Question

If ${\int }_0^1\frac{e^{t}dt}{t+1}=a,then{\int }_{b-1}^b\frac{e^{-t}dt}{(t-b-1)}$ is equal to

• A. $a{e}^{-b}$
• B. $-a{e}^{-b}$
• C. $-b{e}^{-a}$
• D. $a{e}^b$

Answer 1

The correct option is B $-a{e}^{-b}$

$\because a={\int }_0^1\frac{e^{t}dt}{t+1}$ ....(1)
$I={\int }_{b-1}^b\frac{e^{-t}dt}{t-b-1},putt=b-y$
$dt=-dy$
$I=e^{-b}{\int }_0^1\frac{e^{y}dy}{y+1}=-a{e}^{-b}$