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

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

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

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

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

Decimal

54.57983 \dots

Solution steps

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

Rewrite using trig identities

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

Use the common integral:

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

= 2 [sinh (t)]_{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}}

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