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

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

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

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

- e^{- x} + 1

Solution steps

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

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

\frac{d}{d x} ((x - x^{3})^{x})

x \to - \infty lim (- e^{- x})

\int 12 sin^{2} (t) cos (t) d t

\int (12 x + 5)^{2} d x

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