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  • Provide solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 10 textbook solution.

Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 10 textbook solution.

Answers (1)

Answer : \frac{b-a}{2}

Given : \int_{a}^{b} \frac{x^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}} d x, n \in N, n \geq 2

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

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

 

\\I=\int_{a}^{b} \frac{(a+b-x)^{\frac{1}{n}}}{(a+b-x)^{\frac{1}{n}}+x^{\frac{1}{n}}} d x\left[\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]------(2)

Add (1) and (2)

\begin{aligned} &2 I=\int_{a}^{b} \frac{(a+b-x)^{\frac{1}{n}}}{(a+b-x)^{\frac{1}{n}}+x^{\frac{1}{n}}}+\frac{x^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}} d x \\ &2 I=\int_{a}^{b} \frac{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}} \end{aligned}

\begin{aligned} &2 I=\int_{a}^{b} d x \\ &2 I=(x)_{a}^{b} \\ &2 I=b-a \\ &I=\frac{b-a}{2} \end{aligned}

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