Integrals. What are they? What is the difference between a definite and an indefinite integral? Why are they useful?

I have just begun teaching Calculus 2 again, as I stated in Spring Semester. With new, and some familiar, faces, I have to begin anew. I again have to motivate my students to see Calculus as a wholly useful and interesting course of study. While there will be many opportunities to show this throughout the semester, I feel it is important to start doing so immediately. Getting them in the right mindset to be open the learning helps to improve the outcome of the entire semester for everyone involved.

Even though most people will get to integrals in the first semester of Calculus, I have not seen anyone that has been able to finish all the integration techniques by the end of the semester. Therefore, you will inevitably begin second semester Calculus talking about integrals in some way. The section I like to begin with is analytic geometry. Starting here, I am able to review any integration techniques they already have, discuss definite integrals and Riemann sums, and remind the students of the connection that the Fundamental Theorem of Calculus provides. Furthermore, working with geometric figures helps students to really visualize what is happening.

**Area of a triangle**

To begin with, we need to know how to find area. By definition, we say that the area of a rectangle is the width multiplied by the height. That is, it is the product of the length of two intersecting sides. What, then, is the area of some non-rectangular shape? While there are many shapes we know formulas for, these formulas are based off the definition of rectangular shape. Therefore, in order to show that these formulas are correct, we have to refer back to a rectangle.

In order to illustrate the process we will be using, we will start by finding the area of an isosceles right triangle with leg lengths of 1 (see below). We can find the area quickly by reflecting over the hypotenuse. The image and original triangle will then form a square, each with half the area.

However, since we want to work with integrals, we will use a different technique to find the area of the triangle. We will therefore, do the following:

- Draw coordinate axis so that the legs of the triangle correspond to the \(x\) and \(y\)-axis.

- Draw rectangles that cover the width of the triangle.

Note that the rectangles will cover the entire triangle, with some extra area given for each rectangle used. - Sum the areas of the rectangles. This is a Riemann Sum. That is, we are approximating the area of the triangle by finding the sums of areas of rectangles. As stated above, the rectangles we used will result in an over-approximation of the area of the triangle.
- We can then outline the process of taking the limit of Riemann sums if needed.
- Use more rectangles in order to get a better approximation.
- Take the limit of the sums of the rectangles as the width of each rectangles goes to 0, therefore resulting in infinity many rectangles.
- Each time you let the rectangles get thinner, the error will get smaller.
- Furthermore the total error in approximation will also get smaller, eventually going to 0.
- However, at this point I will assume my students have seen the fundamental theorem of Calculus.

- We now focus on one of the rectangular cross sections that we’ve cut out.

- The area of this cross-section is then the width multiplied by the height, or \(A_{cs}=w*h\).
- We want the width to be able to get arbitrarily small.
- The width is moving in the \(x\) direction.
- We will, therefore, call this a change in \(x\) or \(w=\Delta x\).
- The height of the rectangle is then the \(y\) value.
- We note that \(y\) is linearly dependant on \(x\).
- The line goes through the corresponding points \((0,1)\) and \((1,0)\).
- Therefore, the \(y=\frac{-1}{1}(x-0)+1=-x+1\).
- Hence, \(A_{cs}=(1-x)\Delta x\).

- Recalling that the definite integral is just defined to be the limit of the Riemann Sum, we note that if we add up all these rectangles and let the widths approach 0, the total area, \(A_{t}\), will be \[A_{t}=\int_{a}^{b}A_{cs}.\]
- \(\Delta x\) becomes \(dx\) since the change becomes infinitesimally small.
- The \(a\) is the smallest \(x\) value we are interested in, the \(b\) the largest.
- We note that \(a=0\) and \(b=1\).
- We can now find that \[A_{t}&=\int_{0}^{1}(1-x)dx\].
- At this point I remind students that this definite integral is precisely defined to be the limit of Riemann sums. We will use antiderivatives to find this answer because it is easier to do. However, it is because of the fundamental theorem of Calculus that we are able to do this.
- I then proceed to find \[\begin{align*}A_{t}&=\int_{0}^{1}(1-x)dx\\&=x-\frac{x^{2}}{2}|_{0}^{1} \\ &=1-\frac{1}{2}-(0-0) =\frac{1}{2}.\]
- Note that our technique of using integrals resulted in the same area as if we had used the equation for a triangle.

**Conclusion**

Most students will know that formula for the area of the triangle. Even though the result won’t be new, it is often interesting for students to see justification of things they already know. This is because no justification may have previously been given, or that they’ve since forgotten it. By allowing them to check your work, it reinforces that the steps provided are correct, and allows for them to buy into the entire process more when you move onto more complicated shapes.

After showing this example, depending on timing, I will often work with an arbitrary triangle and provide a proof that the formula for the area of a triangle is always correct. This will then allow me to move onto talking about areas of other shapes and, eventually, volumes. By connecting integrals to tangible geometric objects, I find that students will often seem much more motivated to learn integrals.

I hope this provided you with an idea for a lecture on integrals. If you use this idea or something like it, let me know in the comments below how it went for you.