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

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

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integral from 0 to 3 of (e^t+e^{-t})

\int_{0}^{3} (e^{t} + e^{- t}) d t

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

Decimal Notation

20.03574 \dots

Solution steps

\int_{0}^{3} e^{t} + e^{- t} d t

Rewrite using trig identities

= \int_{0}^{3} 2 cosh (t) d t

Take the constant out:

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

= 2 \cdot \int_{0}^{3} cosh (t) d t

Use the common integral:

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

= 2 [sinh (t)]_{0}^{3}

Compute the boundaries:

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

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

Simplify

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

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