You learned integration wrong. That's quite a claim, so let me clarify what we mean by it. Our claim is based on studies of former calculus students that analyzed what students think integration means. Students tend to conceptualize integration to have one of three meanings: Integration is finding an area under a curve, integration is finding an antiderivative of a function, and integration is adding up tiny-bits of a quantity. Now we can think about integration as all of these, but research shows that students who only think of integration as finding areas under curves or finding antiderivatives really struggle to solve real-world problems where integration is required. But which students did succeed in solving these same problems? The students that could think of integration as adding up tiny-bits of a quantity.
So why do we claim you learned integration wrong? Well because most students leave calculus with one of the first two conceptions, areas under curves or anti-derivatives. And, to be fair, these are the conceptions mostly emphasized in calculus classes. But this leaves the vast majority of students without the tools to recognize when integration could be used in a novel situation, and even if they do recognize the use of integration, they still get stumped trying to set up the appropriate functional or numerical integral to solve the problem. So this is what I mean by "you learned integration wrong." You probably don't think about integration in a way that actually makes it useful for real world problem solving. This video will help you understand integration in this more powerful way, adding up tiny bits of a quantity. For first year calculus, the quantities are all calculated with multiplication, so we use these examples in our video. This video is a bit longer than our other videos but we hope you find it worth it. Here are three different problem situations to illustrate integration: 1) When I pull back my bow to shoot an arrow, it gets harder and harder to pull, so how much total energy do I put into the bow if I pull it all the way back? 2) How much more energy could I get from solar panels if I put them on solar trackers instead of laying them flat on a roof? 3) In basketball, my coach would tell me to run wide on the fast break, but wouldn't it be better if I ran straight down the court instead? Doesn't going wide give my defender more time to beat me down the court and set-up defenses? Since I play near the basket, I need
to know how much farther I run going from basket to basket out wide, versus in a straight line. These 3 problem situations are almost 6th grade math problems, what I mean by that, is that we could adjust each of these problems slightly and end up just doing multiplication. In reality the force to pull a bow back increases the farther back I pull. But if the force was constant, like lifting a book from the floor to a table, then calculating the energy would just be a multiplication problem. The force times the distance. Our solar panel is stagnant. But if the solar panel moved to always face the sun, assuming a sunny day, then the power generated is almost constant, so I can multiply the power rating of the solar panel by the amount of time it's in the sun. If I always ran at the same angle relative to my basketball defender, rather than along this arc like I would in reality, then it's
easy to find out how much further I run compared to my defender. I just multiply his distance by a certain ratio We can build on these simple cases to develop a strategy for solving each of these problems. Chop up each situation into a lot of small intervals, where the problems are quite close to the multiplication problems described above, do the multiplication in each of these small intervals to calculate the quantity we're interested in (in these cases energy or distance), then add up the values from each small piece to get a total amount. The idea is that for a short interval in time, or distance, the difference between the function value at the beginning of the interval and the end of the interval is so small the effect on the final answer is negligible so we can treat them as constant and just use multiplication The force of pulling the bow doesn't change much from one millimeter to the next. The power generated from the solar panel doesn't change much from one minute to another. The ratio of my distance to my defenders distance is about the same from any point and a point one centimeter further down the court.
So for each situation, we can break it up into a lot of small intervals, do the multiplication to calculate the amount in each interval, and then add up the amounts from each interval. Let's dive deeper in my bow example. Here is data from a toy compound bow I bought at a garage sale to use in demonstrations in my college math classes. I calculated the force, in lbs, it took to
hold the string back every 1.5 inches of the pull, which is a total of 18 inches. I have 12 intervals, so 13 positions where I collected force values. I am going to use the average value of each end point of the interval to estimate the average lbs of force to pull the string on that interval. I will convert the distance
measurements to feet, so I end up with foot-lbs as the energy unit. Within each 1.5 inch interval, or 1/8th of a foot, I can calculate the energy to pull back the bow by multiplying the average force times the distance. Then I can calculate the total amount of energy by adding up the energy expended in each of the intervals. I got a value of 10.425 foot-lbs. So enough to hurt someone... well not that bad. More like dropping a 1 lb weight from a height of 10 ft on someone's foot. This process we just went through is integration at. its. heart. We first: Chopped up a situation where there isn't a constant relationship between quantities, so that within each interval it is approximately constant. Second: we calculated the quantity we are trying to find within each interval by multiplying, and 3) added up the quantity in each piece for the total value. You may have noticed, we didn't write down an integral sign, though we could have. We didn't find an anti-derivative. We actually don't have a function to even find the anti-derivative of. But that's the case in a lot of real world integration problems when we are working with data.
Now you may be thinking that my answer is not exact. Well, I would say you are correct, but in this case, no one could be exact. We could certainly have used a lot more measurements on smaller intervals to be more accurate with our data, but at some point there will be more error in trying to pull the string with the scale and hold it exactly at the right distance, or in the truncated digits of the scale, than in the mathematical calculations. We could use some assumptions to get around some of the physical limitations of taking measurements, but the assumptions don't mean that we are still going to get the exact answer. For example, we could fit a model to the values I measured earlier. Here is a 4th degree polynomial fit to our points, where the distance the bow is pulled in feet is on the x-axis and the force in lbs on the y-axis. It seems to fit quite well (with an r-squared of over
.98 for those interested). So we could calculate the total energy assuming that the force to pull the bow at any point is the value given by the function. This gets us around the measurement issues, but of course, our model won't fit the real-world phenomenon, exactly, so we are still just approximating, but for different reasons than before.
If you have taken calculus before, you might be more comfortable setting up and calculating the energy based on this model with a function. But the thinking to set up the integral is still the same as before. But, first, let's connect this idea of multiplication and summation to the integral notation you may be familiar with. The structure of integral notation actually suggests the multiplication and summation processes central to integration. We chop up the situation into arbitrarily small intervals of equal width called the differential, often labeled dx or dt. Then we multiply the function value on the interval with the width of the interval, and finally add it all up. The integral sign is actually an elongated s, chosen by Leibniz to symbolize "summa" which means total or sum. It's interesting how once we know what each piece of the integration notation represents, how easy it is to see that the core concept of integration is this process of multiplication and summation. And yet this conception's not emphasized and often overlooked in so many calculus classes. Let's get back to the bow example. So in order to quantitatively understand this situation we'll chop up the distance into a bunch of small pulls. How much energy is any one of the pulls? It is the force you
feel times the distance you pull. Since we have a function that can tell us the force at any value, we can then pick extremely small intervals, infinitesimally small intervals, so the differential, dx, is the length of each interval. What are all of the values for which we accumulated energy? Well, our pull goes from zero feet to 1.5 feet, so that is the interval we are integrating over. And finally, adding up all of those products that we could calculate over that interval and we get our final answer. Since we have a function that connects a force at any point on the pull, we could use a spreadsheet to do the calculations. Or use any one of the many online numerical integration calculators, where people have programmed computers to do it for us. This is also one of the special cases where we could use the fundamental theorem of calculus. Since our function is continuous on our interval of integration, and we can find an antiderivative, then we can use an antiderivative to make the calculation. Why does this antidifferentiation technique work? That's beyond the scope of this video. But perhaps we'll address it in another video in the future. Comment below if you'd like to learn more on why antidifferentiation works! Whether using the Fundamental Theorem of Calculus or using an integration calculator, with this model we get an estimated force of 10.3275, quite close to our first estimate with 12 sub-intervals. Now let's talk about integrals as areas. It is true that each of the 3 integration situations can be connected to a problem about areas. The problem is that thinking about areas doesn't help to recognize any of those problems as an integration problem in the first place. But thinking about integration as the chop, multiply, and Add process, helps with both recognizing an integration problem and setting up the integrand.
Perhaps the problem that is closest to being made sense of as an area in an authentic way, true to the science of the situation, is the one we did with the bow, at least once we had a function to model the force. There are times when thinking about integrals as areas is helpful. Engineers that design compound bows use the area under the force curve of the bow to reason about the properties of an ideal bow. And since any integration problem can be thought of as an area problem, then areas can help us visually understand properties of integrals, and strategies for approximating integrals. I think this is one reason why textbooks emphasize the area conception so much, because it is helpful in making sense of abstract integrals. Thinking about integration as areas unfortunately is not very helpful at the front end of real-world problems where we need to recognize the problem as an integration problem and set up the correct integral. So we hope now you can start to see integration in a new way, if you haven't been thinking of it this way before. If you can develop this way of thinking, you might find, like I have, that you start to notice integration problems all around you. Thank you for watching. Please subscribe and share our videos. Be sure to follow MathTheWorld on Twitter, Instagram, and Facebook. Thank you so much for your support!
