Question

Prove that: ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx,$ and hence evaluate $\underset{0}{\overset{1}{\int }}{x}^2{\left(1-x\right)}^{n}dx$

Answer 1

To Prove: ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$

$$\begin{gathered} I={\int }_0^{a}f\left(x\right)dx \\ \mathrm{Let}x=a-t \\ \Rightarrow dx=0-dt \\ \Rightarrow dx=-dt \\ \mathrm{Also},\mathrm{if}x=0,t=a \\ \mathrm{if}x=a,t=0 \\ \mathrm{Thus}, \\ I={\int }_a^{0}f\left(a-t\right)\left(-dt\right) \\ =-{\int }_a^{0}f\left(a-t\right)dt \\ =-\left(-{\int }_0^{a}f\left(a-t\right)dt\right)\left(\because {\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx\right) \\ ={\int }_0^{a}f\left(a-t\right)dt \\ ={\int }_0^{a}f\left(a-x\right)dx\left(\because {\int }_a^{b}f\left(t\right)dt={\int }_a^{b}f\left(x\right)dx\right) \\ \mathrm{Hence}, \\ {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx \end{gathered}$$


Now,
$$\begin{gathered} \mathrm{Let}J=\underset{0}{\overset{1}{\int }}{x}^2{\left(1-x\right)}^{n}dx \\ =\underset{0}{\overset{1}{\int }}{\left(1-x\right)}^2{\left(1-\left(1-x\right)\right)}^{n}dx\left(\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right) \\ =\underset{0}{\overset{1}{\int }}\left(1+x^2-2x\right){\left(1-1+x\right)}^{n}dx \\ =\underset{0}{\overset{1}{\int }}\left(1+x^2-2x\right){\left(x\right)}^{n}dx \\ =\underset{0}{\overset{1}{\int }}\left(x^n+x^n{x}^2-2x^{n}x\right)dx \\ =\underset{0}{\overset{1}{\int }}\left(x^n+x^{n+2}-2x^{n+1}\right)dx \\ ={\left(\frac{x^{n+1}}{n+1}+\frac{x^{n+3}}{n+3}-2\frac{x^{n+2}}{n+2}\right)}_0^1 \\ =\frac{1^{n+1}}{n+1}+\frac{1^{n+3}}{n+3}-2\frac{1^{n+2}}{n+2} \\ =\frac{1}{n+1}+\frac{1}{n+3}-\frac{2}{n+2} \end{gathered}$$


Hence, $\underset{0}{\overset{1}{\int }}{x}^2{\left(1-x\right)}^{n}dx=\frac{1}{n+1}+\frac{1}{n+3}-\frac{2}{n+2}.$​