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

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

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

\int_{0}^{1} \frac{e ^{x} - e ^{- x}}{2} d x

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

Decimal

0.54308 \dots

Solution steps

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

Rewrite using trig identities

= \frac{1}{2} \cdot \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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

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\int_{0}^{1} sin (x) cos (x) d x

\int_{1}^{1} 1 + x^{4} d x

\int_{1}^{3} 2 (1 + ln (x)) x^{2 x} d x

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