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

The integral of-3e^{-3t} is e^{-3t}+C

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integral of-3e^{-3t}

\int - 3 e^{- 3 t} d t

e^{- 3 t} + C

Solution steps

\int - 3 e^{- 3 t} d t

Take the constant out:

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

= - 3 \cdot \int e^{- 3 t} d t

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

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

Use the common integral:

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

= - 3 (- \frac{1}{3} e^{u})

Substitute back

u = - 3 t

= - 3 (- \frac{1}{3} e^{- 3 t})

Simplify

- 3 (- \frac{1}{3} e^{- 3 t}) : e^{- 3 t}

= e^{- 3 t}

Add a constant to the solution

= e^{- 3 t} + C

\frac{\partial}{\partial x} (∣ x ∣)

\int_{- 8}^{10} (x + 10) d x

\frac{d}{d x} (2 π x)

\int \frac{x + 2}{x ^{2} + 9 x - 10} d x

y = 8 x^{4} - 8 x^{2}, y = 9 x^{2}