What is the integral from 0 to 1 of (e^x)-(e^{-x}) ?

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

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integral from 0 to 1 of (e^x)-(e^{-x})

\int_{0}^{1} (e^{x}) - (e^{- x}) d x

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

Decimal Notation

1.08616 \dots

Solution steps

\int_{0}^{1} e^{x} - e^{- x} d x

Rewrite using trig identities

= \int_{0}^{1} 2 sinh (x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} sinh (x) d x

Use the common integral:

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

= 2 [cosh (x)]_{0}^{1}

Compute the boundaries:

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

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

Simplify

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

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