What is the integral of-4e^{-x} ?

The integral of-4e^{-x} is 4e^{-x}+C

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integral of-4e^{-x}

\int - 4 e^{- x} d x

4 e^{- x} + C

Solution steps

\int - 4 e^{- x} d x

Take the constant out:

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

= - 4 \cdot \int e^{- x} d x

Apply u-substitution

: \int - e^{u} d u

= - 4 \cdot \int - e^{u} d u

Take the constant out:

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

= - 4 (- \int e^{u} d u)

Use the common integral:

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

= - 4 (- e^{u})

Substitute back

u = - x

= - 4 (- e^{- x})

Simplify

= 4 e^{- x}

Add a constant to the solution

= 4 e^{- x} + C

(22 + x^{2}) y^{^{'}} - 2 x y = 0

\int \frac{x ^{2}}{( x ^{3} + 1 )} d x

\frac{d r}{d θ} = π sin (π θ), r (1) = 3

(- \frac{4}{3}) \cdot e^{- 2 t} + (- \frac{2}{3}) \cdot e^{t}

cos (t + 1)