If for n ≥ 1, P_ n=∫_1^ e (log x^ n ) dx, then P_10-90 P_8 is equal to:

$\displaystyle \mathrm{P}_{\mathrm{n}}=\int_1^e(\log x)^n d x$Pn​=∫1e​(logx)ndx put $\log x=t$logx=t then $x=e^t$x=et and $d x=e^t d t$dx=etdt Also, when $x=1$x=1, then $t=\log 1=0$t=log1=0 and when $x=e$x=e, then $t=\log _e e=1$t=loge​e=1 $$\begin{gathered} \therefore \mathrm{P}_{\mathrm{n}}=\int_0^1 t^n \cdot e^t d t \\ \therefore \mathrm{P}_{10}=\int_0^1 t^{10} e^t d t \text { and } \mathrm{P}_8=\int_0^1 t^8 e^t d t \\ \text { Now, } \mathrm{P}_{10}-90 \mathrm{P}_8=\int_0^1 t^{10} e^t d t-90 \int_0^1 t^8 e^t d t \\ \mathrm{P}_{10}-90 \mathrm{P}_8 \\ \mathrm{P}_{10}-90 \mathrm{P}_8 \\ =\left[t^{10} e^t\right]_0^1-10 \int_0^1 t^9 e^t d t-90 \int_0^1 t^8 e^t d t \\ \mathrm{P}_{10}-90 \mathrm{P}_8= \\ e-10\left[t^9 \int_0^1 e^t d t-\int_0^1 \frac{d}{d t}\left(t^9\right) \int e^t d t\right]-90 \int_0^1 t^8 e^t d t \end{gathered}$$∴Pn​=∫01​tn⋅etdt∴P10​=∫01​t10etdt and P8​=∫01​t8etdt Now, P10​−90P8​=∫01​t10etdt−90∫01​t8etdtP10​−90P8​P10​−90P8​=[t10et]01​−10∫01​t9etdt−90∫01​t8etdtP10​−90P8​=e−10[t9∫01​etdt−∫01​dtd​(t9)∫etdt]−90∫01​t8etdt​ $$\begin{gathered} \mathrm{P}_{10}-90 \mathrm{P}_8=e-10\left[e-9 \int_0^1 t^8 e^t d t\right]-90 \int_0^1 t^8 e^t d t \\ \mathrm{P}_{10}-90 \mathrm{P}_8=e-10 e+90 \int t^8 e^t d t-90 \int_0^1 t^8 e^t d t \\ \therefore P_{10}-90 P_8=-9 e \end{gathered}$$P10​−90P8​=e−10[e−9∫01​t8etdt]−90∫01​t8etdtP10​−90P8​=e−10e+90∫t8etdt−90∫01​t8etdt∴P10​−90P8​=−9e​