A) 0
B) e
C) \[\frac{1}{e}\]
D) \[\frac{-1}{e}\]
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, \[I=[(x-1)\{-{{e}^{-x}}\}]_{0}^{1}-\int_{0}^{1}{1.(-{{e}^{-x}})dx}\] \[=-(1-1)\frac{1}{e}+(-1){{e}^{0}}+[-{{e}^{-x}}]_{0}^{1}\] \[=-1-\frac{1}{e}+1=-\frac{1}{e}\]
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