Question

A positive integer $n\le 5$, such that ${\int }_0^1{e}^{2x-1}(x-1{)}^{n}dx=\frac{1}{4}(\frac{7}{e}-e)$ is

Answer 1

Let $I_n={\int }_0^1{e}^{2x-1}(x-1{)}^{n}dx$
For $n=0,$
$I_0={\int }_0^1{e}^{2x-1}(x-1{)}^{0}dx={\frac{1}{2}{e}^{2x-1}]}_0^1=\frac{1}{2}(e-\frac{1}{e})$

For $n\ge 1,$
${I_n=\frac{1}{2}{e}^{2x-1}(x-1{)}^n]}_0^1-\frac{n}{2}{\int }_0^1{e}^{2x-1}(x-1{)}^{n-1}dx$

$=-\frac{1}{2}{e}^{-1}(-1{)}^n-\frac{n}{2}{I}_{n-1}$

$=\frac{1}{2}(-1{)}^{n-1}{e}^{-1}-\frac{n}{2}{I}_{n-1}$

$\therefore I_1=\frac{1}{2}{e}^{-1}-\frac{1}{2}\frac{1}{2}(e-\frac{1}{e})$

$=-\frac{1}{4}(\frac{3}{e}-e)$

$I_2=-\frac{1}{2e}-I_1=-\frac{1}{2e}-\frac{1}{4}(\frac{3}{e}-e)$

$=\frac{1}{4}(e-\frac{5}{e})$

$I_3=\frac{1}{2e}-I_2=\frac{1}{2e}-\frac{1}{4}(\frac{5}{e}-e)$

$=\frac{1}{4}(\frac{7}{e}-e)$

Hence $n=3$