Question

The value of the integral $\underset{0}{\overset{10}{\int }}\frac{x^{10}}{{\left(10-x\right)}^{10}+x^{10}}dx$ is ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{10}{\int }}\frac{x^{10}}{{\left(10-x\right)}^{10}+x^{10}}dx...\left(1\right) \\ \mathrm{We}\mathrm{know}, \\ \underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f\left(a-x\right)dx \\ \mathrm{Thus}, \\ I=\underset{0}{\overset{10}{\int }}\frac{{\left(10-x\right)}^{10}}{{\left(10-\left(10-x\right)\right)}^{10}+{\left(10-x\right)}^{10}}dx \\ =\underset{0}{\overset{10}{\int }}\frac{{\left(10-x\right)}^{10}}{{\left(10-10+x\right)}^{10}+{\left(10-x\right)}^{10}}dx \\ =\underset{0}{\overset{10}{\int }}\frac{{\left(10-x\right)}^{10}}{{\left(x\right)}^{10}+{\left(10-x\right)}^{10}}dx...\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2I=\underset{0}{\overset{10}{\int }}\frac{x^{10}}{{\left(10-x\right)}^{10}+x^{10}}dx+\underset{0}{\overset{10}{\int }}\frac{{\left(10-x\right)}^{10}}{{\left(x\right)}^{10}+{\left(10-x\right)}^{10}}dx \\ =\underset{0}{\overset{10}{\int }}\frac{x^{10}+{\left(10-x\right)}^{10}}{{\left(x\right)}^{10}+{\left(10-x\right)}^{10}}dx \\ =\underset{0}{\overset{10}{\int }}1dx \\ ={\left[x\right]}_0^{10} \\ =10 \\ \Rightarrow I=5 \end{gathered}$$​

​Hence, the value of the integral $\underset{0}{\overset{10}{\int }}\frac{x^{10}}{{\left(10-x\right)}^{10}+x^{10}}dx$ is 5.