The value of the integral int_{frac{1}{n}}^{f | vedclass

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(C) Let $I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} dx$ .....$(i)$Using the property $\int_{A}^{B} f(x) dx = \

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$\frac{na - 2}{2n}$

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(C) Let $I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} dx$ .....$(i)$Using the property $\int_{A}^{B} f(x) dx = \int_{A}^{B} f(A + B - x) dx$,where $A = \frac{1}{n}$ and $B = \frac{an - 1}{n}$,we have $A + B = \frac{1 + an - 1}{n} = a$.So,$I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{a - x}}{\sqrt{a - (a - x)} + \sqrt{a - x}} dx = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} dx$ .....$(ii)$Adding $(i)$ and $(ii)$,we get:$2I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} \left( \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} + \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} \right) dx$$2I = \int_{\frac{1}{n}}^{\frac{an - 1}{n}} 1 dx$$2I = [x]_{\frac{1}{n}}^{\frac{an - 1}{n}}$$2I = \frac{an - 1}{n} - \frac{1}{n} = \frac{an - 2}{n}$$I = \frac{an - 2}{2n}$

The value of the integral $\int_{\frac{1}{n}}^{\frac{an - 1}{n}} \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} dx$ is

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