Question

Evaluate \[ \int_{a-1}^{a}\frac{e^x(a-x)}{(x-a+1)^2}\,dx \]

• A. \(2e^a+e\)
• B. \(\frac{2e^{a+2}}{e-2}\)
• C. \(\frac{e^a(e+2)}{2}\)
• D. \(\frac{e^a(e-2)}{2}\)

Hint:

Before evaluating a definite integral, always check whether the integrand is defined at the limits. If the denominator becomes zero and the numerator is non-zero, the integral may be improper and divergent.

Answer 1

The Correct Option is D

Step 1: Check the denominator.
The given integral is \[ \int_{a-1}^{a}\frac{e^x(a-x)}{(x-a+1)^2}\,dx. \] At the lower limit, \[ x=a-1. \] Then, \[ x-a+1=(a-1)-a+1=0. \] So, the denominator becomes \[ (x-a+1)^2=0. \]

Step 2: Check the numerator at the same point.
At \[ x=a-1, \] the numerator is \[ e^{a-1}\{a-(a-1)\}=e^{a-1}. \] This is non-zero.
Therefore, the integrand becomes unbounded at \[ x=a-1. \]

Step 3: Use substitution to confirm divergence.
Let \[ t=x-a+1. \] Then, \[ x=a-1+t \] and \[ a-x=1-t. \] When \[ x=a-1,\quad t=0, \] and when \[ x=a,\quad t=1. \] Thus, the integral becomes \[ \int_0^1 \frac{e^{a-1+t}(1-t)}{t^2}\,dt. \] Near \[ t=0, \] the integrand behaves like \[ \frac{e^{a-1}}{t^2}. \] But \[ \int_0^1 \frac{1}{t^2}\,dt \] is divergent.

Step 4: Final conclusion.
Hence, the printed integral does not have a finite value.
Therefore, \[ \boxed{\text{The integral is divergent.}} \]