Let 0 . If 0 x x + - x d x = 16 + 20 2 15 , then is equal to [2023]

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Solution

Q.
Let $α > 0$ . If $\int_{0}^{α} \frac{x}{\sqrt{x + α} - \sqrt{x}} d x = \frac{16 + 20 \sqrt{2}}{15},$ then $α$ is equal to [2023]

1 $2$

2 $2 \sqrt{2}$

3 $4$

4 $\sqrt{2}$

Ans.
(1)

After rationalising, $I = \int_{0}^{α} \frac{x}{α} (\sqrt{x + α} + \sqrt{x}) d x$

$= \int_{0}^{α} \frac{1}{α} [{(x + α)}^{3 / 2} - α {(x + α)}^{1 / 2} + x^{3 / 2}] d x$

$= \frac{1}{α} {[\frac{2}{5} {(x + α)}^{5 / 2} - α \frac{2}{3} {(x + α)}^{3 / 2} + \frac{2}{5} x^{5 / 2}]}_{0}^{α}$

$= \frac{1}{α} (\frac{2}{5} {(2 α)}^{5 / 2} - \frac{2 α}{3} {(2 α)}^{3 / 2} + \frac{2}{5} α^{5 / 2} - \frac{2}{5} α^{5 / 2} + \frac{2}{3} α^{5 / 2})$

$= \frac{1}{α} (\frac{2^{7 / 2} α^{5 / 2}}{5} - \frac{2^{5 / 2} α^{5 / 2}}{3} + \frac{2}{3} α^{5 / 2})$ $= α^{3 / 2} (\frac{2^{7 / 2}}{5} - \frac{2^{5 / 2}}{3} + \frac{2}{3})$

$= \frac{α^{3 / 2}}{15} (24 \sqrt{2} - 20 \sqrt{2} + 10) = \frac{α^{3 / 2}}{15} (4 \sqrt{2} + 10)$

Now, $\frac{α^{3 / 2}}{15} (4 \sqrt{2} + 10) = \frac{16 + 20 \sqrt{2}}{15} = \frac{2 \sqrt{2} (4 \sqrt{2} + 10)}{15}$

$\Rightarrow α^{3 / 2} = 2 \sqrt{2} = 2^{3 / 2} \Rightarrow α = 2$

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