What is the integral of 6e^{-0.4x} ?

The integral of 6e^{-0.4x} is -15e^{-0.4x}+C

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integral of 6e^{-0.4x}

\int 6 e^{- 0.4 x} d x

- 15 e^{- 0.4 x} + C

Solution steps

\int 6 e^{- 0.4 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^{- 0.4 x} d x

Apply u-substitution

: \int - \frac{1}{0.4} e^{u} d u

= 6 \cdot \int - \frac{1}{0.4} 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}{0.4} \cdot \int e^{u} d u)

Simplify

= 6 (- 2.5 \cdot \int e^{u} d u)

Use the common integral:

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

= 6 (- 2.5 e^{u})

Substitute back

u = - 0.4 x

= 6 (- 2.5 e^{- 0.4 x})

Simplify

= - 15 e^{- 0.4 x}

Add a constant to the solution

= - 15 e^{- 0.4 x} + C

\int \frac{6}{2} d x

\frac{d}{d x} (x^{2} \cdot sin^{2} (x))

f (x) = \frac{36 x}{x ^{2} + 36}, (- 3, - \frac{12}{5})

\frac{d}{d x} ((sin (x) + cos (x)) sec (x) d x)

\frac{d}{d x} (\frac{6 x}{x ^{2} + 2})

What is the integral of 6e^{-0.4x} ?
The integral of 6e^{-0.4x} is -15e^{-0.4x}+C