What is the integral of 2e^{-3t} ?

The integral of 2e^{-3t} is -2/3 e^{-3t}+C

Popular >

integral of 2e^{-3t}

\int 2 e^{- 3 t} d t

- \frac{2}{3} e^{- 3 t} + C

Solution steps

\int 2 e^{- 3 t} d t

Take the constant out:

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

= 2 \cdot \int e^{- 3 t} d t

Apply u-substitution

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

= 2 \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

= 2 (- \frac{1}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= 2 (- \frac{1}{3} e^{u})

Substitute back

u = - 3 t

= 2 (- \frac{1}{3} e^{- 3 t})

Simplify

2 (- \frac{1}{3} e^{- 3 t}) : - \frac{2}{3} e^{- 3 t}

= - \frac{2}{3} e^{- 3 t}

Add a constant to the solution

= - \frac{2}{3} e^{- 3 t} + C

4 x + 3

(- 4, 2), (6, - 3)

\frac{d}{d x} (\frac{1}{1 + e ^{4 x}})

x \to 0 + lim (\frac{x + ∣ x ∣}{x})

x \to 4 lim (x^{3} + 6 x^{2} - 5 x - 1)