What is the integral of 5e^{-2x+1} ?

The integral of 5e^{-2x+1} is -5/2 e^{-2x+1}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5e^{-2x+1}

\int 5 e^{- 2 x + 1} d x

- \frac{5}{2} e^{- 2 x + 1} + C

Solution steps

\int 5 e^{- 2 x + 1} d x

Take the constant out:

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

= 5 \cdot \int e^{- 2 x + 1} d x

Apply u-substitution

: \int - \frac{1}{2} e^{u} d u

= 5 \cdot \int - \frac{1}{2} e^{u} d u

Take the constant out:

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

= 5 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= 5 (- \frac{1}{2} e^{u})

Substitute back

u = - 2 x + 1

= 5 (- \frac{1}{2} e^{- 2 x + 1})

Simplify

5 (- \frac{1}{2} e^{- 2 x + 1}) : - \frac{5}{2} e^{- 2 x + 1}

= - \frac{5}{2} e^{- 2 x + 1}

Add a constant to the solution

= - \frac{5}{2} e^{- 2 x + 1} + C

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

\int a t e^{- k t} d t

\int 11 x x^{2} + 9 d x

\int \frac{e ^{x} - 5 x ^{2}}{7} d x

\int \frac{x}{x ^{2} + x} d x

What is the integral of 5e^{-2x+1} ?
The integral of 5e^{-2x+1} is -5/2 e^{-2x+1}+C