Question

The natural number $n(\le 5)$ for which
$I_n={\int }_0^1{e}^x(x-1{)}^{n}dx=16-6e$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

We have $I_0={\int }_0^1{e}^{x}dx=e^x{]}_0^1=e-1$
and for $n\ge 1,$
$I_n=e^x(x-1{)}^n{]}_0^1-n{\int }_0^1{e}^x(x-1{)}^{n-1}dx$
$=-(-1{)}^n-n{I}_{n-1}$
$\therefore I_1=1-\left(1\right)I_0=1-(e-1)=2-e$
$I_2=-1-2I_1=-1-2(2-e)=2e-5$
and $I_3=1-3I_2=1-3(2e-5)=16-6e$
$\therefore n=3.$