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

Question

If for $n \geq 1$, $P_n = \displaystyle\int_1^e (\log x)^n\,dx$, then $P_{10}-90P_8$ is equal to:

Accepted Answer

Put $\log x = t$, $x=e^t$, $dx = e^t\,dt$; limits $x=1 \Rightarrow t=0$, $x=e \Rightarrow t=1$

$P_n = \int_0^1 t^n e^t\,dt$

$P_{10} = \int_0^1 t^{10}e^t\,dt$, $P_8 = \int_0^1 t^8e^t\,dt$

$P_{10}-90P_8 = [t^{10}e^t]_0^1 - 10\int_0^1 t^9e^t\,dt - 90\int_0^1 t^8e^t\,dt$

$= e - 10\left([t^9e^t]_0^1-9\int_0^1 t^8e^t\,dt\right)-90\int_0^1 t^8e^t\,dt$

$= e-10e+90\int_0^1 t^8e^t\,dt - 90\int_0^1 t^8e^t\,dt = -9e$

Solutions · 1