What is the integral of 400e^{-0.1x}+8 ?

The integral of 400e^{-0.1x}+8 is -4000e^{-0.1x}+8x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 400e^{-0.1x}+8

\int 400 e^{- 0.1 x} + 8 d x

- 4000 e^{- 0.1 x} + 8 x + C

Solution steps

\int 400 e^{- 0.1 x} + 8 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 400 e^{- 0.1 x} d x + \int 8 d x

\int 400 e^{- 0.1 x} d x = - 4000 e^{- 0.1 x}

\int 8 d x = 8 x

= - 4000 e^{- 0.1 x} + 8 x

Add a constant to the solution

= - 4000 e^{- 0.1 x} + 8 x + C

\int x^{4} e (1 - x^{5}) d x

\int \frac{2}{4 - 9 x ^{2}} d x

\int (2^{x} - 4 sin (x)) d x

\int \frac{5 x - 2}{x ^{3} + x ^{2}} d x

\int \frac{x ^{3}}{x ^{8} - 2} d x

What is the integral of 400e^{-0.1x}+8 ?
The integral of 400e^{-0.1x}+8 is -4000e^{-0.1x}+8x+C