Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals3 Marks
Question
Evaluate the integral $\int_{0}^{2} x \sqrt{x+2} ($Put $x + 2 =t^2)$ using substitution.
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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.
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