int_{3}^{6} frac{sqrt{x}}{sqrt{9-x} + sqrt{x} | vedclass

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(B) Let $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \quad \dots(i)$Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Let $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \quad \dots(i)$Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we get:$I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{9-(9-x)} + \sqrt{9-x}} \, dx$$I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx \quad \dots(ii)$Adding equations $(i)$ and $(ii)$:$2I = \int_{3}^{6} \frac{\sqrt{x} + \sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx$$2I = \int_{3}^{6} 1 \, dx$$2I = [x]_{3}^{6} = 6 - 3 = 3$$I = \frac{3}{2}$

$\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx = $

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