What is the integral from 0 to 10 of te^{-0.6t} ?

The integral from 0 to 10 of te^{-0.6t} is 2.72957…

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integral from 0 to 10 of te^{-0.6t}

\int_{0}^{10} t e^{- 0.6 t} d t

2.72957 \dots

Solution steps

\int_{0}^{10} t e^{- 0.6 t} d t

Apply u-substitution

: \int_{0}^{- 6} \frac{5 e ^{u} u}{1.8} d u

= \int_{0}^{- 6} \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_{- 6}^{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_{- 6}^{0} e^{u} u d u

Simplify

= - 2.77777 \dots \cdot \int_{- 6}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 6}^{0}

= - 2.77777 \dots [e^{u} u - \int e^{u} d u]_{- 6}^{0}

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

= - 2.77777 \dots [e^{u} u - e^{u}]_{- 6}^{0}

Compute the boundaries:

- 1 + \frac{7}{e ^{6}}

= - 2.77777 \dots (- 1 + \frac{7}{e ^{6}})

Simplify

= 2.72957 \dots

x^{^{'}} = r x - 12

6 y^{^{''}} + 9 y^{^{'}} - y = 23

\frac{d}{d x} (y cos (x))

\frac{1}{1 - s}

y = \frac{cos ( x )}{sin ( x )}

What is the integral from 0 to 10 of te^{-0.6t} ?
The integral from 0 to 10 of te^{-0.6t} is 2.72957…