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

The integral from 0 to 1 of (e^{2x}-e^{-x}) is (2+e^3-e)/(2e)-1

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

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

\frac{2 + e ^{3} - e}{2 e} - 1

Decimal Notation

2.56240 \dots

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

\frac{e ^{2} - 1}{2} - (1 - \frac{1}{e}) : \frac{2 + e ^{3} - e}{2 e} - 1

= \frac{2 + e ^{3} - e}{2 e} - 1

x \to 1 lim (\frac{3}{1 - x ^{3}})

\frac{\partial}{\partial x} (x^{4} - y^{3})

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{6} e^{- x} d y = 0

\frac{d ^{2}}{d x ^{2}} (x^{2} + 12)

x^{3} + x^{2}, 2 x^{2} + 2 x

What is the integral from 0 to 1 of (e^{2x}-e^{-x}) ?
The integral from 0 to 1 of (e^{2x}-e^{-x}) is (2+e^3-e)/(2e)-1