Question

If \(n\) is an odd natural number and \[ I_n=\int_0^1 e^x(x-1)^n\,dx \] then \( I_n+nI_{n-1} \) is equal to:

• A. \(1\)
• B. \(-1\)
• C. \(e\)
• D. \(0\)

Hint:

Integration by parts is extremely useful in deriving recurrence relations involving powers and exponentials.

Answer 1

The Correct Option is D

Concept: Use integration by parts: \[ \int u\,dv=uv-\int v\,du \] This helps establish recurrence relations.

Step 1: Apply integration by parts. Given: \[ I_n=\int_0^1 e^x(x-1)^n\,dx \] Take: \[ u=(x-1)^n, \qquad dv=e^x dx \] Then: \[ du=n(x-1)^{n-1}dx, \qquad v=e^x \] Thus: \[ I_n = \left[e^x(x-1)^n\right]_0^1 - n\int_0^1 e^x(x-1)^{n-1}dx \] \[ I_n = \left[e^x(x-1)^n\right]_0^1 - nI_{n-1} \]

Step 2: Evaluate boundary term. At \(x=1\): \[ e^1(0)^n=0 \] At \(x=0\): \[ e^0(-1)^n=-1 \] Since \(n\) is odd: \[ (-1)^n=-1 \] Hence: \[ \left[e^x(x-1)^n\right]_0^1 = 0-(-1) = 1 \] Thus: \[ I_n=1-nI_{n-1} \]

Step 3: Rearrange relation. \[ I_n+nI_{n-1}=1 \] \[ \boxed{ 1 } \] Hence correct option is: \[ \boxed{(1)} \]