What is the integral from 0 to 2 of e^{-3x} ?

The integral from 0 to 2 of e^{-3x} is (e^6-1)/(3e^6)

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integral from 0 to 2 of e^{-3x}

\int_{0}^{2} e^{- 3 x} d x

\frac{e ^{6} - 1}{3 e ^{6}}

Decimal Notation

0.33250 \dots

Solution steps

\int_{0}^{2} e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 6} - \frac{1}{3} e^{u} d u

= \int_{0}^{- 6} - \frac{1}{3} e^{u} d u

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

= - \int_{- 6}^{0} - \frac{1}{3} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{- 6}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{3} [e^{u}]_{- 6}^{0})

Simplify

= \frac{1}{3} [e^{u}]_{- 6}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{6}}

= \frac{1}{3} (1 - \frac{1}{e ^{6}})

Simplify

= \frac{e ^{6} - 1}{3 e ^{6}}

y^{^{''}} + y^{^{'}} - 6 y = 6 t - 7 - (20 t e^{2 t} + 14 e^{2 t})

f (x) = x^{6}, at x = 2

\frac{d y}{d x} = 1 + x y

\frac{d}{d x} (\frac{2}{x} + 1)

x^{^{'}} = 5 (1 - \frac{x}{10})

What is the integral from 0 to 2 of e^{-3x} ?
The integral from 0 to 2 of e^{-3x} is (e^6-1)/(3e^6)