What is the integral of 9e^{-3x} ?

The integral of 9e^{-3x} is -3e^{-3x}+C

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integral of 9e^{-3x}

\int 9 e^{- 3 x} d x

- 3 e^{- 3 x} + C

Solution steps

\int 9 e^{- 3 x} d x

Take the constant out:

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

= 9 \cdot \int e^{- 3 x} d x

Apply u-substitution

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

= 9 \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

= 9 (- \frac{1}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= 9 (- \frac{1}{3} e^{u})

Substitute back

u = - 3 x

= 9 (- \frac{1}{3} e^{- 3 x})

Simplify

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

= - 3 e^{- 3 x}

Add a constant to the solution

= - 3 e^{- 3 x} + C

\frac{1}{1 - x ^{2}}

y^{^{''}} = - y + 0.1 y^{^{'}}

x \to 0 lim (81 x^{4} - 216 x^{3} + 216 x^{2} - 96 x + 16)

y^{^{'}} - y = 11 t e^{2 t}

n \to \infty lim (\frac{1}{4 n})