Mathematics - Definite Integration Question with Solution | TestHub

${\mathrm{P}}_{\mathrm{n}}={\int }_1^e(\log x{)}^{n}dx$ put $\log x=t$ then $x=e^t$ and $dx=e^{t}dt$ Also, when $x=1$, then $t=\log 1=0$ and when $x=e$, then $t={\log }_{e}e=1$ $$\begin{array}{c}\therefore {\mathrm{P}}_{\mathrm{n}}={\int }_0^1{t}^n\cdot e^{t}dt\\ \therefore {\mathrm{P}}_{10}={\int }_0^1{t}^{10}{e}^{t}dt\text{ and }{\mathrm{P}}_8={\int }_0^1{t}^8{e}^{t}dt\\ \text{ Now, }{\mathrm{P}}_{10}-90{\mathrm{P}}_8={\int }_0^1{t}^{10}{e}^{t}dt-90{\int }_0^1{t}^8{e}^{t}dt\\ {\mathrm{P}}_{10}-90{\mathrm{P}}_8\\ {\mathrm{P}}_{10}-90{\mathrm{P}}_8\\ ={\left[t^{10}{e}^t\right]}_0^1-10{\int }_0^1{t}^9{e}^{t}dt-90{\int }_0^1{t}^8{e}^{t}dt\\ {\mathrm{P}}_{10}-90{\mathrm{P}}_8=\\ e-10\left[t^9{\int }_0^1{e}^{t}dt-{\int }_0^1\frac{d}{dt}\left(t^9\right)\int e^{t}dt\right]-90{\int }_0^1{t}^8{e}^{t}dt\end{array}$$ $$\begin{array}{c}{\mathrm{P}}_{10}-90{\mathrm{P}}_8=e-10\left[e-9{\int }_0^1{t}^8{e}^{t}dt\right]-90{\int }_0^1{t}^8{e}^{t}dt\\ {\mathrm{P}}_{10}-90{\mathrm{P}}_8=e-10e+90\int t^8{e}^{t}dt-90{\int }_0^1{t}^8{e}^{t}dt\\ \therefore P_{10}-90P_8=-9e\end{array}$$