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

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

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

s l o p e y = 3 x^{2} + 2 xx a

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

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

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n + 2}

t an g e n t f (x) = 5 x^{3} - 2 x, at x = - 1

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