What is the integral of 3e^{-0.4x} ?

The integral of 3e^{-0.4x} is -7.5e^{-0.4x}+C

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integral of 3e^{-0.4x}

\int 3 e^{- 0.4 x} d x

- 7.5 e^{- 0.4 x} + C

Solution steps

\int 3 e^{- 0.4 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^{- 0.4 x} d x

Apply u-substitution

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

= 3 \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

= 3 (- \frac{1}{0.4} \cdot \int e^{u} d u)

Simplify

= 3 (- 2.5 \cdot \int e^{u} d u)

Use the common integral:

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

= 3 (- 2.5 e^{u})

Substitute back

u = - 0.4 x

= 3 (- 2.5 e^{- 0.4 x})

Simplify

= - 7.5 e^{- 0.4 x}

Add a constant to the solution

= - 7.5 e^{- 0.4 x} + C

x^{2}, (0, 2)

\int x^{2} x^{3} + 5 d x

\frac{4}{s ^{2} ( s + 2 )} + \frac{1}{s}

7 lo g_{7} (x)

f (x) = e^{6 x}

What is the integral of 3e^{-0.4x} ?
The integral of 3e^{-0.4x} is -7.5e^{-0.4x}+C