Evaluate the definite integral: int_{0}^{1} l | vedclass

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Question

(B) The given expression inside the integral is the Maclaurin series expansion for $e^{-x}$.Thus,the integral becomes $\int_{0}^{1} e^{-x} \cdot e^{2x

Vedclass · Accepted Answer

$e - 1$

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(B) The given expression inside the integral is the Maclaurin series expansion for $e^{-x}$.Thus,the integral becomes $\int_{0}^{1} e^{-x} \cdot e^{2x} \, dx$.Using the property of exponents,$e^{-x} \cdot e^{2x} = e^{-x + 2x} = e^{x}$.So,the integral is $\int_{0}^{1} e^{x} \, dx$.Evaluating the definite integral,we get $[e^{x}]_{0}^{1}$.Substituting the limits,we have $e^{1} - e^{0} = e - 1$.

Evaluate the definite integral: $\int_{0}^{1} \left(1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots \infty\right) e^{2x} \, dx$.

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