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

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

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

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

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

Decimal Notation

0.86466 \dots

Solution steps

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

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}}

x \to \infty lim (8 + \frac{5}{x})

\frac{d}{d x} (- \frac{5}{( x - 2 ) ^{2}})

\frac{\partial}{\partial y} (\frac{y}{y - 2 x})

\frac{d v}{d t} = 9.8 - \frac{v}{5}, v (0) = 0

- 2 7 s^{2} + 8

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