The value of int_{0}^{4042} frac{sqrt{x} , dx | vedclass

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Question

(B) Let $I = \int_{0}^{4042} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4042-x}} \, dx \quad \dots (i)$ Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a}

Vedclass · Accepted Answer

$2021$

Vedclass · Answer

(B) Let $I = \int_{0}^{4042} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4042-x}} \, dx \quad \dots (i)$ Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get: $I = \int_{0}^{4042} \frac{\sqrt{4042-x}}{\sqrt{4042-x}+\sqrt{x}} \, dx \quad \dots (ii)$ Adding equations $(i)$ and $(ii)$: $2I = \int_{0}^{4042} \frac{\sqrt{x}+\sqrt{4042-x}}{\sqrt{x}+\sqrt{4042-x}} \, dx$ $2I = \int_{0}^{4042} 1 \, dx$ $2I = [x]_{0}^{4042} = 4042$ $I = \frac{4042}{2} = 2021$

The value of $\int_{0}^{4042} \frac{\sqrt{x} \, dx}{\sqrt{x}+\sqrt{4042-x}}$ is equal to

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