What is the integral from 0 to 5 of xe^{-2x} ?

The integral from 0 to 5 of xe^{-2x} is -(-e^{10}+11)/(4e^{10)}

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integral from 0 to 5 of xe^{-2x}

\int_{0}^{5} x e^{- 2 x} d x

- \frac{- e ^{10} + 11}{4 e ^{10}}

Decimal Notation

0.24987 \dots

Solution steps

\int_{0}^{5} x e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 10} \frac{e ^{u} u}{4} d u

= \int_{0}^{- 10} \frac{e ^{u} u}{4} d u

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

= - \int_{- 10}^{0} \frac{e ^{u} u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{- 10}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 10}^{0}

= - \frac{1}{4} [e^{u} u - \int e^{u} d u]_{- 10}^{0}

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

= - \frac{1}{4} [e^{u} u - e^{u}]_{- 10}^{0}

Compute the boundaries:

- 1 + \frac{11}{e ^{10}}

= - \frac{1}{4} (- 1 + \frac{11}{e ^{10}})

Simplify

= - \frac{- e ^{10} + 11}{4 e ^{10}}

y^{^{'}} = e^{4 x} - 6 x

x \to 0 lim (\frac{- 6}{e ^{3} - x})

\frac{8 s ^{2} + 12 s + 7}{s ^{2} + s}

\frac{d}{d x} (ln ((2 x + 5)^{3}))

x \to \infty lim (∣ x - 5 ∣)

What is the integral from 0 to 5 of xe^{-2x} ?
The integral from 0 to 5 of xe^{-2x} is -(-e^{10}+11)/(4e^{10)}