Q115 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

$\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x=$

  • A
    $\frac{1}{2}$
  • Correct answer
    B
    $\frac{3}{2}$
  • C
    2
  • D
    1

Solution

$I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ ...(i) $\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$ $I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$ ...(ii) Add (i) and (ii), $\begin{aligned} & 2 I=\int_3^6 1 d x=[x]_3^6 \\ & \Rightarrow 2 I=3 \Rightarrow I=\frac{3}{2}\end{aligned}$

Hence, the correct answer is option: 'B'.

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