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

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

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

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

- e^{- x} + 1

Solution steps

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

Apply u-substitution

: \int_{0}^{- x} - e^{v} d v

= \int_{0}^{- x} - e^{v} d v

Take the constant out:

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

= - \int_{0}^{- x} e^{v} d v

Use the common integral:

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

= - [e^{v}]_{0}^{- x}

Compute the boundaries:

e^{- x} - 1

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

Simplify

= - e^{- x} + 1

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