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

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

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

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

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

Decimal Notation

1.28594 \dots

Solution steps

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

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

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

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

\int_{- 2}^{2} 4 + x^{2} d x

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

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

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