If ∫01 etdtt+1=a then, ∫b−1b e−tdtt−b−1 is equal to

# If ${\int }_{0}^{1} \frac{{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}}{\mathrm{t}+1}=\mathrm{a}$ then, ${\int }_{\mathrm{b}-1}^{\mathrm{b}} \frac{{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}}{\mathrm{t}-\mathrm{b}-1}$ is equal to

1. A

${\mathrm{ae}}^{-\mathrm{b}}$

2. B

$-{\mathrm{ae}}^{-\mathrm{b}}$

3. C

$-{\mathrm{be}}^{-\mathrm{a}}$

4. D

${\mathrm{ae}}^{\mathrm{b}}$

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### Solution:

$\begin{array}{r}{\int }_{\mathrm{b}-1}^{\mathrm{b}} \frac{{\mathrm{e}}^{-\mathrm{t}}}{\mathrm{t}-\mathrm{b}-1}\mathrm{dt}={\int }_{-1}^{0} \frac{{\mathrm{e}}^{-\left(\mathrm{t}+\mathrm{b}\right)}}{\left(\mathrm{t}+\mathrm{b}\right)-\mathrm{b}-1}\mathrm{dt}\\ ={\mathrm{e}}^{-\mathrm{b}}{\int }_{-1}^{0} \frac{{\mathrm{e}}^{-\mathrm{t}}}{\mathrm{t}-1}\mathrm{dt}={\mathrm{e}}^{-\mathrm{b}}{\int }_{0}^{1} \frac{{\mathrm{e}}^{-\left(\mathrm{t}-1\right)}}{\mathrm{t}-2}\mathrm{dt}\end{array}$

Put,

$\begin{array}{l}\mathrm{t}-1=-\mathrm{s}⇒\mathrm{dt}=-\mathrm{ds}\\ =-{\mathrm{e}}^{-\mathrm{b}}{\int }_{1}^{0} \frac{{\mathrm{e}}^{\mathrm{s}}}{-\left(\mathrm{s}+1\right)}\mathrm{ds}={\mathrm{e}}^{-\mathrm{b}}{\int }_{1}^{0} \frac{{\mathrm{e}}^{\mathrm{s}}}{\mathrm{s}+1}\mathrm{ds}\\ =-{\mathrm{e}}^{-\mathrm{b}}{\int }_{0}^{1} \frac{{\mathrm{e}}^{\mathrm{t}}}{\mathrm{t}+1}\mathrm{dt}=-{\mathrm{ae}}^{-\mathrm{b}}\end{array}$

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