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

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

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

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

- e^{- y} + 1

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{- y} - 1

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

Simplify

= - e^{- y} + 1

\int (x + 3) ln (x) d x

\frac{d}{d x} (x^{4} y^{3})

\int sin (π) t d t

x^{2} y \cdot \frac{d y}{d x} - 9 = 0

x^{^{''}} - x^{^{'}} + x = 0

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