What is the integral from 0 to 5 of xe^{-0.6x} ?

The integral from 0 to 5 of xe^{-0.6x} is 2.22458…

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integral from 0 to 5 of xe^{-0.6x}

\int_{0}^{5} x e^{- 0.6 x} d x

2.22458 \dots

Solution steps

\int_{0}^{5} x e^{- 0.6 x} d x

Apply u-substitution

: \int_{0}^{- 3} \frac{5 e ^{u} u}{1.8} d u

= \int_{0}^{- 3} \frac{5 e ^{u} u}{1.8} d u

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

= - \int_{- 3}^{0} \frac{5 e ^{u} u}{1.8} d u

Take the constant out:

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

= - \frac{5}{1.8} \cdot \int_{- 3}^{0} e^{u} u d u

Simplify

= - 2.77777 \dots \cdot \int_{- 3}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 3}^{0}

= - 2.77777 \dots [e^{u} u - \int e^{u} d u]_{- 3}^{0}

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

= - 2.77777 \dots [e^{u} u - e^{u}]_{- 3}^{0}

Compute the boundaries:

- 1 + \frac{4}{e ^{3}}

= - 2.77777 \dots (- 1 + \frac{4}{e ^{3}})

Simplify

= 2.22458 \dots

\frac{d}{d x} (cot (x))

y = \frac{e ^{x}}{x + e ^{x}}

\frac{\partial}{\partial x} (e^{x y} x^{2})

\frac{d}{d x} (x^{3} (x^{2} + 4 + \frac{1}{x}))

\frac{d}{d x} (x^{2} + 2 y^{2} + z^{2} - 6 x + 3 y - 2 z + 5)

What is the integral from 0 to 5 of xe^{-0.6x} ?
The integral from 0 to 5 of xe^{-0.6x} is 2.22458…