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

The integral from 0 to 2 of e^x-e^{-x} is (e^4+1)/(e^2)-2

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integral from 0 to 2 of e^x-e^{-x}

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

\frac{e ^{4} + 1}{e ^{2}} - 2

Decimal Notation

5.52439 \dots

Solution steps

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

Rewrite using trig identities

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

Use the common integral:

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

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

Compute the boundaries:

\frac{e ^{4} + 1}{2 e ^{2}} - 1

= 2 (\frac{e ^{4} + 1}{2 e ^{2}} - 1)

Simplify

= \frac{e ^{4} + 1}{e ^{2}} - 2

g (x) = \frac{4 x ^{2} - x + 2}{3 - 4 x}

y = 3 x^{2} + 2 xx a

x \to 0 lim (x^{6} (cos (\frac{9}{x})))

y = e^{- 8 x}, x \geq 0

\frac{d}{d x} (sin (7 x) + sin^{2} (7 x))

What is the integral from 0 to 2 of e^x-e^{-x} ?
The integral from 0 to 2 of e^x-e^{-x} is (e^4+1)/(e^2)-2