What is the integral from 0 to 2 of e^{-x} ?

The integral from 0 to 2 of e^{-x} is 1-1/(e^2)

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integral from 0 to 2 of e^{-x}

\int_{0}^{2} e^{- x} d x

1 - \frac{1}{e ^{2}}

Decimal Notation

0.86466 \dots

Solution steps

\int_{0}^{2} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 2} - e^{u} d u

= \int_{0}^{- 2} - e^{u} d u

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

= - \int_{- 2}^{0} - e^{u} d u

Take the constant out:

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

= - (- \int_{- 2}^{0} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{- 2}^{0})

Simplify

= [e^{u}]_{- 2}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{2}}

= 1 - \frac{1}{e ^{2}}

\int 3 + x^{2} d x

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\frac{d}{d x} (arcsin (5 x))

What is the integral from 0 to 2 of e^{-x} ?
The integral from 0 to 2 of e^{-x} is 1-1/(e^2)