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$\int_{0}^{1}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\cdots\right.$ upto $\left.\infty\right) e^{2 x} d x=$
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Verified Answer
The correct answer is:
$e-1$
(B)
$\int_{0}^{1}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots \infty\right) e^{2 x} d x$
$=\int_{0}^{1} e^{-x} e^{2 x} d x=\int_{0}^{1} e^{x} d x$
$=\left[e^{x}\right]_{0}^{1}=e^{1}-e^{0}=e-1$
$\int_{0}^{1}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots \infty\right) e^{2 x} d x$
$=\int_{0}^{1} e^{-x} e^{2 x} d x=\int_{0}^{1} e^{x} d x$
$=\left[e^{x}\right]_{0}^{1}=e^{1}-e^{0}=e-1$
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