Question

The value of ${\int }_2^3\frac{\sqrt{x}}{\sqrt{5-x}+\sqrt{x}}.dx$ is equal to

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{2}$

Here from inspecting the question, we need to sense the application of the following property. You might get this intuition by looking at the peculiarity of the limits and functions given in the integral.
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x).dx$
Applying it here,
$I={\int }_2^3\frac{\sqrt{x}}{\sqrt{5-x}+\sqrt{x}}.dx$
$={\int }_2^3\frac{\sqrt{5-x}}{\sqrt{5-(5-x)}x+\sqrt{5-x}}.dx={\int }_2^3\frac{\sqrt{5-x}}{\sqrt{x}+\sqrt{5-x}}.dx$
If we add this expression with the initial form of the expression then we get
$2l={\int }_2^3\frac{\sqrt{x}}{\sqrt{5-x}+\sqrt{x}}.dx+{\int }_2^3\frac{\sqrt{5-x}}{\sqrt{5-x}+\sqrt{x}}.dx$
${\int }_2^3\frac{\sqrt{x}+\sqrt{5-x}}{\sqrt{5-x}+\sqrt{x}}.dx$
${\int }_2^{3}1dx$
$=x{|}_2^3$
=1
$l=\frac{1}{2}$