What is the integral of 10e^{1-5x} ?

The integral of 10e^{1-5x} is -2e^{1-5x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 10e^{1-5x}

\int 10 e^{1 - 5 x} d x

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

Solution steps

\int 10 e^{1 - 5 x} d x

Take the constant out:

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

= 10 \cdot \int e^{1 - 5 x} d x

Apply u-substitution

: \int - \frac{1}{5} e^{u} d u

= 10 \cdot \int - \frac{1}{5} e^{u} d u

Take the constant out:

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

= 10 (- \frac{1}{5} \cdot \int e^{u} d u)

Use the common integral:

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

= 10 (- \frac{1}{5} e^{u})

Substitute back

u = 1 - 5 x

= 10 (- \frac{1}{5} e^{1 - 5 x})

Simplify

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

= - 2 e^{1 - 5 x}

Add a constant to the solution

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

\frac{\partial}{\partial y} (3 x sin (y))

\int 7 x^{\frac{2}{5}} + 2 x^{- \frac{4}{5}} d x

\frac{d}{d x} (tan (x^{2} + y))

\frac{d}{d x} (3 arcsin (x))

\int \frac{1}{( x ^{2} - 81 ) ^{\frac{3}{2}}} d x