Question
Evaluate the integral $\int_{0}^{2} x \sqrt{x+2}$ (Put $x + 2 =t^2$) using substitution.
✓
Answer
Given integral is: $\int_{0}^{2} x \sqrt{x+2} d x$
Let $x + 2 = t^2 \Rightarrow$ dx = 2t dt
And $x = t^2 - 2$
when, x = 0, t = $\sqrt{2}$ and when x = 2, t = 2
So, $\int_{0}^{2} x \sqrt{x+2} d x=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) \sqrt{t^{2}} 2 t d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t \cdot t d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t^{2} d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{4}-2 t^{2}\right) d t$
= $2\left[\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]_{\sqrt{2}}^2$
= $2\left[\frac{(2)^{5}}{5}-\frac{2(2)^{3}}{3}-\frac{(\sqrt{2})^{5}}{5}+\frac{2(\sqrt{2})^{3}}{3}\right]$
= $2\left[\frac{32}{5}-\frac{16}{3}-\frac{4 \sqrt{2}}{5}+\frac{4 \sqrt{2}}{3}\right]$
= $2\left[\frac{96-80-12 \sqrt{2}+20 \sqrt{2}}{15}\right]$
= $2\left[\frac{16+8 \sqrt{2}}{15}\right]$
= $\left[\frac{16(2+\sqrt{2})}{15}\right]$
= $\frac{16 \sqrt{2}(\sqrt{2}+1)}{15}$ .Which is the required solution.
Let $x + 2 = t^2 \Rightarrow$ dx = 2t dt
And $x = t^2 - 2$
when, x = 0, t = $\sqrt{2}$ and when x = 2, t = 2
So, $\int_{0}^{2} x \sqrt{x+2} d x=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) \sqrt{t^{2}} 2 t d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t \cdot t d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t^{2} d t$
= $2 \int_{\sqrt{2}}^{2}\left(t^{4}-2 t^{2}\right) d t$
= $2\left[\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]_{\sqrt{2}}^2$
= $2\left[\frac{(2)^{5}}{5}-\frac{2(2)^{3}}{3}-\frac{(\sqrt{2})^{5}}{5}+\frac{2(\sqrt{2})^{3}}{3}\right]$
= $2\left[\frac{32}{5}-\frac{16}{3}-\frac{4 \sqrt{2}}{5}+\frac{4 \sqrt{2}}{3}\right]$
= $2\left[\frac{96-80-12 \sqrt{2}+20 \sqrt{2}}{15}\right]$
= $2\left[\frac{16+8 \sqrt{2}}{15}\right]$
= $\left[\frac{16(2+\sqrt{2})}{15}\right]$
= $\frac{16 \sqrt{2}(\sqrt{2}+1)}{15}$ .Which is the required solution.
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