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Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 13 textbook solution.

Answers (1)

Answer : \frac{5}{2}

Given : \int_{0}^{5} \frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}+\sqrt[4]{9-x}} d x

Hint :  Use the formula of \int_{0}^{a} f(x) d x

Solution : I=\int_{0}^{5} \frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}+\sqrt[4]{9-x}} d x-------(1)

\begin{aligned} &I=\int_{0}^{5} \frac{\sqrt[4]{(5-x)+4}}{\sqrt[4]{(5-x)+4}+\sqrt[4]{9-(5-x)}} d x\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right) \\ &I=\int_{0}^{5} \frac{\sqrt[4]{9-x}}{\sqrt[4]{9-x}+\sqrt[4]{4+x}} d x------(2) \end{aligned}

Add (1) and (2)

\begin{aligned} &2 I=\int_{0}^{5} \frac{\sqrt[4]{x+4}+\sqrt[4]{9-x}}{\sqrt[4]{9-x}+\sqrt[4]{4+x}} d x \\ &2 I=\int_{0}^{5} 1 \cdot d x \\ &2 I=(x)_{0}^{5} \end{aligned}

\begin{aligned} &2 I=5 \\ &I=\frac{5}{2} \end{aligned}

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