What is the integral from 0 to t of e^{-4x} ?

The integral from 0 to t of e^{-4x} is -1/4 (e^{-4t}-1)

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integral from 0 to t of e^{-4x}

\int_{0}^{t} e^{- 4 x} d x

- \frac{1}{4} (e^{- 4 t} - 1)

Solution steps

\int_{0}^{t} e^{- 4 x} d x

Apply u-substitution

: \int_{0}^{- 4 t} - \frac{1}{4} e^{u} d u

= \int_{0}^{- 4 t} - \frac{1}{4} e^{u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{0}^{- 4 t} e^{u} d u

Use the common integral:

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

= - \frac{1}{4} [e^{u}]_{0}^{- 4 t}

Compute the boundaries:

e^{- 4 t} - 1

= - \frac{1}{4} (e^{- 4 t} - 1)

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What is the integral from 0 to t of e^{-4x} ?
The integral from 0 to t of e^{-4x} is -1/4 (e^{-4t}-1)