If for $\mathrm{n} \geq 1, \mathrm{P}_{\mathrm{n}}=\int_1^{\mathrm{e}}\left(\log…
If for $\mathrm{n} \geq 1, \mathrm{P}_{\mathrm{n}}=\int_1^{\mathrm{e}}\left(\log \mathrm{x}^{\mathrm{n}}\right) \mathrm{dx}$, then $\mathrm{P}_{10}-90 \mathrm{P}_8$ is equal to:
$-9$
$10 \mathrm{e}$
$-9 \mathrm{e}$
10
Solution
$\mathrm{P}_{\mathrm{n}}=\int_1^e(\log x)^n d x$ put $\log x=t$ then $x=e^t$ and $d x=e^t d t$
Also, when $x=1$, then $t=\log 1=0$ and when $x=e$, then $t=\log _e e=1$
$
\begin{gathered}
\therefore \mathrm{P}_{\mathrm{n}}=\int_0^1 t^n \cdot e^t d t \\
\therefore \mathrm{P}_{10}=\int_0^1 t^{10} e^t d t \text { and } \mathrm{P}_8=\int_0^1 t^8 e^t d t \\
\text { Now, } \mathrm{P}_{10}-90 \mathrm{P}_8=\int_0^1 t^{10} e^t d t-90 \int_0^1 t^8 e^t d t \\
\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 d t-90 \int_0^1 t^8 e^t d t \\
\mathrm{P}_{10}-90 \mathrm{P}_8= \\
e-10\left[t^9 \int_0^1 e^t d t-\int_0^1 \frac{d}{d t}\left(t^9\right) \int e^t d t\right]-90 \int_0^1 t^8 e^t d t
\end{gathered}
$
$
\begin{gathered}
\mathrm{P}_{10}-90 \mathrm{P}_8=e-10\left[e-9 \int_0^1 t^8 e^t d t\right]-90 \int_0^1 t^8 e^t d t \\
\mathrm{P}_{10}-90 \mathrm{P}_8=e-10 e+90 \int t^8 e^t d t-90 \int_0^1 t^8 e^t d t \\
\therefore P_{10}-90 P_8=-9 e
\end{gathered}
$