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

The integral from 0 to 4 of e^{-2x}+4 is (e^8-1)/(2e^8)+16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{e ^{8} - 1}{2 e ^{8}} + 16

Decimal Notation

16.49983 \dots

Solution steps

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

\int_{0}^{4} e^{- 2 x} d x = \frac{e ^{8} - 1}{2 e ^{8}}

\int_{0}^{4} 4 d x = 16

= \frac{e ^{8} - 1}{2 e ^{8}} + 16

\int 1 + e^{x} d x

\frac{d y}{d x} = \frac{( y - 3 x ^{2} + y ^{2} )}{x}

x \to 0 lim (x (cos (x)))

x \to 0 lim (\frac{3 - 3 cos ( x )}{x})

\frac{d}{d x} (e^{e^{3 x^{2}}})

What is the integral from 0 to 4 of e^{-2x}+4 ?
The integral from 0 to 4 of e^{-2x}+4 is (e^8-1)/(2e^8)+16