What is the integral from 0 to 2 of e^{-2x}+4 ?

The integral from 0 to 2 of e^{-2x}+4 is (e^4-1)/(2e^4)+8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of e^{-2x}+4

\int_{0}^{2} e^{- 2 x} + 4 d x

\frac{e ^{4} - 1}{2 e ^{4}} + 8

Decimal

8.49084 \dots

Solution steps

\int_{0}^{2} e^{- 2 x} + 4 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^{- 2 x} d x + \int_{0}^{2} 4 d x

\int_{0}^{2} e^{- 2 x} d x = \frac{e ^{4} - 1}{2 e ^{4}}

\int_{0}^{2} 4 d x = 8

= \frac{e ^{4} - 1}{2 e ^{4}} + 8

\frac{d}{d x} (\frac{6}{2 x + 1})

tan^{3} (x)

\frac{d}{d x} (\frac{150}{x})

\frac{d y}{d x} = 4 x

e^{1}

What is the integral from 0 to 2 of e^{-2x}+4 ?
The integral from 0 to 2 of e^{-2x}+4 is (e^4-1)/(2e^4)+8