What is the integral of 3e^{-0.1x} ?

The integral of 3e^{-0.1x} is -30e^{-0.1x}+C

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integral of 3e^{-0.1x}

\int 3 e^{- 0.1 x} d x

- 30 e^{- 0.1 x} + C

Solution steps

\int 3 e^{- 0.1 x} d x

Take the constant out:

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

= 3 \cdot \int e^{- 0.1 x} d x

Apply u-substitution

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

= 3 \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

= 3 (- \frac{1}{0.1} \cdot \int e^{u} d u)

Simplify

= 3 (- 10 \cdot \int e^{u} d u)

Use the common integral:

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

= 3 (- 10 e^{u})

Substitute back

u = - 0.1 x

= 3 (- 10 e^{- 0.1 x})

Simplify

= - 30 e^{- 0.1 x}

Add a constant to the solution

= - 30 e^{- 0.1 x} + C

x^{2} y d x + y (x^{3} + e^{3 y}) d y = 0

x \to 2 lim (\frac{( 7 x ^{5} - 10 x ^{4} - 13 x + 6 )}{( 3 x ^{2} - 6 x - 8 )})

y = \frac{5}{( 2 x ) ^{3}}

\int 1 - \frac{8}{x ^{3}} d x

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

What is the integral of 3e^{-0.1x} ?
The integral of 3e^{-0.1x} is -30e^{-0.1x}+C