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

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

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integral of-9e^{-x}

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

9 e^{- x} + C

Solution steps

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

Apply u-substitution

: \int - e^{u} d u

= - 9 \cdot \int - e^{u} d u

Take the constant out:

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

= - 9 (- \int e^{u} d u)

Use the common integral:

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

= - 9 (- e^{u})

Substitute back

u = - x

= - 9 (- e^{- x})

Simplify

= 9 e^{- x}

Add a constant to the solution

= 9 e^{- x} + C

\frac{d}{d x} (\frac{25}{x ^{2}} - \frac{5}{x})

y^{^{''}} - 2 y^{^{'}} + y = e^{3 x}

\frac{d}{d x} (r^{2} - 19 x^{2})

x \to 3 lim (x^{2} + 3 x)

x \to 2 lim ((x^{2} - 3) e^{x})