What is the integral from 0 to 4 of 1-e^{-2x} ?

The integral from 0 to 4 of 1-e^{-2x} is 4-(e^8-1)/(2e^8)

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integral from 0 to 4 of 1-e^{-2x}

\int_{0}^{4} 1 - e^{- 2 x} d x

4 - \frac{e ^{8} - 1}{2 e ^{8}}

Solution steps

\int_{0}^{4} 1 - e^{- 2 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{4} 1 d x - \int_{0}^{4} e^{- 2 x} d x

\int_{0}^{4} 1 d x = 4

\int_{0}^{4} e^{- 2 x} d x = \frac{e ^{8} - 1}{2 e ^{8}}

= 4 - \frac{e ^{8} - 1}{2 e ^{8}}

\int_{1}^{12} 4 e^{0.1 x} d x

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\int_{0}^{2} x^{3} - 2 x d x

What is the integral from 0 to 4 of 1-e^{-2x} ?
The integral from 0 to 4 of 1-e^{-2x} is 4-(e^8-1)/(2e^8)