What is the integral from 0 to x of ae^{-ax} ?

The integral from 0 to x of ae^{-ax} is -e^{-ax}+1

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integral from 0 to x of ae^{-ax}

\int_{0}^{x} a e^{- a x} d x

- e^{- a x} + 1

Solution steps

\int_{0}^{x} a e^{- a x} d x

Take the constant out:

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

= a \cdot \int_{0}^{x} e^{- a x} d x

Apply u-substitution

: \int_{0}^{- a x} - \frac{e ^{u}}{a} d u

= a \cdot \int_{0}^{- a x} - \frac{e ^{u}}{a} d u

Take the constant out:

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

= a (- \frac{1}{a} \cdot \int_{0}^{- a x} e^{u} d u)

Use the common integral:

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

= a (- \frac{1}{a} [e^{u}]_{0}^{- a x})

Simplify

a (- \frac{1}{a} [e^{u}]_{0}^{- a x}) : - [e^{u}]_{0}^{- a x}

= - [e^{u}]_{0}^{- a x}

Compute the boundaries:

e^{- a x} - 1

= - (e^{- a x} - 1)

Simplify

= - e^{- a x} + 1

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

2 x^{2} + 3, (- 2, 1)

\frac{d}{d x} (\frac{x ^{3}}{x})

x \to 0 + lim (sin (3 x) ln (x))

\int \frac{1}{2} u d u

What is the integral from 0 to x of ae^{-ax} ?
The integral from 0 to x of ae^{-ax} is -e^{-ax}+1