How do you integrate $\int e^{- 2 x} d x$ $int e^(-2x)dx$ from $[0, 1]$ $[0,1]$ ?

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Answer 1 · 2016-11-28T13:35:29

In general $\int (e^{α x}) d x = \frac{1}{α} e^{α x}$ $int(e^(alphax))dx = 1/alphae^(alphax)$

$\int_{0}^{1} e^{- 2 x} d x = - \frac{1}{2} e^{- 2 x} ∣_{0}^{1} = - \frac{1}{2} (e^{- 2} - 1)$ $int_0^1e^(-2x)dx = -1/2e^(-2x)|_0^1=-1/2 (e^-2-1)$

How do you integrate int e^(-2x)dx∫e−2xdxint e^(-2x)dx from [0,1][0,1][0,1]?