The positive integer n leq 5 for which int_0^ | vedclass

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(C) Let $I_n = \int_0^1 e^x(x-1)^n dx$. Using integration by parts, we have $\int u dv = uv - \int v du$. Let $u = (x-1)^n$ and $dv = e^x dx$. Then $d

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$3$

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(C) Let $I_n = \int_0^1 e^x(x-1)^n dx$. Using integration by parts, we have $\int u dv = uv - \int v du$. Let $u = (x-1)^n$ and $dv = e^x dx$. Then $du = n(x-1)^{n-1} dx$ and $v = e^x$. $I_n = [e^x(x-1)^n]_0^1 - n \int_0^1 e^x(x-1)^{n-1} dx = (0 - (-1)^n) - n I_{n-1} = (-1)^{n+1} - n I_{n-1}$. For $n=1$: $I_1 = (-1)^2 - 1 I_0 = 1 - \int_0^1 e^x dx = 1 - (e-1) = 2-e$. For $n=2$: $I_2 = (-1)^3 - 2 I_1 = -1 - 2(2-e) = -1 - 4 + 2e = 2e-5$. For $n=3$: $I_3 = (-1)^4 - 3 I_2 = 1 - 3(2e-5) = 1 - 6e + 15 = 16-6e$. Comparing this with the given value $16-6e$, we find $n=3$.

The positive integer $n \leq 5$ for which $\int_0^1 e^x(x-1)^n dx = 16-6e$ is

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