What is the integral from 0 to 6 of e^t-e^{-t} ?

The integral from 0 to 6 of e^t-e^{-t} is (e^{12}+1)/(e^6)-2

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integral from 0 to 6 of e^t-e^{-t}

\int_{0}^{6} e^{t} - e^{- t} d t

\frac{e ^{12} + 1}{e ^{6}} - 2

Decimal

401.43127 \dots

Solution steps

\int_{0}^{6} e^{t} - e^{- t} d t

Rewrite using trig identities

= \int_{0}^{6} 2 sinh (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}^{6} sinh (t) d t

Use the common integral:

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

= 2 [cosh (t)]_{0}^{6}

Compute the boundaries:

\frac{e ^{12} + 1}{2 e ^{6}} - 1

= 2 (\frac{e ^{12} + 1}{2 e ^{6}} - 1)

Simplify

= \frac{e ^{12} + 1}{e ^{6}} - 2

\int_{0}^{6} (x) d x

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What is the integral from 0 to 6 of e^t-e^{-t} ?
The integral from 0 to 6 of e^t-e^{-t} is (e^{12}+1)/(e^6)-2