int_{a}^{b} frac{sqrt{x}}{sqrt{x} + sqrt{a + | vedclass

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(B) Let $I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} dx$ . . . .$(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b

Vedclass · Accepted Answer

$\frac{b - a}{2}$

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(B) Let $I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} dx$ . . . .$(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we get:$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a + b - (a + b - x)}} dx$$I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{x}} dx$ . . . .$(ii)$Adding equations $(i)$ and $(ii)$:$2I = \int_{a}^{b} \frac{\sqrt{x} + \sqrt{a + b - x}}{\sqrt{x} + \sqrt{a + b - x}} dx$$2I = \int_{a}^{b} 1 dx = [x]_{a}^{b} = b - a$$I = \frac{b - a}{2}$

$\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a + b - x}} dx = . . . . . .$

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