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

The integral from 0 to 2 of xe^{-3x} is -(-e^6+7)/(9e^6)

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

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

- \frac{- e ^{6} + 7}{9 e ^{6}}

Decimal

0.10918 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{- 6} \frac{e ^{u} u}{9} d u

= \int_{0}^{- 6} \frac{e ^{u} u}{9} d u

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

= - \int_{- 6}^{0} \frac{e ^{u} u}{9} d u

Take the constant out:

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

= - \frac{1}{9} \cdot \int_{- 6}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 6}^{0}

= - \frac{1}{9} [e^{u} u - \int e^{u} d u]_{- 6}^{0}

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

= - \frac{1}{9} [e^{u} u - e^{u}]_{- 6}^{0}

Compute the boundaries:

- 1 + \frac{7}{e ^{6}}

= - \frac{1}{9} (- 1 + \frac{7}{e ^{6}})

Simplify

= - \frac{- e ^{6} + 7}{9 e ^{6}}

\int \frac{1}{1 - x ^{2}} \cdot ln (\frac{1 + x}{1 - x}) d x

f (x) = 5 x^{2}, at x = 1

\frac{d y}{d x} = 4 (y^{2} + 1)

\frac{1}{4 z - 5}

y = (x^{2} - 35)^{6}, at x = 6

What is the integral from 0 to 2 of xe^{-3x} ?
The integral from 0 to 2 of xe^{-3x} is -(-e^6+7)/(9e^6)