$
• C. $\frac{5\pi }{8}$
• D. $\frac{7\pi }{8}$

Answer 1

The correct option is C $\frac{5\pi }{8}$

${\int }_0^{2}x5/2\sqrt{2-x}dx$

$x=2si{n}^2\theta .$

$dx=4sin\theta cos\theta d\theta$

Thus ${\int }_0^{7/2}\left({\sqrt{2}}^5\right)si{n}^5\theta \left(\sqrt{2}\right)cos\theta .4sin\theta cos\theta d\theta$

$=8\times 4{\int }_0^{7/2}si{n}^6\theta co{s}^{2}d\theta$

$=8\times 4{\int }_0^{7/2}si{n}^6\theta (1-si{n}^2\theta )d\theta$

$=8\times 4\left[{\int }_0^{7/2}si{n}^{6}d\theta -{\int }_0^{7/2}si{n}^8\theta d\theta \right]$

$=8\times 4(\frac{5}{6}\times \frac{3}{4}\times \frac{1}{2}\times 7/2-\frac{7}{8}\times \frac{5}{6}\times \frac{3}{4}\times \frac{1}{2}\times 7/2)$

$=5/6\times 3/4\times 1/2\times 7/2\Rightarrow \frac{57}{32}\times =\frac{57}{8}$