The value of integral, _3^6 x√9-x+ x,dx is:

Question

The value of integral, $\displaystyle\int_3^6 \dfrac{\sqrt x}{\sqrt{9-x}+\sqrt x}\,dx$ is:

Accepted Answer

$$I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x}  + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x}  + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$\left[ \, \right.$$ using $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } $$ $$\left. \, \right]$$

Adding equation $(1)$ and $(2)$

$$2I = \int\limits_3^6 {dx}  = 3 \Rightarrow I = {3 \over 2}$$

Solutions · 1