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

The integral of e^{-3t} is -1/3 e^{-3t}+C

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integral of e^{-3t}

\int e^{- 3 t} d t

- \frac{1}{3} e^{- 3 t} + C

Solution steps

\int e^{- 3 t} d t

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = - 3 t

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

Add a constant to the solution

= - \frac{1}{3} e^{- 3 t} + C

\frac{\partial}{\partial x} (z \cdot e^{x yz})

y^{^{'}} = 6 x y^{2}

\frac{d}{d x} (x^{3} (x - 5)^{2})

5 y y^{^{'}} = x

h \to \infty lim (\frac{5 h ^{4} - 2 h ^{2} + 3}{3 h ^{3} + 2 h ^{2} + h})