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

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

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integral of 2e^{-2t}

\int 2 e^{- 2 t} d t

- e^{- 2 t} + C

Solution steps

\int 2 e^{- 2 t} d t

Take the constant out:

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

= 2 \cdot \int e^{- 2 t} d t

Apply u-substitution

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

= 2 \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

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

Use the common integral:

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

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

Substitute back

u = - 2 t

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

Simplify

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

= - e^{- 2 t}

Add a constant to the solution

= - e^{- 2 t} + C

x^{5}

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

\frac{\partial}{\partial y} (5 y^{2} e^{5 x y})

\int sin (wt) d t

x y^{^{'}} = \frac{1}{y ^{3}}