What is the integral of e^{3y} ?

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

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integral of e^{3y}

\int e^{3 y} d y

\frac{1}{3} e^{3 y} + C

Solution steps

\int e^{3 y} d y

Apply u-substitution

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

= \int e^{u} \frac{1}{3} 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 y

= \frac{1}{3} e^{3 y}

Add a constant to the solution

= \frac{1}{3} e^{3 y} + C

e^{- 3 t} sin (t)

\frac{d y}{d x} = (\frac{- 2 y}{( x - 2 )})

x \to 0 + lim (- 1)

\int 6 (x + 2) d x

x \to 0 lim (\frac{x}{x ^{2}})