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Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 12 textbook solution.

Answers (1)

Answer : \frac{a}{2}

Given : \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x

Hint : Use the formula of \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x

Solution : I=\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x-------(1)

I=\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x-------(2)\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)

Add (1) and (2)

\begin{aligned} &2 I=\int_{0}^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x \\ &2 I=\int_{0}^{a} 1 . d x \\ &2 I=(x)_{0}^{a} \end{aligned}

\begin{aligned} &2 I=(x)_{0}^{a} \\ &2 I=a \\ &I=\frac{a}{2} \end{aligned}

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