The value of the integral int_3^6 frac{sqrt{x | vedclass

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(B) Let $I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$  $(1)$ Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$, we get: $I = \i

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$\frac{3}{2}$

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(B) Let $I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$  $(1)$ Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$, we get: $I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{9-(9-x)}+\sqrt{9-x}} d x$ $I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$  $(2)$ Adding equations $(1)$ and $(2)$: $2I = \int_3^6 \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} d x$ $2I = \int_3^6 1 d x$ $2I = [x]_3^6 = 6-3 = 3$ $I = \frac{3}{2}$

The value of the integral $\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ is

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