Question

If for $n\ge 1,P_n={\int }_1^e{\left(\log x\right)}^{n}dx$, then $P_{10}-90P_8$ is equal to :

• A. $-9$
• B. $10$
• C. $-9e$
• D. $10e$

Answer 1

The correct option is C $-9e$

$P_n={\int }_1^e{\left(\log x\right)}^n$
$P_{10}={\int }_1^e{\left(\log x\right)}^{10}$
$={\int }_0^1{t}^{10}{e}^{t}dt$
$={\left[t^{10}{e}^t\right]}_0^1-10{\int }_0^1{t}^9{e}^{t}dt$
$=e-10{{[t}^9{e}^t]}_0^1+90{\int }_0^1{t}^8{e}^{t}dt$
$\Rightarrow P_{10}=e-10e+90P_{10}$
$\Rightarrow P_{10}-90P_{10}=-9e$