What is the integral from 0 to a of e^{-ax} ?

The integral from 0 to a of e^{-ax} is -(e^{-a^2}-1)/a

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integral from 0 to a of e^{-ax}

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

- \frac{e ^{- a^{2}} - 1}{a}

Solution steps

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

Apply u-substitution

: \int_{0}^{- a^{2}} - \frac{e ^{u}}{a} d u

= \int_{0}^{- a^{2}} - \frac{e ^{u}}{a} d u

Take the constant out:

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

= - \frac{1}{a} \cdot \int_{0}^{- a^{2}} e^{u} d u

Use the common integral:

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

= - \frac{1}{a} [e^{u}]_{0}^{- a^{2}}

Simplify

- \frac{1}{a} [e^{u}]_{0}^{- a^{2}} : - \frac{[ e ^{u} ] _{0}^{- a^{2}}}{a}

= - \frac{[ e ^{u} ] _{0}^{- a^{2}}}{a}

Compute the boundaries:

e^{- a^{2}} - 1

= - \frac{e ^{- a^{2}} - 1}{a}

\int_{1}^{2} \frac{x}{x + y} d y

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

\int_{- 2}^{2} (1 - 2 x)^{2} d x

\int_{0}^{1} (3 x^{2} + 4 x - \frac{1}{2}) d x

\int_{- 1}^{1} - x^{2} + 9 d x

What is the integral from 0 to a of e^{-ax} ?
The integral from 0 to a of e^{-ax} is -(e^{-a^2}-1)/a