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

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

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

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

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

Decimal Notation

6.30122 \dots

Solution steps

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

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

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

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

\int t e^{- 6 t} d t

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 6}{x})

n = 1 \sum \infty 4 (0.8)^{n - 1}

d x + e^{3 x} d y = 0

\frac{d ^{2}}{d x ^{2}} (- \frac{5}{x})

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