What is the integral from 0 to 7 of xe^{-4x} ?

The integral from 0 to 7 of xe^{-4x} is -(-e^{28}+29)/(16e^{28)}

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integral from 0 to 7 of xe^{-4x}

\int_{0}^{7} x e^{- 4 x} d x

- \frac{- e ^{28} + 29}{16 e ^{28}}

Decimal Notation

0.06249 \dots

Solution steps

\int_{0}^{7} x e^{- 4 x} d x

Apply u-substitution

: \int_{0}^{- 28} \frac{e ^{u} u}{16} d u

= \int_{0}^{- 28} \frac{e ^{u} u}{16} d u

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

= - \int_{- 28}^{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_{- 28}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 28}^{0}

= - \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 28}^{0}

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

= - \frac{1}{16} [e^{u} u - e^{u}]_{- 28}^{0}

Compute the boundaries:

- 1 + \frac{29}{e ^{28}}

= - \frac{1}{16} (- 1 + \frac{29}{e ^{28}})

Simplify

= - \frac{- e ^{28} + 29}{16 e ^{28}}

\frac{d}{d x} (a x^{3})

f (x) = (2 x + 1)^{\frac{1}{2}}, at x = 2

\int \frac{x ^{4}}{x ^{4} - 1} d x

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

g (t) = t^{2} t

What is the integral from 0 to 7 of xe^{-4x} ?
The integral from 0 to 7 of xe^{-4x} is -(-e^{28}+29)/(16e^{28)}