int_{1}^{3} frac{sqrt{4-x}}{sqrt{x}+sqrt{4-x} | vedclass

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(A) Let $I = \int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx \quad \dots(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx

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(A) Let $I = \int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx \quad \dots(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = 1+3 = 4$.Substituting $x$ with $(4-x)$:$I = \int_{1}^{3} \frac{\sqrt{4-(4-x)}}{\sqrt{4-x}+\sqrt{4-(4-x)}} dx$$I = \int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} dx \quad \dots(ii)$Adding equations $(i)$ and $(ii)$:$2I = \int_{1}^{3} \left( \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} + \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} \right) dx$$2I = \int_{1}^{3} \frac{\sqrt{4-x}+\sqrt{x}}{\sqrt{x}+\sqrt{4-x}} dx$$2I = \int_{1}^{3} 1 dx$$2I = [x]_{1}^{3} = 3 - 1 = 2$$I = \frac{2}{2} = 1$

$\int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx$ is equal to

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