Question

Evaluate each of the following integrals:

${\int }_a^b\frac{x^{{\displaystyle \frac{1}{n}}}}{x^{{\displaystyle \frac{1}{n}}}+{\left(a+b-x\right)}^{{\displaystyle \frac{1}{n}}}}dx,n\in \mathbf{N},n\ge 2$

Answer 1

Let I = ${\int }_a^b\frac{x^{{\displaystyle \frac{1}{n}}}}{x^{{\displaystyle \frac{1}{n}}}+{\left(a+b-x\right)}^{{\displaystyle \frac{1}{n}}}}dx.....\left(1\right)$
Then,
$$\begin{gathered} I={\int }_a^b\frac{{\left(a+b-x\right)}^{{\displaystyle \frac{1}{n}}}}{{\left(a+b-x\right)}^{{\displaystyle \frac{1}{n}}}+{\left[a+b-\left(a+b-x\right)\right]}^{{\displaystyle \frac{1}{n}}}}dx\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right] \\ ={\int }_a^b\frac{{\left(a+b-x\right)}^{{\displaystyle \frac{1}{n}}}}{{\left(a+b-x\right)}^{{\displaystyle \frac{1}{n}}}+x^{{\displaystyle \frac{1}{n}}}}dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_a^b\frac{x^{{\displaystyle \frac{1}{n}}}+{\left(a+b-x\right)}^{{\displaystyle \frac{1}{n}}}}{x^{\frac{1}{n}}+{\left(a+b-x\right)}^{\frac{1}{n}}}dx \\ \Rightarrow 2I={\int }_a^{b}dx \\ \Rightarrow 2I={\overline{)x}}_a^b=\left(b-a\right) \\ \Rightarrow I=\frac{b-a}{2} \end{gathered}$$