#### Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 16.

Answer: $\frac{16\sqrt{2}(\sqrt{2}+1)}{15}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: $\int_{0}^{2}x\sqrt{x+2}dx$

Solution: $I=\int_{0}^{2}x\sqrt{x+2}dx$

Put $x+2 =t^2$

$dx=2t\; \; dt$

When x=0  then t=2  and

when x=2  then t=2

\begin{aligned} &I=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) \sqrt{t^{2}} 2 t d t \\ &=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t 2 t d t \end{aligned}

\begin{aligned} &=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) 2 t^{2} d t \\ &=2 \int_{\sqrt{2}}^{2}\left(t^{4}-2 t^{2}\right) d t \\ &=2 \int_{\sqrt{2}}^{2} t^{4} d t-2 \times 2 \int_{\sqrt{2}}^{2} t^{2} d t \\ &=2\left[\frac{t^{4+1}}{4+1}\right]_{\sqrt{2}}^{2}-4\left[\frac{t^{2+1}}{2+1}\right]_{\sqrt{2}}^{2} \end{aligned}

\begin{aligned} &=\frac{2}{5}\left[t^{5}\right]_{\sqrt{2}}^{2}-\frac{4}{3}\left[t^{3}\right]_{\sqrt{2}}^{2} \\ &=\frac{2}{5}\left[2^{5}-(\sqrt{2})^{5}\right]-\frac{4}{3}\left[2^{3}-(\sqrt{2})^{3}\right] \\ &=\frac{2}{5}[32-2 \times 2 \times \sqrt{2}]-\frac{4}{3}[8-2 \sqrt{2}] \\ &=2\left[\frac{32}{5}-\frac{4 \sqrt{2}}{5}\right]-\frac{4 \times 8}{3}+\frac{8 \sqrt{2}}{3} \end{aligned}

\begin{aligned} &=2\left[\left(\frac{32}{5}-\frac{4 \sqrt{2}}{5}\right)-\frac{16}{3}+\frac{4 \sqrt{2}}{3}\right] \\ &=2\left[\frac{3(32-4 \sqrt{2})-5(16-4 \sqrt{2})}{15}\right] \\ &=2\left[\frac{96-12 \sqrt{2}-80+20 \sqrt{2}}{15}\right] \\ &=2\left[\frac{16+8 \sqrt{2}}{15}\right] \\ &=2 \times \frac{8}{15}[2+\sqrt{2}] \\ &=\frac{16 \sqrt{2}}{15}(\sqrt{2}+1) \end{aligned}