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

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

Answers (1)

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}

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