Introducing integration
There are two approaches to introducing integration. One is its application to computing area and, by extension, volume and length; and the second is that it is the inverse operation of differentiation whereby we can determine a function that has a given rate of change. Which of these is the best way to introduce integration? This a critical question to think about before starting integral calculus with your students. There are plenty of textbooks that utilize either approach. I believe it is better – both conceptually and pedagogically – to introduce integration with finding indefinite integrals (which I initially refer to as an antiderivative) by reversing the process of differentiation. Students’ understanding is supported by doing this in the context of finding the distance an object travels during a given interval of time from knowing the object’s velocity; or, alternatively, finding an object’s future position (displacement) given its velocity and initial position or given its acceleration and initial velocity and initial position.
Although the computation of areas and volumes is a worthwhile application of integration, the more valuable application (in the long-term) for most students is using integration to solve differential equations. Thus, in my opinion, it is best to initially focus on integration as a process by which we proceed back to a function from knowing about its derivative – i.e. antidifferentiation.
I’m certainly not saying to ignore teaching the connection between integration and area or to not deal with the integral as a limit of a (Riemann) sum. I’m just suggesting that there are good reasons to begin the study of integration emphasizing it as the inverse process of differentiation. I believe that, in general, students find this aspect of integration a bit easier to understand along with highlighting a very important (perhaps most important) application of integration – i.e. solving differential equations. Using integration to compute areas can eventually develop from using antidifferentiation to solve certain kinds of differential equations, as I outline below.
During the early days of integral calculus, I would not tell students that we are “solving a differential equation” even though that is really what’s going on. I would not use the phrase 'differential equation' in early days of integral calculus. Let’s start by looking at a simple example.
Example 1: The velocity, in cm per second, at time t of an object moving along a line is given by \(v\left( t \right) = 0.3{t^2}\). Given that the initial position, s, of the object is \( + 10\) cm, find the position of the object at \(t = 2\) seconds.
Clearly, velocity is the rate of change (derivative) of position. So, position (displacement) is the antiderivative of velocity.
\(s\left( t \right) = \int {v\left( t \right)dt} = \int {0.3{t^2}dt = 0.1{t^3} + C}\)
Since \(s\left( 0 \right) = 10\) then \(C = 10\). Thus, \(s\left( t \right) = 0.1{t^3} + 10\)
Therefore, the position of the object at \(t = 2\) seconds is \(s\left( 2 \right) = 0.1{\left( 2 \right)^3} + 10 = 10.8\)
The problem above could be re-stated in terms of solving a differential equation as follows (but not calling it a 'differential equation'):
Given \(\frac{{ds}}{{dt}} = 0.3{t^2}\) and initial condition \(s\left( 0 \right) = 10\), find the value of \(s\left( 2 \right)\).
In this simple example, it was straightforward to find the antiderivative of \(0.3{t^2}\) by knowing basic differentiation rules. Hence, we can find an exact answer. The answer can be interpreted as the 2nd coordinate of a point on the solution curve for the differential equation. So, another way of stating the problem above is:
Find the s-coordinate of the point \(\left( {2,s} \right)\) on the graph of the function whose derivative is \(\frac{{ds}}{{dt}} = 0.3{t^2}\) given that \(\left( {0,10} \right)\) lies on the graph.
Now, let’s look at a similar but very different example.
Example 2: The velocity, in cm per second, of an object moving along a line at time t is given by \(v\left( t \right) = {{\textrm{e}}^{{t^2}}}\). Given that the initial position, s, of the object is at the origin, i.e. \(s\left( 0 \right) = 0\), find the position of the object at \(t = 2\) seconds.
\(s\left( t \right) = \int {v\left( t \right)dt} = \int {{{\textrm{e}}^{{t^2}}}dt} = ?\)
We immediately encounter an obstacle. The function \(v\left( t \right) = {{\textrm{e}}^{{t^2}}}\) does not have an antiderivative. In other words, there is no function whose derivative is \({{\textrm{e}}^{{t^2}}}\). Students will never have come across a function that could not be differentiated, so realizing that functions exist that cannot be integrated makes them aware that integration is often more problematic than differentiation.
Back to Example 2 ... How can we use the facts that \(s'\left( t \right) = {{\textrm{e}}^{{t^2}}}\) and \(s\left( 0 \right) = 0\) to approximate the value of \(s\left( 2 \right)\)? Again, I would not say this to students yet, but we are solving a differential equation. Not an exact solution because we cannot find an antiderivative (solution curve), so the best we can do is find an approximate solution which is often the case with differential equations.
The way to proceed is illustrated in the diagram. In general, two points \(\left( {{x_0},{y_0}} \right)\) and \(\left( {{x_1},{y_1}} \right)\) lie on the graph of a function f. The difference between the x-coordinates is and the difference between the y-coordinates is . Thus, the average rate of change between the two points is . Given the values of \({x_0}\), \({y_0}\), , and the average rate of change , computing the value of \({y_1}\) is straightforward. If the average rate of change is approximate, then clearly this will lead to an approximate value for \({y_1}\). The accuracy of this process is improved by using more ‘steps’ to move from \(\left( {{x_0},{y_0}} \right)\) to the point whose y-coordinate is needed. That is, by reducing the step size (a constant portion of ) and finding intermediates points between the starting point and the point whose y-coordinate is being approximated, the accuracy of the result can be significantly improved.
If we know the average rate of change of s as t moves from 0 to 2, then we can multiply it by 2 (the step size or change in t) to obtain the value of \(s\left( 2 \right)\). However, the rate of change of s, i.e. \(s'\left( t \right) = {{\textrm{e}}^{{t^2}}}\), is not constant; so, it’s not possible to compute an exact value for the average rate of change of s between two points on its graph. One reasonable approach to approximate the average rate of change between two points is to average the values for the rates of change at the two points. For example, \(s'\left( 0 \right) = 1\) and \(s'\left( 2 \right) = 54.6\). The average of these two rates is about 27.8, giving \(s\left( 2 \right) \approx 55.6\). Clearly, we can get a better approximation if we find an average rate of change for s over the interval \(0 < t < 1\) and another average rate of change for \(1 < t < 2\). In other words, reducing the step size to 1 and then ‘stepping’ twice – first from the starting point \(\left( {t = 0} \right)\) to an intermediate point \(\left( {t = 1} \right)\) and then from the intermediate point to the final point \(\left( {t = 2} \right)\). At each step we compute the y-coordinate and add it to the previous y-coordinate until we reach the y-coordinate of the final point.
Given that \(s'\left( 1 \right) \approx 2.72\), an approximation for the average rate of change for \(0 < t < 1\) is \(\frac{{1 + 2.72}}{2} = 1.86\) and for \(1 < t < 2\) the average rate of change is approximately \(\frac{{2.72 + 54.6}}{2} = 28.66\). Thus, \(s\left( 2 \right) \approx \frac{{s'\left( 0 \right) + s'\left( 1 \right)}}{2} \cdot 1 + \frac{{s'\left( 1 \right) + s'\left( 2 \right)}}{2} \cdot 1 \approx 1.86 + 28.66 \approx 30.52\)
If we approximate \(s\left( 2 \right)\) using a step size of 0.25 (requiring 8 steps), then our technique tells us that
\(s\left( 2 \right) \approx \frac{{s'\left( 0 \right) + s'\left( {0.25} \right)}}{2} \cdot 0.25 + \frac{{s'\left( {0.25} \right) + s'\left( {0.5} \right)}}{2} \cdot 0.25 + \; \cdots \; + \frac{{s'\left( {1.75} \right) + s'\left( 2 \right)}}{2} \cdot 0.25\)
This is equivalent to
\(s\left( 2 \right) \approx \frac{{v\left( 0 \right) + v\left( {0.25} \right)}}{2} \cdot 0.25 + \frac{{v\left( {0.25} \right) + v\left( {0.5} \right)}}{2} \cdot 0.25 + \; \cdots \; + \frac{{v\left( {1.0} \right) + v\left( {1.25} \right)}}{2} \cdot 0.25\)
Note that each expression in the sum is the area of a trapezoid, and these trapezoids make up the area under the graph of \(v\left( t \right)\) (see diagrams below). Decreasing the step size and thereby increasing the number of trapezoids will give us a better and better approximation of the area under the graph of \(v\left( t \right)\).
We can now express the area under the graph of \(v\left( t \right)\) from \(t = 0\) to \(t = 2\) as the definite integral \(\int_0^2 {v\left( t \right)dt} \).
