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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= - \frac{1}{k} \cdot \int_{0}^{- k t} e^{u} d u

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- k t} - 1

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

x \to \frac{π}{2} lim (\frac{sec ( x )}{sec ( 3 x )})

\int (x^{2} - 1)^{3} (2 x) d x

f (x) = 3 + 4 x^{2} - 2 x^{3}, at x = a

(y + 4) y^{^{'}} = 2 x y + 4 y, y (2) = 1

\int (1 + 30 w) w d w

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