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

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

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

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

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

Decimal

1.08616 \dots

Solution steps

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

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

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

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

Simplify

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

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