I'm having trouble articulating my difficulty, but here's my best shot. Thanks in advance for any light you can shed.

I understand that a definite integral is the limit of a Riemann sum, and can represent an area between a curve and the x-axis.

I have found a couple of problems in my first-year calculus textbook that have led me to believe that I do not have a full understanding of definite integrals. These are applications problems, and although I have full solutions to them, I don't understand why the solutions produce answers that make sense, particular in regards to the units of the answers.

These are from the Larson Calculus text, eighth edition.

This first one, I understand, but I need to quote it for context to frame what it is I don't understand. (I am not giving the actual functions because they aren't relevant to my misunderstanding.)

Water Supply: A model for the flow rate of water at a pumping station on a given day is: R(t). R is the flow rate in thousands of gallons per hour, and t is the time in hours. Approximate the total volume of water pumped in 1 day.

This is my understanding: R(t) is a rate in thousands of gallons per hour, and t is in hours. If I integrate R(t) over the interval [0, 24], then the answer I get from the integral, in addition to being the area under R(t) from 0 to 24, will represent a volume of water. This makes sense to me because if I were to graph R(t) and look at it as an area, the Riemann rectangles would have a width of hours and a height of thousands of gallons per hour and multiplying them together means the area will have units of gallons, because (gal/hr) times hours would leave gallons. So the integral will evaluate to a value representing gallons.

In general, integrating a rate of change should result in the net change over the interval, yes? This makes sense to me.

Now, the second problem is this:

Temperature: The temperature in degrees Fahrenheit in a house is T(t) where t is the time in hours, with t=0 representing midnight. The hourly cost of cooling a house is $0.10 per degree. Find the cost of cooling the house if its thermostat is set at 72 F. by evaluating the integral

$$C=0.1\displaystyle \int_{8}^{20}{T(t)-72} \ {\rm d}t$$

Now here, the function T is not a rate of change, it gives the actual temperature of the house at time t, and not the rate at which the temperature changes.

So I guess I have some conceptual misunderstanding of the definite integral because I can't reconcile the units or the meaning of what's going on here.

I can see that the 0.1 in front of the integral is there because in order to get the cost in $$ we need to multiply the $0.1 times the number of degrees, so it appears from the question that I am to interpret the integral itself as calculating a number of degrees. But I don't understand how it could.

That is, if the units of T(t) are degrees, and the time is hours, then my area understanding of the integral should give the area calculated by this integral in units of degree times hours, or degree-hours? Since the units of the 0.1 are dollars/degree, then multiplying dollars/degree by degree times hours results in dollars times hours, but it should come out in dollars alone, and I don't see how.

The integral would have to have units of degrees in order for the cost to come out in dollars. How can it? If the function T is not a rate, but an actual amount of something, like degrees, then what meaning do I give to integrating it over an interval? I will get an area, yes, but would that mean anything necessarily?

If I integrate velocity, which is a rate of change of position, then I get the total change of position, or distance. What if I have a position function, which is not a rate of change, and integrate that? Does the result have any meaning? That's what this temperature function seems to be doing. T(t) is not a rate of change function, it's like a position function. If I integrate it, why do I end up with total degrees?

  • 1
    $\begingroup$ (.1 dollars per degree per hour) $\times (T - 72)$ degrees gives you dollars per hour. $\endgroup$
    – littleO
    Jan 6, 2016 at 1:42
  • 1
    $\begingroup$ "The hourly cost ... is $0.10 per degree" $\endgroup$
    – user137731
    Jan 6, 2016 at 1:43
  • 1
    $\begingroup$ Disregarding units for a moment, you should note that your cost is also dependent on how much time your thermostat is set at 72 F. For instance, if your temperature is 82F for an hour the cost will be equivalent to 77F for 2 hours. By the way, you may get a better physical understanding in physics.stackexchange.com. For mathematicians the definite integral is simply the limit of the Riemann sum. $\endgroup$
    – B. Freitas
    Jan 6, 2016 at 1:45
  • $\begingroup$ Overall, don't let one question phase you. An integral is an infinite summation of infinitesimal parts, like you said. There's nothing more to it than that. $\endgroup$
    – Kaynex
    Jan 6, 2016 at 1:48

2 Answers 2


As a(n aspiring) geometer, the conceptual meaning of the integral for me is that integration is the process of adding little flats volumes (or areas, or...) to get big generally curved volumes (...).

enter image description here

In $1$-d calculus the regions you integrate over may not be as interesting but the idea is the same.

enter image description here

That's really all there is to integration -- add up a lot of little things to get a (potentially) big thing.

The fact that integration over an interval of a single-variable rate of change function is equal to the total change of that interval is actually somewhat of a minor miracle. That inverse relationship between the derivative and the integral doesn't work nearly as nicely in higher dimensions. Even so, it really is useful ... in fact I'd say it's almost fundamental to calculus. :)

Mathematicians don't normally worry about units, but physicists do and they do tend to always work out. So it is a good idea to use dimensional analysis to figure out if the quantities you're dealing with actually make sense.

Speaking of physics, here's one more image just for fun.

enter image description here

The thing you seem to have misunderstood in your problem is that the quantity \$0.10 actually has units of dollars per degree per hour (equivalently dollars per degree-hour). You can see this from the beginning of that line which reads "The hourly cost ... is \$0.10 per degree".

The integral $\int_8^{20} (T(t) -72)dt$ itself doesn't give the total temperature change over the interval $8\le t\le 20$ -- it gives the value of the average temperature (measured from $72^\circ$) over that time interval times the length of the time interval. For instance if the temperature were $85^\circ$ for the first 6 hours and then $83^\circ$ for the last 6 hours then the value of the integral would be $(84-72)\times 12 = 144$. The units of this are then degree-hours.

The above comes essentially from the definition of an average. To calculate the average value of a function $f$ over the interval $a\le t\le b$ we find that $f_{av} = \frac{1}{b-a}\int_a^b f(t)dt$. Multiplying by $b-a$ on both sides we see that the integral $\int_a^b f(t)dt$ is equal to the average value of the function $f$ over the interval times the length of the interval.

enter image description here

Then we can see that dollars per degree-hour $\times$ degree-hour $=$ dollars, as expected.

  • $\begingroup$ This makes sense, thank you for pointing it out. My continued difficulty, though, is I am hung up on why integrating T(t) should give degrees at all. If I say d(t)=10t, where t = hours and d is the miles I drove, and integrate d(t) from 0 to 3, the answer is 45. But that does not mean I have driven a total of 45 miles. So why does integrating a function that gives the temperature of the house in degrees mean an answer of total degrees? I can't wrap my head around that and I have got to be missing something very fundamental. $\endgroup$ Jan 6, 2016 at 2:56
  • $\begingroup$ The integral $\int T(t)dt$ doesn't have units of degrees. It has units of degree-hours. Then because the constant out front has units of dollars per degree per hour (equivalently dollars per degree-hour) the quantity $C$ has units of dollars as expected. $\endgroup$
    – user137731
    Jan 6, 2016 at 3:01
  • $\begingroup$ The way to understand [blank]-hours (as opposed to [blank] per hour) is that 1 [blank]-hour corresponds to holding 1 [blank] over the course of 1 hour. Whereas 1 [blank] per hour corresponds to 1 [blank] for every 1 hour. IDK if that makes sense, but assuming you've taken classical mechanics try to think of something like impulse which has units of Newton-seconds. So an impulse of 1 Newton-second is the amount of impulse you get when you apply a 1 Newton force for 1 second. $\endgroup$
    – user137731
    Jan 6, 2016 at 3:04
  • $\begingroup$ That is very helpful, thank you. I have not had classical mechanics in many years, but I think it is clearer now. 1 Newton-second is a 1 Newton force applied for 1 second of time, and would that mean that 1 Newton per second would mean 1 Newton applied every second? But possibly for less time? No, I guess if a 1 Newton force applied for half a second, every one second, then you'd have 0.5 Newton-second per second. So 1 Newton per second wouldn't really be meaningful without context? $\endgroup$ Jan 7, 2016 at 2:38
  • $\begingroup$ Newtons per second would be the units of a rate. This would tell you how fast the force on an object is changing. In general divisions represent rates. I don't know of a general term for the thing that multiplication represents. But it's always something like this, Newton-seconds are the units you get when you want a quantity which is proportional not only to the amount of (average) force applied to an object, but also the amount of time that it was applied. A rate of force change doesn't give you any indication of how long the force was applied, but impulse does. $\endgroup$
    – user137731
    Jan 7, 2016 at 3:07

At the very least the second question is poorly phrased. T(t) can't be the temperature in a house if the thermostat is set for 72 degrees. The question only makes sense if T(t) is the outside temperature. Also, the integral should have parentheses surrounding "T(t) - 72"

T(t) is temperature expressed as a function of time. T(t)-72 is the spending rate in dimes per hour at time t. Integrating the spending rate over the integration time gives the cost in dimes. Dividing dimes by 10 gives dollars.


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