Question

Evaluate: ${\int }_{\sqrt{8}}^{\sqrt{15}}x\sqrt{1+x^2}.dx$

• A. $\frac{15}{8}$
• B. $\frac{37}{3}$
• C. $\frac{37}{6}$
• D. $\frac{37}{9}$

Answer 1

Answer: (B)

${\int }_{\sqrt{8}}^{\sqrt{15}}x\sqrt{1+x^2}dx$

Let $z=1+x^2$

$⟹dz=2xdx$

$⟹xdx=\frac{dz}{2}$

For, $x=\sqrt{8},z=9$ and for $x=\sqrt{15},z=16$

Hence, integration becomes-

${\int }_9^{16}\sqrt{z}\frac{dz}{2}$

$=\frac{1}{2}{\left[\frac{z^{(\frac{1}{2}+1)}}{\frac{1}{2}+1}\right]}_9^{16}$

$=\frac{1}{2}\times \frac{2}{3}{\left[z^{\left(\frac{3}{2}\right)}\right]}_9^{16}$

$=\frac{1}{3}[{16}^{\left(\frac{3}{2}\right)}-9^{\left(\frac{3}{2}\right)}]$

$=\frac{1}{3}[64-27]$

$=\frac{37}{3}$