A) 0
B) \[e-1\]
C) 1
D) e
E) none of these
Correct Answer: B
Solution :
\[\int_{0}^{1}{\left[ 1-\frac{x}{1!}+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+..... \right]}{{e}^{2x}}dx\] \[=\int_{0}^{1}{{{e}^{-x}}.}{{e}^{2x}}dx=[{{e}^{x}}]_{0}^{1}=e-1\]You need to login to perform this action.
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