What is the integral from 0 to 2 of xe^{-4x} ?

The integral from 0 to 2 of xe^{-4x} is -(-e^8+9)/(16e^8)

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integral from 0 to 2 of xe^{-4x}

\int_{0}^{2} x e^{- 4 x} d x

- \frac{- e ^{8} + 9}{16 e ^{8}}

Decimal Notation

0.06231 \dots

Solution steps

\int_{0}^{2} x e^{- 4 x} d x

Apply u-substitution

: \int_{0}^{- 8} \frac{e ^{u} u}{16} d u

= \int_{0}^{- 8} \frac{e ^{u} u}{16} d u

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

= - \int_{- 8}^{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_{- 8}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 8}^{0}

= - \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 8}^{0}

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

= - \frac{1}{16} [e^{u} u - e^{u}]_{- 8}^{0}

Compute the boundaries:

- 1 + \frac{9}{e ^{8}}

= - \frac{1}{16} (- 1 + \frac{9}{e ^{8}})

Simplify

= - \frac{- e ^{8} + 9}{16 e ^{8}}

f (t) = t^{2} e^{t}

y^{^{'}} = (cos^{2} (t)) (cos^{2} (2 y))

(\frac{y ^{3}}{3}) + \frac{1}{4 y}

f (x) = \frac{x \cdot , ( x )}{1 + x}

x \to - 1 lim (x^{2} - 1)

What is the integral from 0 to 2 of xe^{-4x} ?
The integral from 0 to 2 of xe^{-4x} is -(-e^8+9)/(16e^8)