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

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

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

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

2 - e - \frac{1}{e}

Decimal Notation

- 1.08616 \dots

Solution steps

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

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

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

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

Simplify

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

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