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# Water flows from the bottom of a storage tank at a rate of $r(t) = 200 - 4t$ liters per minute, where $0 \le t \le 50$. Find the amount of water that flows from the tank during the first 10 minutes.

## 1800 liters

Integrals

Integration

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

So since we're trying to find the total amount of water flowing from the tank, we can just think of this as the area underneath this curve of the rate equation were given. Um So we can write integral. So time starts to zero. So a lot of ground zero. We're looking for the water Flowing during the 1st 10 minutes. So our upper bound will be 10. So it's plug in our equation. Find the anti derivative Which will be 200 times T -4 times t squared over to. We're going to simplify this to two times t squared profound of 10. Lower bound of zero, Plugging in 10, we get 200 times 10. Last two times 10 squared. Not plugging in zero, 200 times zero -2 times zero squared. This also evaluation zero. So we can cancel that. 200 times 10 is 2000 -2 times 10 squared, Which is just 200, 2000 -200. And we have 1800 L flowing from the tank in the 1st 10 minutes.

#### Topics

Integrals

Integration

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp