What is the integral from 0 to 4 of xe^{-3x} ?

The integral from 0 to 4 of xe^{-3x} is -(-e^{12}+13)/(9e^{12)}

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integral from 0 to 4 of xe^{-3x}

\int_{0}^{4} x e^{- 3 x} d x

- \frac{- e ^{12} + 13}{9 e ^{12}}

Decimal Notation

0.11110 \dots

Solution steps

\int_{0}^{4} x e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 12} \frac{e ^{u} u}{9} d u

= \int_{0}^{- 12} \frac{e ^{u} u}{9} d u

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

= - \int_{- 12}^{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_{- 12}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 12}^{0}

= - \frac{1}{9} [e^{u} u - \int e^{u} d u]_{- 12}^{0}

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

= - \frac{1}{9} [e^{u} u - e^{u}]_{- 12}^{0}

Compute the boundaries:

- 1 + \frac{13}{e ^{12}}

= - \frac{1}{9} (- 1 + \frac{13}{e ^{12}})

Simplify

= - \frac{- e ^{12} + 13}{9 e ^{12}}

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

x^{- \frac{3}{5}} + x^{\frac{3}{5}}

\frac{\partial}{\partial x} (v sin (x))

\int_{\frac{1}{2}}^{7} 16 x ln (2 x) d x

f (x) = 2 + x

What is the integral from 0 to 4 of xe^{-3x} ?
The integral from 0 to 4 of xe^{-3x} is -(-e^{12}+13)/(9e^{12)}