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

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

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

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

- e^{- t} + 1

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{- t} - 1

= - (e^{- t} - 1)

Simplify

= - e^{- t} + 1

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