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  4. The value of ∫ limits04042 (√x d x/√x+√4042-x) is equal to

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Q. The value of $\int\limits_{0}^{4042} \frac{\sqrt{x} \,d x}{\sqrt{x}+\sqrt{4042-x}}$ is equal to

KCETKCET 2021Integrals

Solution:

$\int\limits_{a}^{b} \frac{f(x)}{f(x)+f(a +b-x)} d x=\frac{b-a}{2}$