Explanation:
(B) : Given,
\(I=\int_{a}^{b} \frac{\sqrt{x} d x}{\sqrt{x}+\sqrt{a+b-x}}\)
\(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\)
\(\therefore I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{x}} d x\)
Adding equation (i) and (ii),
\(2 I=\int_{a}^{b} \frac{\sqrt{x} d x}{\sqrt{x}+\sqrt{a+b-x}}+\int_{a}^{b} \frac{\sqrt{a+b-x} d x}{\sqrt{a+b-x}+\sqrt{x}}\)
\(=\int_{a}^{b} \frac{\sqrt{x}+\sqrt{a+b-x} d x}{\sqrt{a+b+x}-\sqrt{x}}=\int_{a}^{b} 1 . d x=[x]_{a}^{b}\)
\(2 \mathrm{I}=\mathrm{b}-\mathrm{a}\)
\(\mathrm{I}=\frac{\mathrm{b}-\mathrm{a}}{2}\)