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

The integral from 0 to 1 of (e^{-x}) is 1-1/e

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integral from 0 to 1 of (e^{-x})

\int_{0}^{1} (e^{- x}) d x

1 - \frac{1}{e}

Decimal Notation

0.63212 \dots

Solution steps

\int_{0}^{1} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 1} - e^{u} d u

= \int_{0}^{- 1} - e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{0} - e^{u} d u

Take the constant out:

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

= - (- \int_{- 1}^{0} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- 1}^{0})

Simplify

= [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= 1 - \frac{1}{e}

f (x) = x^{2} (1 - 9 x)

\int sin (3 x) cos (5 x) d x

(3 x^{2} y - 6 x) d x + (x^{3} + 2 y) d y = 0

\int_{4}^{6} \frac{1}{x ^{2} - 1} d x

\frac{\partial}{\partial x} (t^{8} e^{- x})

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