What is the integral from 0 to 2 of 8000te^{-0.6t} ?

The integral from 0 to 2 of 8000te^{-0.6t} is 7497.17186…

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integral from 0 to 2 of 8000te^{-0.6t}

\int_{0}^{2} 8000 t e^{- 0.6 t} d t

7497.17186 \dots

Solution steps

\int_{0}^{2} 8000 t e^{- 0.6 t} d t

Take the constant out:

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

= 8000 \cdot \int_{0}^{2} t e^{- 0.6 t} d t

Apply u-substitution

: \int_{0}^{- 1.2} \frac{5 e ^{u} u}{1.8} d u

= 8000 \cdot \int_{0}^{- 1.2} \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

= 8000 (- \int_{- 1.2}^{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

= 8000 (- \frac{5}{1.8} \cdot \int_{- 1.2}^{0} e^{u} u d u)

Simplify

= 8000 (- 2.77777 \dots \cdot \int_{- 1.2}^{0} e^{u} u d u)

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 1.2}^{0}

= 8000 (- 2.77777 \dots [e^{u} u - \int e^{u} d u]_{- 1.2}^{0})

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

= 8000 (- 2.77777 \dots [e^{u} u - e^{u}]_{- 1.2}^{0})

Simplify

= - 22222.22222 \dots [e^{u} u - e^{u}]_{- 1.2}^{0}

Compute the boundaries:

- 0.33737 \dots

= - 22222.22222 \dots (- 0.33737 \dots)

Simplify

= 7497.17186 \dots

\int \frac{6 x ^{2} + 3 x + 10}{x ^{3} + 2 x ^{2} + 5 x} d x

x^{2} + 40

\frac{d}{d x} (\frac{x ^{5} - 2 x}{- 5})

7 u^{e}

e^{- x^{2}}

What is the integral from 0 to 2 of 8000te^{-0.6t} ?
The integral from 0 to 2 of 8000te^{-0.6t} is 7497.17186…