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

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

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integral of 2e^{-3x}

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

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

Solution steps

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

Take the constant out:

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

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

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 x

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (cos^{\frac{1}{x}} (x))

sin^{2} (y) d x + cos^{2} (x) d y = 0

x \to - 5 lim (2 x + 3)

\frac{1 + e ^{- π s}}{s ^{2} + 1}

f (x) = 2 sec (x), at x = \frac{π}{3}