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An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of r(t) = 110e−0.05t liters per minute. How much oil leaks out during the first hour? (Round your answer to the nearest liter.)

Question

An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of r(t) = 110e−0.05t liters per minute. How much oil leaks out during the first hour? (Round your answer to the nearest liter.)

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Solution

To solve this problem, we need to integrate the rate function from 0 to 60 (since we're looking for the amount of oil leaked in the first hour and there are 60 minutes in an hour).

The integral of r(t) from 0 to 60 gives us the total amount of oil leaked during this time period.

The integral ∫ from 0 to 60 of 110e^(-0.05t) dt represents the total amount of oil leaked.

To solve this integral, we can use the formula for the integral of e^(at) which is (1/a)e^(at). Here, a is -0.05.

So, the antiderivative of our function is - (110/0.05) e^(-0.05t) = -2200e^(-0.05t).

We evaluate this from 0 to 60:

-2200[e^(-0.0560) - e^(-0.050)] = -2200[e^(-3) - 1] = -2200[-0.0498 - 1] = -2200[-1.0498] = 2315.6 liters

So, approximately 2316 liters of oil leak out during the first hour.

This problem has been solved

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