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

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

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

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

- e^{- x} + 1

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{- x} - 1

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

Simplify

= - e^{- x} + 1

f (x) = 4 x - 4 e^{x}

x^{2} y^{^{''}} + x y^{^{'}} - y = 1

\int e^{\frac{- x}{2}} cos (3 x) d x

\frac{d}{d x} (\frac{1 - x}{2})

\int 4 x (1 + x) d x

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