smc-5134-50-Recursive

If \(I_n=\displaystyle {\int_0^1\left((1-x)^n e^x\right) d x}\), where \(n \in N\), then for \(n \geq 1, I_n\) equals

\(-1+n I_{n-1}\)
\(n I_{n-1}\)
\(-1-n I_{n-1}\)
\(-n I_{n-1}\)
\((1-x)^n e^x+n I_{n-1}\)

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\(A\)

Show Worked Solution

\(\text{Using integration by parts:}\)

\begin{aligned}
u &=(1-x)^n, & u^{′} &=n(1-x)^{n-1}\\
v^{′} & = e^x, & v & = e^x\ \\
\end{aligned}

\begin{aligned}
\int_0^1 ((1-x)^n e^x)\,dx\ &= uv-\int u^{′} v \ dx\\
& = \Bigr[(1-x)^n\,e^x\Bigr]_0 ^1- \int_0 ^1 n(1-x)^{n-1}\,e^x\ dx \\
& =-1-n\int_0 ^1 (1-x)^{n-1}\,e^x\ dx \\
& =-1-nI_{n-1} \\
\end{aligned}

\(\Rightarrow A\)