What is the integral from 0 to 5 of te^{-(5-t)/(20)} ?

The integral from 0 to 5 of te^{-(5-t)/(20)} is (400)/(e^{1/4)}-300

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integral from 0 to 5 of te^{-(5-t)/(20)}

\int_{0}^{5} t e^{- \frac{5 - t}{20}} d t

\frac{400}{e ^{\frac{1}{4}}} - 300

Decimal

11.52031 \dots

Solution steps

\int_{0}^{5} t e^{- \frac{5 - t}{20}} d t

Apply u-substitution

: \int_{5}^{0} - e^{- \frac{u}{20}} (- u + 5) d u

= \int_{5}^{0} - e^{- \frac{u}{20}} (- u + 5) d u

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

= - \int_{0}^{5} - e^{- \frac{u}{20}} (- u + 5) d u

Take the constant out:

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

= - (- \int_{0}^{5} e^{- \frac{u}{20}} (- u + 5) d u)

Apply Integration By Parts

: [- 20 e^{- \frac{u}{20}} (- u + 5) - \int 20 e^{- \frac{u}{20}} d u]_{0}^{5}

= - (- [- 20 e^{- \frac{u}{20}} (- u + 5) - \int 20 e^{- \frac{u}{20}} d u]_{0}^{5})

\int 20 e^{- \frac{u}{20}} d u = - 400 e^{- \frac{u}{20}}

= - (- [- 20 e^{- \frac{u}{20}} (- u + 5) - (- 400 e^{- \frac{u}{20}})]_{0}^{5})

Simplify

- (- [- 20 e^{- \frac{u}{20}} (- u + 5) - (- 400 e^{- \frac{u}{20}})]_{0}^{5}) : [20 e^{- \frac{u}{20}} u + 300 e^{- \frac{u}{20}}]_{0}^{5}

= [20 e^{- \frac{u}{20}} u + 300 e^{- \frac{u}{20}}]_{0}^{5}

Compute the boundaries:

\frac{400}{e ^{\frac{1}{4}}} - 300

= \frac{400}{e ^{\frac{1}{4}}} - 300

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What is the integral from 0 to 5 of te^{-(5-t)/(20)} ?
The integral from 0 to 5 of te^{-(5-t)/(20)} is (400)/(e^{1/4)}-300