Integral Calculus Question 142
Question: What is the value of $ \int\limits_0^{1}{(x-1){e^{-x}}dx} $ ?
Options:
A) 0
B) e
C) $ \frac{1}{e} $
D) $ \frac{-1}{e} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given integral is $ I=\int_0^{1}{(x-1){e^{-x}}dx} $ Integrating by parts taking $ (x-1) $ as first function We get, $ I=[(x-1){-{e^{-x}}}]_0^{1}-\int_0^{1}{1.(-{e^{-x}})dx} $ $ =-(1-1)\frac{1}{e}+(-1)e^{0}+[-{e^{-x}}]_0^{1} $ $ =-1-\frac{1}{e}+1=-\frac{1}{e} $
BETA
