The value of int_{2}^{8} frac{sqrt{10-x}}{sq | vedclass

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(D) Let $ I = \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \quad (1) $ Using the property $ \int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x)

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$ 03 $

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(D) Let $ I = \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \quad (1) $ Using the property $ \int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x $,where $ a=2 $ and $ b=8 $,we have $ a+b-x = 2+8-x = 10-x $. Substituting this into the integral,we get: $ I = \int_{2}^{8} \frac{\sqrt{10-(10-x)}}{\sqrt{10-x}+\sqrt{10-(10-x)}} d x $ $ I = \int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x \quad (2) $ Adding equations $(1)$ and $(2)$: $ 2I = \int_{2}^{8} \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}} d x $ $ 2I = \int_{2}^{8} 1 d x $ $ 2I = [x]_{2}^{8} = 8 - 2 = 6 $ $ I = \frac{6}{2} = 3 $

The value of $ \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x $ is

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