WBJEE · Maths · Definite Integration
The value of the integral \(\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\) is
- A \(\frac{1}{2}\)
- B \(\frac{3}{2}\)
- C \(2\)
- D 1
Answer & Solution
Correct Answer
(B) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\) Applying King's property \(I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x\) Adding \(2 \mathrm{I}=\int_3^6 \mathrm{dx}=3 \quad \Rightarrow \mathrm{I}=\frac{3}{2}\)
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