Question

Oil is leaking from a tanker at the rate of R(t) = 2000e^(-0.2t) gallons per hour where t is measured in hours. How much oil leaks out of the tanker from the time t = 0 to t = 10

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Answer 1

To find the amount of oil that leaks out of the tanker from t = 0 to t = 10, integrate the rate function R(t) = 2000e^(-0.2t) with respect to t over the interval [0, 10].

Step-by-step explanation:

To find the amount of oil that leaks out of the tanker from t = 0 to t = 10, we need to integrate the rate function R(t) with respect to t over the interval [0, 10].

The integral of R(t) = 2000e^(-0.2t) with respect to t is:

∫R(t) dt = -10000e^(-0.2t) + C

Now, evaluate the definite integral from 0 to 10:

∫(0 to 10) R(t) dt = [-10000e^(-0.2t)]|010 = [-10000e^(-0.2*10)] - [-10000e^0] = -10000e^(-2) + 10000

So, the amount of oil that leaks out of the tanker from t = 0 to t = 10 is approximately -10000e^(-2) + 10000 gallons.

Answer 2

The calculation of the integral will give us the answer.

Amount = Int (2000e^(-0,2t))dt

With t = 0 to t = 10

A = 2000.In(e^(-0,2t))dt


Making U = -0,2t

U = -0,2t

dU/dt = d(-0,2t)/dt

dU/dt = -0,2

dU = -0,2dt

dt = dU/-0,2
_______________


Then,

A = 2000.Int (e^U).dU/-0,2

A = -10.000.Int(e^U)dU

A = -10.000.e^U

A = -10.000.e^(-0,2t) | (0,10)

A = -10.000.(e^(-2)) - [ -10.000.e^(0)]

A = -10.000.e^(-2) + 10.000

A ~ 8,646 Gallons