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

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

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

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

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

Decimal

4.70080 \dots

Solution steps

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

Rewrite using trig identities

= \int_{- 1}^{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_{- 1}^{1} cosh (x) d x

Use the common integral:

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

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

Compute the boundaries:

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

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

Simplify

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

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