# Evaluate the integral using substitution $\int\limits_0^2 x\sqrt{x+2}\;(Put\;x+2=t^2)$

$\begin{array}{1 1} \large\frac{16 \sqrt 2}{15}+\large\frac{32}{15} \\\large\frac{16 \sqrt 2}{15}-\large\frac{32}{15} \\ \large\frac{16 \sqrt 2}{15}+\large\frac{30}{15} \\\large\frac{16 \sqrt 3}{15}+\large\frac{32}{15} \end{array}$

Toolbox:
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} Step 1: \int_0^2x\sqrt{x+2}dx Add and subtract 2 we get, \Rightarrow \int_0^2\{(x+2)-2\}\sqrt{x+2}dx \qquad\qquad=\int_0^2\{(x+2)^{\Large\frac{3}{2}}-2(x+2)^{\Large\frac{1}{2}}\}dx Step 2: Differentiating with respect to x we get, \qquad\qquad=\bigg[\large\frac{(x+2)^{(\large\frac{3}{2}+1)}}{\large\frac{3}{2}+1}-\large\frac{2(x+2)^{\large\frac{1}{2}+1}}{\large\frac{1}{2}+1}\bigg]_0^2 By substituting the upper limit and lower limit \qquad\qquad=\large\frac{2}{5}$$\bigg[(x+2)^{\Large\frac{5}{2}}\bigg]_0^2-\large\frac{4}{3}$$\bigg[(x+2)^{\Large\frac{3}{2}}\bigg]_0^2 \qquad\qquad=\large\frac{2}{5}$$\bigg[4^{\Large\frac{5}{2}}-2^{\Large\frac{5}{2}}\bigg]_0^2-\large\frac{4}{3}$$\bigg[4^{\Large\frac{3}{2}}-2^{\Large\frac{3}{2}}\bigg] \qquad\qquad=\large\frac{2}{5}$$\bigg[2^5-2^{\Large\frac{5}{2}}\bigg]_0^2-\large\frac{4}{3}$$\bigg[2^3-2^{\Large\frac{3}{2}}\bigg] \qquad\qquad=\large\frac{2}{5}$$\bigg[32-4\sqrt 2\bigg]-\large\frac{4}{3}$$\bigg[8-2\sqrt 2\bigg] \qquad\qquad=\large\frac{8}{5}$$\bigg[8-\sqrt 2\bigg]-\large\frac{8}{3}$$\bigg[4-\sqrt 2\bigg]$
$\qquad\qquad=(\large\frac{64}{5}-\frac{32}{3})+(\frac{8\sqrt 2}{3}-\frac{8\sqrt 2}{5})$
$\qquad\qquad=\large\frac{32}{15}+\frac{16\sqrt 2}{15}$