The Fundamental Theorem of Calculus (FTC) is one of the cornerstones of the course. It is a deep theorem that relates the processes of integration and differentiation, and shows that they are in a certain sense inverse processes.
The FTC has two parts. We will first discuss the first form of the theorem:
Fundamental Theorem of Calculus, Part I
If \(F(x)\) is an antiderivative of \(f(x)\) on \([a, b]\), i.e. \(F'(x) = f(x)\) on \([a, b]\), then we have: \[\int_a^bf(x)dx = F(b) - F(a)\]
In essence, we can directly compute an integral if we know of an antiderivative for the integrand.
Shorthand notation: The difference \(F(b)-F(a)\) is often written as: \[\bigl.F(x)\bigr|_a^b\]
Examples:
\[\int_1^4 x^2dx = \left.\frac{x^3}{3}\right|_1^4 = \frac{4^3}{3}-\frac{1^3}{3} = \frac{63}{3} = 21\] \[\int_0^\pi \cos xdx = \bigl.\sin x\bigr|_0^\pi = \sin\pi - \sin 0 = 0\]
Make sure to understand that second integral graphically, and why it should indeed equal \(0\).
The idea of the theorem is to relate the difference \(F(b) - F(a)\) to the Riemann sums.
We start with a partition \(P\) of the interval \([a, b]\): \[\left\{ a = x_0 < x_1 < x_2 < x_3 < \cdots < x_N = b \right\}\] On each interval, we want to estimate the difference in the endpoints, \(F(x_i) - F(x_{i-1})\). To do that, we use the mean value theorem, which says that this difference should equal \(F'(c_i)(x_i-x_{i-1})\) for some point \(c_i\in[x_{i-1}, x_i]\)L \[F(x_i) - F(x_{i-1}) = F'(c_i)(x_i-x_{i-1})\] Since \(F'=f\), we have: \[F(x_i) - F(x_{i-1}) = f(c_i)\Delta x_i\] If we write this for every \(i\), then add up all the equations, the left-hand-side has many terms cancelling out: \[\left[F(x_1)-F(x_0)\right] + \left[F(x_2)-F(x_1)\right] + \left[F(x_3)-F(x_2)\right] + \cdots + \left[F(x_N)-F(x_{N-1})\right]\] Note that every term except for \(F(x_0)=F(a)\) and \(F(x_N)=F(b)\) appears twice, once with a plus sign and once with a minus sign. So they all cancel out. We end up with the formula: \[F(b) - F(a) = \sum_{i=1}^N f(c_i)\Delta x_i\] The right-hand-side is exactly a Riemann sum with those specific sample points. In other words:
For every partition \(P\) there is a choice \(C\) of sample points such that: \[R(f, P, C) = \sum_{i=1}^N f(c_i)\Delta x_i = F(b) - F(a)\]
Since this happens for every partition, and since the right-hand-side is a constant independent of the partition, it follows that the limit should have the same relation, so: \[\int_a^b f(x)dx = F(b) - F(a)\] as desired.
The Fundamental Theorem of Calculus has a second formulation, that is in a way the “other direction” than that described in the first part. In that part we started with a function \(F(x)\), looked at its derivative \(f(x) = F'(x)\), then took an integral of that, and landed back to \(F\). This version goes the other way.
Fundamental Theorem of Calculus, Part II
Let \(f(t)\) be a continuous function on \([a, b]\). Then for each \(x\in[a,b]\) we can define: \[G(x) = \int_a^xf(t)dt\] This defines a function \(G\) on \([a, b]\). The theorem is that this function is differentiable, and its derivative is: \[G'(x) = f(x)\] So in a single formula we would write: \[\frac{d}{dx}\left(\int_a^x f(t)dt\right) = f(x)\]
As an example, consider the function: \[g(x) = \int_1^x \frac{1}{t}dt\] Here’s some things we can say about it:
All this information can give us a good handle on the function, and we can for instance do a decent job graphing the function.
Let us look at some variations of the theorem, where the endpoints are more complicated. Suppose we have the following integral function: \[h(x)=\int_1^{x^2}\frac{1}{t}dt\] We have no theorem that deals with this function directly. Instead we will think of the function: \[g(u) = \int_1^u \frac{1}{t}dt\] For that function we know that \(g'(u) = \frac{1}{u}\).
Now \(g\) and \(h\) are related: \[h(x) = g(x^2)\] Therefore we can use the chain rule to find the derivative of \(h\). We would have: \[h'(x) = g'(x^2)(2x) = \frac{1}{x^2}(2x) = \frac{2}{x}\] So “compute the integrand function at the upper endpoint, then multiply by the derivative of the upper endpoint”.
Similarly we can work with cases where the lower endpoint has \(x\) in it: \[h_2(x) = \int_x^1\frac{1}{t}dt\] We can then write the integral as: \[h_2(x) = -\int_1^x\frac{1}{t}dt\] so the derivative would be: \[h_2'(x) = -\frac{1}{x}\]
If we have both endpoints containing \(x\), we can break the problem up in two: \[h_3(x) = \int_{x^2}^{x^3}\frac{1}{t}dt = \int_1^{x^3}\frac{1}{t}dt - \int_1^{x^2}\frac{1}{t}dt\] The derivative would then be, as before: \[h_3'(x) = \frac{1}{x^3}(3x^2) - \frac{1}{x^2}(2x) = \frac{3}{x} - \frac{2}{x} = \frac{1}{x}\] This is rather interesting, the derivative of this function turned out to be the same as the derivative of our original \(\int_1^x \frac{1}{t}dt\). This won’t happen in general of course, but it does happen for this particular integrand.
Practice: Compute the derivatives of \(\int_0^{x^2}\sin(t^2)dt\) and \(\int_x^{2x}\frac{\sin t}{t}dt\).
We start with the function \(f(t)\) and its integral function \(G(x) = \int_a^x f(t) dt\). In order to determine if \(G(x)\) is differentiable at a point \(x\), we need to consider the expression: \[\lim_{h\to 0}\frac{G(x+h) - G(x)}{h}\] Throughout the rest of this section, we will treat this point \(x\) as a constant.
In order to compute this limit, we start by considering the difference: \[G(x+h) - G(x) = \int_a^{x+h}f(t)dt - \int_a^x f(t)dt = \int_x^{x+h}f(t)dt\] We therefore need to get a hold on this integral \(\int_x^{x+h}f(t)dt\). In order to do that, we will make a simpifying assumption that \(f\) is an increasing function on the interval from \(x\) to \(x+h\). With that in mind, we can say that for every \(t\in[x, x+h]\) we have the inequalities: \[f(x)\leq f(t)\leq f(x+h)\] These inequalities are presented when we integrate: \[\int_x^{x+h}f(x)dt\leq \int_x^{x+h}f(t)dt\leq \int_x^{x+h}f(x+h)dt\] The first and third integral are the integrals of constants, as our variable is \(t\). Therefore they become (since \((x+h)-x=h\): \[hf(x) \leq \int_x^{x+h}f(t)dt \leq f(x+h)\] And dividing by \(h\) we get: \[f(x) \leq \frac{\int_x^{x+h}f(t)dt}{h} \leq f(x+h)\] Going back to \(G\), we have: \[f(x) \leq \frac{G(x+h) - G(x)}{h} \leq f(x+h)\] We next need to take the limit as \(h\to 0\). Since \(f\) is a continuous function, then: \[\lim_{h\to 0}f(x+h) = f(x)\] So by the squeeze theorem we get that: \[\lim_{h\to 0}\frac{G(x+h) - G(x)}{h} = f(x)\] This proves that \(G(x)\) is differentiable and its derivative if \(G'(x) = f(x)\).