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

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

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

\inline \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}

\inline \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}