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A bacteria population starts with 400 bacteria and grows at a rate of $ r(t) = (450.268)e^{1.12567t} $ bacteria per hour. How many bacteria will there be after three hours?

11713 bacteria

01:39

Frank L.

01:26

Amrita B.

Calculus 1 / AB

Chapter 5

Integrals

Section 5

The Substitution Rule

Integration

Oregon State University

University of Michigan - Ann Arbor

Boston College

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So if we want to figure out how many bacteria there are after three hours, since we're given the rate here, remember, if we integrate the rate, we get the net change over that time period. So just a to b our rate function. And then if we want the population at the end of this time, we could just take our net change editor initial. And that would be what it is after three hours. Um, so let's just go ahead and plug are written and then integrate. So this would be from 0 to 3 protection. Let me pull out that constant here. So 4 52 68 integral from 0 to 3 of E to the 1.1256 17 DT. And now to integrate this so it's still be e to the whatever. But then we just divide by the constant and the power. So it be or 52 68 divided by 1.12567 and then he to the 1.1256 17. Then we evaluate from 0 to 3. Um, so now we plug it three plug in zero, and that would give. So 1.12 567 times three. So that's he raised 3.37701 minus 80 which is just one. And then let's just go ahead and plug this into a calculator and see what we get. Eso on the inside The proceeds That's around like 28 thin times 4. 50.268 divided by 1.12567 Um and that is saying we have a change of around 11,313 on I'm just going to round down, um, for these because it wouldn't make sense for us to have, like, half of the bacteria floating. Um, yeah. So this is the net change. And so to figure out what is after three hours, what we can do is add our initial, which was equal Thio 400 to this and so that that will be pop factor three hours, which would be 11,713. So this is going to be our bacteria population after those three hours

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