JEE MAIN - Mathematics (2005 - No. 25)
The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx $$ is
$${1 \over 2}$$
$${3 \over 2}$$
$$2$$
$$1$$
Explanation
$$I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
$$\left[ \, \right.$$ 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 $$(2)$$
$$2I = \int\limits_3^6 {dx} = 3 \Rightarrow I = {3 \over 2}$$
$$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
$$\left[ \, \right.$$ 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 $$(2)$$
$$2I = \int\limits_3^6 {dx} = 3 \Rightarrow I = {3 \over 2}$$
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