Reference. I will refer to the textbook for this class using indicators such as section numbers or page numbers.
(Section 5.2) What is a definite integral? A definite integral of a function $f(x)$ in $x$ is a signed area of the region between the graph $y = f(x)$ of the function and the $x$-axis. For example, the integral $$\int_{0}^{1}x dx$$ means the area given by the region surrounded by
- the graph $y = x$ and
- the $x$-axis (which also can be written as $y = 0$) with the bounds
- $x=0$ and $x=1.$
Thus, the integral is given by the area of a triangle whose base is $b = 1$ and height is $h = 1.$ This lets us compute $$\int_{0}^{1}x dx = \frac{1}{2} \cdot bh = \frac{1}{2}.$$ However, note that $$\int_{-1}^{0}x dx = -\frac{1}{2},$$ because the triangle is underneath the $x$-axis. This is why I said that the definite integral is the signed area. Positive and negative areas can cancel each other out: $$\int_{-1}^{1}x dx = 0.$$
Exercise. Can you see why the following is true? Draw a picture (p.124, Figure 2.61) to understand: $$\int_{-1}^{1}|x| dx = 1.$$
Pedantry. For the functions $f(x),$ we will mostly consider continuous functions, or at the worst piece-wise continuous functions. Often, the function $f(x)$ will be assumed to be differentiatiable, but recall that this actually implies that $f(x)$ is continuous. (Section 2.6 Theorem 2.1.)
(Section 5.3) 1st Fundamental Theorem of Calculus. Sure, you know how to compute the area of a triangle, but what about non-triangular regions? For example, how can we compute $$\int_{0}^{1} x^{2} dx,$$ which is not given by a region bounded by straight lines? If you try to draw a picture, you can see that $$0 < \int_{0}^{1}x^{2} dx < \frac{1}{2},$$ but it is quite challenging to actually compute this. The following theorem helps in this situation.
Theorem (5.1, 1st FTC). For any continuous function $f(x)$ defined on an interval $[a, b]$ on the real line, if you find a differentiable function $F(x)$ such that $F'(x) = f(x),$ then $$\int_{a}^{b}f'(x)dx = F(b) - F(a).$$
Let's use this 1st FTC to compute the integral $$\int_{0}^{1} x^{2} dx.$$ First, our function $f(x) = x^{2}$ is continuous on $[0, 1]$ because it is given by a polynomial. We need to find $F(x)$ such that if we differentiate it, we get $f(x) = x^{2}$ back. If you think hard, one candidate is $$F(x) = \frac{{x}^{3}}{3}.$$ Let's check! We have $$F'(x) = 3 \cdot \frac{x^{2}}{3} = x^{2} = f(x).$$ Bingo, so we can use the 1st FTC to have $$\int_{0}^{1} x^{2} dx = \int_{0}^{1} f(x) dx = F(1) - F(0) = \frac{1}{3} - \frac{0}{3} = \frac{1}{3}.$$
Exercise. What goes wrong if we use $$F(x) = \frac{x^{3}}{3} + 3000$$ instead? (Hint: nothing goes wrong, but you really need to see this by repeating the process.)
Exercise. Compute $\int_{-1}^{-1}\sin(x)dx.$ (Hint: draw a good graph of $y = \sin(x).$ You will see some cancellation.)
Exercise. Compute $\int_{-1}^{-1}e^{x}dx.$
Exercise. Compute $\int_{-1}^{-1}3000^{x}dx.$ (Hint: for 1st FTC, you cannot use $F(x) = 3000^{x}$ because $F'(x) = 3000^{x} \ln(3000).$ Thus, you should modify this $F(x)$ by dividing it by $\ln(3000).$)
Read. Some properties about the integrals are summarized as
- Theorem 5.3 (p.304) and
- Theorem 5.4 (p.307)
I think these are quite natural once you accept the definition of "signed area", so I won't refer to this when I teach. It is much better to practice using these properties when you solve problems rather than memorizing them.
Quote. "The best way to learn mathematics is to solve problems with it."
Area between two graphs (p.305). We can compute the area of a more complicated region. For instance, consider the region surrounded by
- $y = x,$
- $y = x^{2},$
- from $x = 0$ to $x = 1.$
Since the graph $y = x$ is above the graph $y = x^{2}$ from $x = 0$ to $x = 1$ (i.e., on the interval $[0, 1]$), the area of the given region can be computed as follows: $$\int_{0}^{1} x dx - \int_{0}^{1} x^{2} dx = \int_{0}^{1} x - x^{2} dx = \left.\left(\frac{x^{2}}{2} - \frac{x^{3}}{3}\right)\right|_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$$ Note the new notation I used to indicate that I am using the 1st FTC. If I take an exam, I would still write "FTC" nearby to let the grader know that I know what I am doing. (Not always, this sometimes counts as a point.)
Exercise. Suppose that we are doing that same problem but change the last condition defining the region to the following:
- from $x = 0$ to $x = 2.$
1. Explain why the following integral is NOT going to compute the correct area: $$\int_{0}^{2} x - x^{2} dx.$$ 2. Write out an expression involving two integrals that wil compute the correct area.
3. Compute the integral.
Exercise. Compute the area of the region surrounded by
- $y = |x|,$
- $y = x^{2},$
- from $x = 0$ to $x = 1.$
(Hint: draw a picture. See p.124, Figure 2.61.)
Exercise. Compute $$\int_{0}^{\pi}\sin(x)dx.$$ (Hint: 1st FTC with $f(x) = \sin(x)$ and $F(x) = -\cos(x).$ Can you tell why you need the minus sign?)
Exercise. Compute $$\int_{0}^{2\pi}\sin(x)dx.$$ (Hint: draw picture and cancel things out!)
Exercise. Compute $$\int_{0}^{\pi}\cos(x)dx.$$ (Hint: draw picture and cancel things out!)
Exercise. Compute $$\int_{0}^{\pi/2}\cos(x)dx.$$
Exercise. Compute $$\int_{0}^{\pi/2}\cos(2x)dx.$$ (Possible hint: draw picture and cancel things out! If the drawing is too difficult, then use 1st FTC with $f(x) = \cos(2x)$ and $F(x) = \sin(2x)/2.$ Can you see why you need to divide by $2$?)
Exercise. Compute the following integrals:
- $\int_{-1}^{1}xdx,$
- $\int_{-1}^{1}x^{3}dx,$
- $\int_{-1}^{1}x^{5}dx,$
- $\int_{-1}^{1}x^{7}dx.$
(Hint: draw picture and cancel things out!)
