What is the integral of-4e^{-2x} ?

The integral of-4e^{-2x} is 2e^{-2x}+C

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integral of-4e^{-2x}

\int - 4 e^{- 2 x} d x

2 e^{- 2 x} + C

Solution steps

\int - 4 e^{- 2 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^{- 2 x} d x

Apply u-substitution

: \int - \frac{1}{2} e^{u} d u

= - 4 \cdot \int - \frac{1}{2} e^{u} d u

Take the constant out:

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

= - 4 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= - 4 (- \frac{1}{2} e^{u})

Substitute back

u = - 2 x

= - 4 (- \frac{1}{2} e^{- 2 x})

Simplify

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

= 2 e^{- 2 x}

Add a constant to the solution

= 2 e^{- 2 x} + C

\int_{0}^{1} 3 x^{2} + 1 d x

2 + 7 t

y^{^{'}} + 2 x^{5} y = \frac{x ^{5}}{y ^{2}}

\frac{10 x}{- 3 y ^{2}}

x \to 0 + lim (\frac{ln ( x )}{x ^{3}})