If for n geq 1,P_n = intlimits_1^e (log x)^n | vedclass

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(D) Given $P_n = \int\limits_1^e (\log x)^n \, dx$. Let $\log x = t$,then $x = e^t$ and $dx = e^t \, dt$. When $x = 1$,$t = 0$ and when $x = e$,$t = 1

Vedclass · Accepted Answer

$-9e$

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(D) Given $P_n = \int\limits_1^e (\log x)^n \, dx$. Let $\log x = t$,then $x = e^t$ and $dx = e^t \, dt$. When $x = 1$,$t = 0$ and when $x = e$,$t = 1$. Thus,$P_n = \int\limits_0^1 t^n e^t \, dt$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = t^n$ and $dv = e^t \, dt$,then $du = nt^{n-1} \, dt$ and $v = e^t$. $P_n = [t^n e^t]_0^1 - n \int\limits_0^1 t^{n-1} e^t \, dt = e - n P_{n-1}$. So,$P_{10} = e - 10 P_9$. Also,$P_9 = e - 9 P_8$. Substituting $P_9$ into the equation for $P_{10}$: $P_{10} = e - 10(e - 9 P_8) = e - 10e + 90 P_8$. $P_{10} = -9e + 90 P_8$. Therefore,$P_{10} - 90 P_8 = -9e$.

If for $n \geq 1$,$P_n = \int\limits_1^e (\log x)^n \, dx$,then $P_{10} - 90P_8$ is equal to

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