int_{3}^{5} frac{sqrt{x}}{sqrt{8-x}+sqrt{x}} | vedclass

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(B) Let $I = \int_{3}^{5} \frac{\sqrt{x}}{\sqrt{8-x}+\sqrt{x}} \, dx$  Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we g

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(B) Let $I = \int_{3}^{5} \frac{\sqrt{x}}{\sqrt{8-x}+\sqrt{x}} \, dx$  Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we get:  $I = \int_{3}^{5} \frac{\sqrt{3+5-x}}{\sqrt{8-(3+5-x)}+\sqrt{3+5-x}} \, dx$  $I = \int_{3}^{5} \frac{\sqrt{8-x}}{\sqrt{x}+\sqrt{8-x}} \, dx$  Adding the two expressions for $I$:  $2I = \int_{3}^{5} \frac{\sqrt{x}+\sqrt{8-x}}{\sqrt{x}+\sqrt{8-x}} \, dx$  $2I = \int_{3}^{5} 1 \, dx$  $2I = [x]_{3}^{5} = 5 - 3 = 2$  $I = 1$

$\int_{3}^{5} \frac{\sqrt{x}}{\sqrt{8-x}+\sqrt{x}} \, dx =$

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