<p>Let \(A_n=\int_0^1 e^x(x-1)^n\,dx\). Show \(A_n+nA_{n-1}=?\). [JEE Advanced 2014]</p>
Step-by-Step Solution
Key Concept: IBP: Aₙ = [eˣ(x-1)ⁿ]_0^1 - n\int_0^1 eˣ(x-1)^(n-1)dx = 0-1 \cdot ... leads to Aₙ+n \cdot Aₙ₋_1=1.
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<p>IBP ($u=(x-1)^n$, $dv=e^x dx$):</p>
<p>$A_n=[e^x(x-1)^n]_0^1-n\int_0^1 e^x(x-1)^{n-1}dx = [0-e^0(-1)^n]-nA_{n-1}=(-1)^{n+1}-nA_{n-1}$</p>
<p>Hmm. For $n$ even: $A_n=-1-nA_{n-1}\Rightarrow A_n+nA_{n-1}=-1$. For $n$ odd: $A_n=1-nA_{n-1}\Rightarrow A_n+nA_{n-1}=1$.</p>
<p>General: $A_n+nA_{n-1}=(-1)^{n+1}$. For $n=1$: $A_1+A_0=1$. $A_0=\int_0^1 e^x dx=e-1$. $A_1=[e^x(x-1)]_0^1-\int_0^1 e^x dx=0+1-(e-1)=2-e$. $A_1+A_0=(2-e)+(e-1)=1$. ✓(B)</p>
Correct Answer: B