What is the integral from 0 to 3 of (e^x+e^{-x})/2 ?

The integral from 0 to 3 of (e^x+e^{-x})/2 is (e^6-1)/(2e^3)

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integral from 0 to 3 of (e^x+e^{-x})/2

\int_{0}^{3} \frac{e ^{x} + e ^{- x}}{2} d x

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

Decimal

10.01787 \dots

Solution steps

\int_{0}^{3} \frac{e ^{x} + e ^{- x}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{3} e^{x} + e^{- x} d x

Rewrite using trig identities

= \frac{1}{2} \cdot \int_{0}^{3} 2 cosh (x) d x

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int_{0}^{3} cosh (x) d x

Use the common integral:

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

= \frac{1}{2} \cdot 2 [sinh (x)]_{0}^{3}

Simplify

\frac{1}{2} \cdot 2 [sinh (x)]_{0}^{3} : [sinh (x)]_{0}^{3}

= [sinh (x)]_{0}^{3}

Compute the boundaries:

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

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

What is the integral from 0 to 3 of (e^x+e^{-x})/2 ?
The integral from 0 to 3 of (e^x+e^{-x})/2 is (e^6-1)/(2e^3)