What is the integral from 0 to 10 of 1200te^{-0.5t} ?

The integral from 0 to 10 of 1200te^{-0.5t} is -4800(-1+6/(e^5))

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integral from 0 to 10 of 1200te^{-0.5t}

\int_{0}^{10} 1200 t e^{- 0.5 t} d t

- 4800 (- 1 + \frac{6}{e ^{5}})

Decimal

4605.94712 \dots

Solution steps

\int_{0}^{10} 1200 t e^{- 0.5 t} d t

Take the constant out:

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

= 1200 \cdot \int_{0}^{10} t e^{- 0.5 t} d t

Apply u-substitution

: \int_{0}^{- 5} \frac{2}{0.5} e^{u} u d u

= 1200 \cdot \int_{0}^{- 5} \frac{2}{0.5} e^{u} u d u

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

= 1200 (- \int_{- 5}^{0} \frac{2}{0.5} e^{u} u d u)

Take the constant out:

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

= 1200 (- \frac{2}{0.5} \cdot \int_{- 5}^{0} e^{u} u d u)

Simplify

= 1200 (- 4 \cdot \int_{- 5}^{0} e^{u} u d u)

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 5}^{0}

= 1200 (- 4 [e^{u} u - \int e^{u} d u]_{- 5}^{0})

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

= 1200 (- 4 [e^{u} u - e^{u}]_{- 5}^{0})

Simplify

= - 4800 [e^{u} u - e^{u}]_{- 5}^{0}

Compute the boundaries:

- 1 + \frac{6}{e ^{5}}

= - 4800 (- 1 + \frac{6}{e ^{5}})

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What is the integral from 0 to 10 of 1200te^{-0.5t} ?
The integral from 0 to 10 of 1200te^{-0.5t} is -4800(-1+6/(e^5))