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

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

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

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

\frac{1}{e} + 1

Solution steps

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

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

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

= 2 - (1 - \frac{1}{e})

Simplify

= \frac{1}{e} + 1

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