What is the integral of 2e^{-5x} ?

The integral of 2e^{-5x} is -2/5 e^{-5x}+C

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integral of 2e^{-5x}

\int 2 e^{- 5 x} d x

- \frac{2}{5} e^{- 5 x} + C

Solution steps

\int 2 e^{- 5 x} d x

Take the constant out:

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

= 2 \cdot \int e^{- 5 x} d x

Apply u-substitution

: \int - \frac{1}{5} e^{u} d u

= 2 \cdot \int - \frac{1}{5} 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}{5} \cdot \int e^{u} d u)

Use the common integral:

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

= 2 (- \frac{1}{5} e^{u})

Substitute back

u = - 5 x

= 2 (- \frac{1}{5} e^{- 5 x})

Simplify

2 (- \frac{1}{5} e^{- 5 x}) : - \frac{2}{5} e^{- 5 x}

= - \frac{2}{5} e^{- 5 x}

Add a constant to the solution

= - \frac{2}{5} e^{- 5 x} + C

- 2 x^{\frac{2}{3}} - 4

\int \frac{5 x ^{2}}{x + 7} d x

x \to 3 - lim (\frac{7}{x - 3})

x \to 0 lim (lo g_{e} (x^{2}))

x \to 1 lim (x - 1 + 4)