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

The integral from-3 to+3 of (e^x+e^{-x}) is (2(e^6-1))/(e^3)

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

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

\frac{2 ( e ^{6} - 1 )}{e ^{3}}

Decimal Notation

40.07149 \dots

Solution steps

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

Rewrite using trig identities

= \int_{- 3}^{3} 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_{- 3}^{3} cosh (x) d x

Use the common integral:

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

= 2 [sinh (x)]_{- 3}^{3}

Compute the boundaries:

\frac{e ^{6} - 1}{e ^{3}}

= 2 \cdot \frac{e ^{6} - 1}{e ^{3}}

Simplify

= \frac{2 ( e ^{6} - 1 )}{e ^{3}}

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What is the integral from-3 to+3 of (e^x+e^{-x}) ?
The integral from-3 to+3 of (e^x+e^{-x}) is (2(e^6-1))/(e^3)