Definite Integration Question 19
Question 19 - 31 January - Shift 2
Let $\alpha>0$. If $\int_0^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} dx=\frac{16+20 \sqrt{2}}{15}$, then $\alpha$ is equal to :
(1) 2
(2) 4
(3) $\sqrt{2}$
(4) $2 \sqrt{2}$
Show Answer
Answer: (1)
Solution:
Formula: Standard formulas for Indefinite Integration
$I=\int_0^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} dx$
on rationalising
$I=\int_0^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} \times \frac{\sqrt{x+\alpha}+\sqrt{x}}{\sqrt{x+\alpha}+\sqrt{x}} dx$
$ \begin{aligned} & I=\int_0^{\alpha} \frac{x}{\alpha}(\sqrt{x+\alpha}+\sqrt{x}) dx \\ & I=\int_0^{\alpha} \frac{(x+\alpha -\alpha)}{\alpha}(\sqrt{x+\alpha}+\sqrt{x})dx \\ & I=\int_0^{\alpha} \frac{1}{\alpha}[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}]dx \\ & I=\frac{1}{\alpha}[\frac{2}{5}(x+\alpha)^{5 / 2}-\alpha \frac{2}{3}(x+\alpha)^{3 / 2}+\frac{2}{5} x^{5 / 2}]_0 ^{\alpha} \\ & I =\frac{1}{\alpha}(\frac{5}{2}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}) \\ & I =\frac{1}{\alpha}(\frac{2^{7 / 2} \alpha^{5 / 2}}{5} \frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}) \\ & I =\alpha^{3 / 2}(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}) \\ & I =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10) \\ &I=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10) \end{aligned} $
Now,
$ \begin{aligned} & \frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\frac{16+20 \sqrt{2}}{15} \\ & \Rightarrow \alpha=2 \end{aligned} $
