How can changing the units of a rate affect an accumulated quantity?

An integral accumulates a rate across an interval. Its units come from multiplying the rate units by the units of the integration variable. For instance, gallons per hour multiplied by hours gives gallons: the “per hour” and “hours” cancel. This is the unit relationship used when integrating a rate to find accumulated change. [S1]

The rate’s units and the time interval’s units need to agree. If a rate is in gallons per hour but the interval is entered in minutes, convert one of them first. Otherwise, the calculation combines mismatched units. The units are a useful check on whether the answer represents a rate or an accumulated quantity.

Assume, just for a math example, that a rate is described by \(r(t)=3+2t\) gallons per hour, where \(t\) is measured in hours. Find the accumulated amount from hour 0 to hour 2:

\[ \int_0^2 (3+2t)\,dt = [3t+t^2]_0^2 = 10. \]

The result is 10 gallons. Check each term’s units: 3 has units of gallons per hour, while the coefficient 2 has units of gallons per hour squared, so that \(2t\) is also gallons per hour when \(t\) is in hours. Multiplying the rate by \(dt\), measured in hours, gives gallons. This is a hypothetical calculation, not a claim about a real process.

Make the time units match before integrating. You can convert minutes to hours, or convert the rate to units per minute. For example, 30 minutes is 0.5 hours. If the hypothetical rate were constant at 4 gallons per hour, then over 30 minutes the accumulation would be \(4\text{ gallons/hour}\times0.5\text{ hours}=2\text{ gallons}\).

If you instead use minutes as the integration variable, rewrite the rate as 4/60 gallons per minute. Then multiplying by 30 minutes gives the same 2 gallons. Don’t multiply a per-hour rate directly by a number of minutes and label the result gallons: the time units have not canceled.

Write the units beside the rate, variable, interval, and final result. Then check that the rate’s denominator cancels with the integration variable’s unit. For a rate in quantity-units per time-unit, integrating with respect to that time produces quantity-units. [S1]

A unit check can reveal a mismatch, but it cannot by itself show that the rate formula or its assumptions correctly describe a real situation. The cited calculus explanation establishes the rate-times-time unit relationship; it does not establish design or operational claims. For a classroom calculation, state the assumed rate formula and interval so someone else can follow your work.