What is the integral of 8*2^{-x} ?

The integral of 8*2^{-x} is -(2^{-x+3})/(ln(2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 8*2^{-x}

\int 8 \cdot 2^{- x} d x

- \frac{2 ^{- x + 3}}{ln ( 2 )} + C

Solution steps

\int 8 \cdot 2^{- x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int 2^{- x} d x

Apply u-substitution

= 8 \cdot \int - 2^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 (- \int 2^{u} d u)

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 8 (- \frac{2 ^{u}}{ln ( 2 )})

Substitute back

u = - x

= 8 (- \frac{2 ^{- x}}{ln ( 2 )})

Simplify

8 (- \frac{2 ^{- x}}{ln ( 2 )}) : - \frac{2 ^{- x + 3}}{ln ( 2 )}

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

Add a constant to the solution

= - \frac{2 ^{- x + 3}}{ln ( 2 )} + C

What is the integral of 8*2^{-x} ?
The integral of 8*2^{-x} is -(2^{-x+3})/(ln(2))+C