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The pump produces 18 gallons per minute during the first 5 hours of its operation. The output of the pump is calculated in an easy manner by multiplying the rate of production by the time. 5 hours is equivalent to 5*60=300 minutes. As 18 gallons are produced per minute, the output in 300 minutes is 18*300 = 5400 gallons.
As an integral the output is `O = int_0^300 f(x) dx` where `(dO)/dx = f(x)` = 18 gallons per minute.
O = `int_0^300 18 dx`
= `18*(300 - 0) `
The output of the pump in 5 hours is 5400 gallons.
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