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

The integral from 0 to 3 of xe^{-4x} is -(-e^{12}+13)/(16e^{12)}

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

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

- \frac{- e ^{12} + 13}{16 e ^{12}}

Decimal

0.06249 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{- 12} \frac{e ^{u} u}{16} d u

= \int_{0}^{- 12} \frac{e ^{u} u}{16} 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}{16} d u

Take the constant out:

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

= - \frac{1}{16} \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}{16} [e^{u} u - \int e^{u} d u]_{- 12}^{0}

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

= - \frac{1}{16} [e^{u} u - e^{u}]_{- 12}^{0}

Compute the boundaries:

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

= - \frac{1}{16} (- 1 + \frac{13}{e ^{12}})

Simplify

= - \frac{- e ^{12} + 13}{16 e ^{12}}

y = 5 e^{x} + 7 ln (x)

\frac{\partial}{\partial x} (1 + x^{3} + x^{4})

\frac{d}{d x} (\frac{7}{x ^{2} - x + 2})

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{t} - t^{2} - 2 t + 3

\int 20 - 0.1

What is the integral from 0 to 3 of xe^{-4x} ?
The integral from 0 to 3 of xe^{-4x} is -(-e^{12}+13)/(16e^{12)}