What is the integral from 0 to t of e^{-at} ?

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

Popular >

integral from 0 to t of e^{-at}

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- a t} - 1

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

\int_{0}^{x} e^{x^{3}} d x

\int_{0}^{4} x + x d x

\int_{1}^{2} \frac{6 x + 5}{x ^{2} + x} d x

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

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

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