Question

Determine a positive integer $n\le 5,$ such that ${\int }_0^1{e}^x{(x-1)}^{n}dx=16-6e$

• A. $n=1$
• B. $n=2$
• C. $n=3$
• D. $n=-3$

Answer 1

Answer: (C)

$n=3$

$l_n={\int }_0^1{e}^x\cdot {(x-1)}^{n}dx$
$=e^x\cdot {(x-1)}^n{\int }_0^1-n{\int }_0^1{e}^x{(x-1)}^{n-1}dx$
$l_n=-{(-1)}^n-n{l}_{n-1}={(-1)}^{n+1}-n{l}_{n-1}$
$n=1,l={\int }_0^1{e}^x(x-1)dx$
$=(x-1)e^x{\int }_0^1-{\int }_0^1{e}^{x}dx=2-e$
$l_2=-1-2(2-e)=2e-5$
$l_3=1-3.(2e-5)=16-6e$
So, $n=3$