What is the integral of (6e^{-0.1x}) ?

The integral of (6e^{-0.1x}) is -60e^{-0.1x}+C

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integral of (6e^{-0.1x})

\int (6 e^{- 0.1 x}) d x

- 60 e^{- 0.1 x} + C

Solution steps

\int 6 e^{- 0.1 x} d x

Take the constant out:

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

= 6 \cdot \int e^{- 0.1 x} d x

Apply u-substitution

: \int - \frac{1}{0.1} e^{u} d u

= 6 \cdot \int - \frac{1}{0.1} e^{u} d u

Take the constant out:

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

= 6 (- \frac{1}{0.1} \cdot \int e^{u} d u)

Simplify

= 6 (- 10 \cdot \int e^{u} d u)

Use the common integral:

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

= 6 (- 10 e^{u})

Substitute back

u = - 0.1 x

= 6 (- 10 e^{- 0.1 x})

Simplify

= - 60 e^{- 0.1 x}

Add a constant to the solution

= - 60 e^{- 0.1 x} + C

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

x \to - 4 lim (\frac{x ^{2} - 6}{4 - x})

\frac{d}{d x} ((x^{3} + 3 x) (2 x + 4))

\int y y d y

\frac{d y}{d x} = x (5 - y)

What is the integral of (6e^{-0.1x}) ?
The integral of (6e^{-0.1x}) is -60e^{-0.1x}+C