A foundational course for understanding mathematical modeling.
This course introduces the fundamental ideas behind integral calculus, including topics such as sequences, sums and integration.
As you progress through this course, you will find links to related topics in other courses (such as algebra). This interconnected structure is designed to help you quickly revisit prerequisite ideas without breaking your workflow.
We must begin with a discussion of what a Riemann sum is is, why we care about it, and examples of when it is useful.
If you want the fast version, below is a short video introducing limits.
Assuming that you are landing here already knowing about derivatives, as well as rules such as the power rule and chain rule, a natural question that arises is the following: what if you want to "undo" the derivative? This is what is explored in the topic of antiderivatives. Later we will dive deeper into the theoretical foundations of a topic called integration which almost by happenstance has a direct link to antidifferention, but for now we will focus on the antiderivative. So what exactly is the antiderivative? It is easiest to work by example here, suppose we have the following function, $$f(x) = x^3,$$ and we wish to find its derivative. Clearly we may use the power rule, and we arrive at, $$g(x) = f'(x) = 3x^2.$$ Nothing too surprising at this point. But now, what if we want to go the other way? Clearly we need a way to take \(g(x)\) back to \(f(x)\) (at least on some interval where both \(g(x)\) and \(f(x)\) exist). Clearly if you hand us the function \(3x^2\), from our knowledge of the power rule we can easily work backwards to figure out how to "undo" the derivative that has been taken, and we from the power rule could find a rule for going in the opposite direction, something like \(\frac{1}{n+1}x^{n+1}\) seems like it would work. However there is a slight issue with this, because it turns out that the derivative is not unique. For instance, note that the following functions all have the same derivative, \(f(x) = x^3, k(x) = x^3+10, l(x) = x^3-3\). In other words, if I add an arbitrary constant to my function, it does not change the derivative. So the derivative is unique only up to an additive constant. So now I can fully propose my rule to find the antiderivative for power rule like derivatives, if I let \(g(x) = f'(x)\) where \(g(x)\) is the type of funtion that the power rule could be used on, then, \(f(x) = \frac{1}{n+1}x^{n+1}+C\) where \(C\) is an additive constant.
Another low hanging fruit in terms of the antiderivative, is the natural log. We know that \(\frac{d}{dx}\ln(|x|) = \frac{1}{x}\) (see derivative of the natural log
). Thus the antiderivative of \(\frac{1}{x}\) must be \(\ln(|x|)+C\), (the constant term will appear in all of our antiderivatives for the same reason as stated above).Placeholder
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Perhaps the most important realization in calculus is that there is a fundamental link between two seemingly different branches, differential and integral calculus. Integral calculus was originally developed as a way to measure areas for curved surfaces, and on its face there doesn't appear to be any reason by derivatives would be linked to integrals. However, it was recognized very quickly that there is indeed a deep link between the two. Differentiation and integration can be viewed as inverses to one another (and thus as we shall see, antiderivatives are linked to integrals) which is what the fundamental theorem of calculus tells us. The fundamental theorem of calculus breaks into two parts, both of which are important, so we will explore both independently. The first part explicitely links the antiderivative (and thereby the derivative as well) and the integral.
Fundamental theorem of calculus part 1: Suppose we have an equation of the form, $$g(x) = \int \limits_a^x f(t) dt,$$ and furthermore that \(f\) is continuous across some interval \([a,b]\) and finally that \(x \in [a,b]\). Then \(g(x)\) is an antiderivative of \(f\), that is \(g'(x) = f(x)\) for \(a < x < b\). For any fixed value \(x\), since \(g(x)\) depends only on \(x\), \(g(x)\) will also be a fixed value. To prove this, we will need to recall a property of the definite integral the definite integral. Specifically we will need to recall the following properties, $$ \int \limits_{a}^c f(x)dx+ \int \limits_{c}^{b}f(x)dx = \int \limits_{a}^{b}f(x)dx,$$ and, $$m(b-a) \leq \int \limits_{a}^b f(x)dx \leq M(b-a),$$ so long as \(m \leq f(x) \leq M\) for \(a \leq x \leq b\). Now for the proof, if we let \(x\) and \(x+h\) be on the open interval \((a,b)\), then $$g(x+h)-g(x) = \int \limits_{a}^{x+h} f(t)dt - \int \limits_{a}^x f(t)dt = \Big(\int \limits_{a}^x f(t) dt+ \int \limits_{x}^{x+h} f(t) dt \Big) - \int \limits_{a}^{x} f(t) dt = \int \limits_{x}^{x+h}f(t)dt.$$ Note that the portion in parenthesis was a result of the first property noted above. Thus we can see that for \(h \neq 0\), $$\frac{g(x+h)-g(x)}{h} = \frac{1}{h}\int \limits_{x}^{x+h}f(t)dt.$$ Since we are assuming that \(f\), and for simplicity we will assume \(h>0\) is continuous on the closed interval \([x, x+h]\), by the extreme value theorem, there are numbers \(u,v \in [x,x+h]\) such that \(f(u) = m, f(v)=M\) where \(m\) is the absolute minimum value of \(f\) and \(M\) is the absolute maximum on the interval \([x,x+h]\). Using the second property of integrals we listed above (in this section), we can see that, $$mh \leq \int \limits_{x}^{x+h}f(t)dt \leq Mh,$$ and from extreme value theorem, $$f(u)h \leq \int \limits_{x}^{x+h}f(t)dt \leq f(v)h.$$ Dividing through by \(h\) (recall that we are assuming \(h>0\)), $$f(u) \leq \frac{1}{h}\int \limits_{x}^{x+h}f(t)dt \leq f(v).$$ Using the equality established above, we may rewrite this as, $$f(u) \leq \frac{g(x+h)-g(x)}{h}\leq f(v).$$ Note, we can follow this same line of logic for \(h<0\), so this holds for any (real) \(h \neq 0\). If we take the limit of the above expression as \(h \rightarrow 0\) we will find that since \(f\) is continuous, then \(u,v \rightarrow x\) because \(u,v\) lie in between \(x\) and \(x+h\) and we find the expression, $$f(x) \leq g'(x) \leq f(x),$$ (Note I used the limit above) and whenever something is upper and lower bounded by the same value that implies equality (by the squeeze theorem) $$g'(x) = f(x),$$ which concludes the proof.
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