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 (e^2-1)/e

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

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

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

Decimal

2.35040 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{1} 2 cosh (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{1} cosh (x) d x

Use the common integral:

\int cosh (x) d x = sinh (x)

= 2 [sinh (x)]_{0}^{1}

Compute the boundaries:

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

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

Simplify

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

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 (e^2-1)/e