Let A = _0^1 , e^t , , ,d ,t 1 , , + , ,t then _ a - 1 ^a , , e^ … - Vidyadip

MCQ

Let $A = \int\limits_0^1 \, \frac{{{e^t}\,\,\,d\,t}}{{1\,\, + \,\,t}}$ then $\int\limits_{a - 1}^a {\,\,\frac{{{e^{ - t}}\,dt}}{{t\, - \,a\, - \,1}}} $ has the value

A
$Ae^{-a}$
✓
$-Ae^{-a}$
C
$-ae^{-a}$
D
$Ae^a$

✓

Answer

Correct option: B.

$-Ae^{-a}$

b
$I = \int\limits_{a - 1}^a {\frac{{{e^{ - t}}}}{{t - a - 1}}\,\,dt} $ put $t = a-1+ y$ (so that lower limit becomes zero)
$\therefore$ $I =$ $\int\limits_0^1 {\frac{{{e^{1 - a - y}}}}{{y - 2}}\,\,\,dy} $ (now using king)
$I =\int\limits_0^1 {\frac{{{e^{1 - a - 1 + y}}}}{{1 - y - 2}}\,\,\,dy} $ $= - {e^{ - a}}\int\limits_0^1 {\frac{{{e^y}}}{{1 + y}}\,\,dy} $ $= - e^{-a} A$

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