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प्रश्न
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उत्तर
\[Let\ I = \int_0^2 x\sqrt{x + 2}\ d x . \]
\[Let\ x + 2 = t^2 . Then, dx = 2t\ dt\]
\[When\ x = 0, t = \sqrt{2}\ and\ x\ = 2, t = 2\]
\[ \therefore I = \int_\sqrt{2}^2 \left( t^2 - 2 \right) t\ 2t\ dt\]
\[ \Rightarrow I = 2 \int_\sqrt{2}^2 \left( t^4 - 2 t^2 \right) dt\]
\[ \Rightarrow I = 2 \left[ \frac{t^5}{5} - \frac{2}{3} t^3 \right]_\sqrt{2}^2 \]
\[ \Rightarrow I = 2\left[ \left( \frac{32}{3} - \frac{16}{3} \right) - \left( \frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} \right) \right]\]
\[ \Rightarrow I = 2\left( \frac{16}{15} + \frac{8\sqrt{2}}{15} \right)\]
\[ \Rightarrow I = \frac{16}{15}\left( 2 + \sqrt{2} \right)\]
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