What is the integral from 0 to 4 of e^x+e^{-x} ?

The integral from 0 to 4 of e^x+e^{-x} is (e^8-1)/(e^4)

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integral from 0 to 4 of e^x+e^{-x}

\int_{0}^{4} e^{x} + e^{- x} d x

\frac{e ^{8} - 1}{e ^{4}}

Decimal

54.57983 \dots

Solution steps

\int_{0}^{4} e^{x} + e^{- x} d x

Rewrite using trig identities

= \int_{0}^{4} 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}^{4} cosh (x) d x

Use the common integral:

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

= 2 [sinh (x)]_{0}^{4}

Compute the boundaries:

\frac{e ^{8} - 1}{2 e ^{4}}

= 2 \cdot \frac{e ^{8} - 1}{2 e ^{4}}

Simplify

= \frac{e ^{8} - 1}{e ^{4}}

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What is the integral from 0 to 4 of e^x+e^{-x} ?
The integral from 0 to 4 of e^x+e^{-x} is (e^8-1)/(e^4)