Question

${\int }_0^{\lambda }\frac{y}{{\left(\sqrt{y+\lambda }\right)}^3}dy=$

• A. $\frac{2}{3}\left(2-\sqrt{2}\right)\lambda \sqrt{\lambda }$
• B. $\frac{2}{3}\left(2+\sqrt{2}\right)\lambda \sqrt{\lambda }$
• C. $\frac{1}{3}\left(2-\sqrt{2}\right)\lambda \sqrt{\lambda }$
• D. $\frac{1}{3}\left(2+\sqrt{2}\right)\lambda \sqrt{\lambda }$

Answer 1

The correct option is A

$\frac{2}{3}\left(2-\sqrt{2}\right)\lambda \sqrt{\lambda }$

Explanation for the Correct Answer:

Finding the value for the given integration:

$\text{I}={\int }_0^{\lambda }\frac{y}{{\left(\sqrt{y+\lambda }\right)}^3}dy$

Using the substitution method, we get

$\begin{array}{rcl}{z}^2& =& y+\lambda \\ & \Rightarrow & \left(2z\right)dz=dy\end{array}$

limits

$y$	$0$	$\lambda$
$z$	$$\begin{gathered} z=\sqrt{y+\lambda } \\ =\sqrt{0+\lambda } \\ =\sqrt{\lambda } \end{gathered}$$	$$\begin{gathered} z=\sqrt{\lambda +\lambda } \\ =\sqrt{2\lambda } \end{gathered}$$

$\begin{array}{rcl}& =& {\int }_{\sqrt{\lambda }}^{\sqrt{2\lambda }}\frac{2z\left(z^2-\lambda \right)dz}{\sqrt{{{\displaystyle z}}^2}}\\ & =& {\int }_{\sqrt{\lambda }}^{\sqrt{2\lambda }}\frac{2z\left(z^2-\lambda \right)dz}{z}\\ & =& 2{\int }_{\sqrt{\lambda }}^{\sqrt{2\lambda }}\left(z^2-\lambda \right)dz\\ & =& 2{\left[\frac{z^3}{3}-z\lambda \right]}_{\sqrt{\lambda }}^{\sqrt{2\lambda }}\\ & =& 2\left[\left(\frac{{\left(\sqrt{2\lambda }\right)}^3}{3}-\sqrt{2\lambda }.\lambda \right)-\left(\frac{{\left(\sqrt{\lambda }\right)}^3}{3}-\sqrt{\lambda }.\lambda \right)\right]\\ & =& 2\left[\left(\frac{\left(\sqrt{2\lambda }\right)2\lambda }{3}-\sqrt{2\lambda }.\lambda \right)-\left(\frac{\lambda \sqrt{\lambda }}{3}-\sqrt{\lambda }.\lambda \right)\right]\\ & =& 2\lambda \sqrt{\lambda }\left[\frac{2\sqrt{2}-3\sqrt{2}-1+3}{3}\right]\\ & =& 2\lambda \sqrt{\lambda }\left[\frac{-\sqrt{2}+2}{3}\right]\\ & =& \frac{2\lambda \sqrt{\lambda }}{3}\left(2-\sqrt{2}\right)\\ & & \end{array}$

Hence, option (A) is the correct answer.