What is the integral from 1 to 0 of (e^x+e^{-3}) ?

The integral from 1 to 0 of (e^x+e^{-3}) is -e+1-1/(e^3)

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integral from 1 to 0 of (e^x+e^{-3})

\int_{1}^{0} (e^{x} + e^{- 3}) d x

- e + 1 - \frac{1}{e ^{3}}

Decimal

- 1.76806 \dots

Solution steps

\int_{1}^{0} e^{x} + e^{- 3} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} e^{x} + e^{- 3} 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^{- 3} d x)

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

\int_{0}^{1} e^{- 3} d x = \frac{1}{e ^{3}}

= - (e - 1 + \frac{1}{e ^{3}})

Simplify

= - e + 1 - \frac{1}{e ^{3}}

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What is the integral from 1 to 0 of (e^x+e^{-3}) ?
The integral from 1 to 0 of (e^x+e^{-3}) is -e+1-1/(e^3)