Integrating a Derivative Function (Half of FTC)

Any theorem comes best with an intuition behind it.

Let's consider a position function $s(t)$, a function of time. We'll plot it here and label some points:

Since this is a position function, what is the total change in position if one travels from $a$ to $b$? That's simple, it's just $s(b) - s(a)$.

Now, let's look at the derivative of this function, which is just the velocity function! Let's graph that! (We're assuming that this is a parabola for example's sake.)

Let's try dividing up the area between $a$ and $b$ into smaller rectangles. Look at the first rectangle--what happens when you multiply $v(a)$ and $\Delta t$? You get $s(a)$, since velocity is change in position over time! This also gives the area of the triangle!

A formal way of dividing up the area into smaller rectangles and finding the sum of their areas is a Riemann sum. If we want to find the sum of these areas, we write it like this:

In other words, this sum should give us the change in position from $a$ to $b$!

However, this sum is not approximate. To get a better sense of the area, we need the rectangles to be infinitesimally small, that is, the sum has to be $$\lim_{n \rightarrow \infty} \sum_{i=1}^n{v(t_{i-1})\Delta t}$$ This will give us really small rectangles allowing us to calculate the exact area. But this is just the definite integral from $a$ to $b$!

So technically, our change in position is:

Pause right here. Look back at what we found. What can we conclude?

Earlier, we found that the change in position is $s(b) - s(a)$. And since $s'(t) = v(t)$, putting all our information together gives us

This is known as the second part of the fundamental theorem of calculus. This works for any function that can be integrated, not just position functions.

This was not a formal proof, but it is a good intuitive way to know why the fundamental theorem of calculus is true. If you would like to see a formal proof, look at the next section!

Theorem (FTC part 2): Let $F$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $F'$ is integrable on $[a,b]$, then $$\int_a^b F'(x)\,dx=F(b)-F(a).$$ Proof: The key ideas in the proof are the Mean Value Theorem and telescoping sums. Let $\varepsilon>0$.

$F'$ is integrable, so there exists a $\delta>0$ such that for any partition $P$ with $\operatorname{mesh}(P):=\max_{k=1,\ldots n}\{x_k-x_{k-1}\}<\delta$, then all the Riemann sums $S$ for this partition satisfy $|S-\int_a^bF'(x)\,dx|<\varepsilon$. For a refresher of this material, see formal Riemann sums.

Let $P=\{x_0,x_1,\ldots,x_{n-1},x_n\}$ be a partition of $[a,b]$, i.e. $a=x_0<x_1<\ldots<x_{n-1}<x_n=b$. Suppose $\operatorname{mesh}(P)<\delta$, so any Riemann sum for this partition is within $\varepsilon$ of the value of the integral.

We will use the Mean Value Theorem to construct a Riemann sum that exactly equals $F(b)-F(a)$. Then, we will have $F(b)-F(a)$ within $\varepsilon$ of $\int_a^bF'(x)\,dx$ for any $\varepsilon>0$, which implies the theorem.

Applying the MVT to $F$ on each interval $[x_{k-1},x_k]$ gives us a $x_k^*\in[x_{k-1},x_k]$ such that $$F'(x_k^*)(x_k-x_{k-1})=F(x_k)-F(x_{k-1}).$$ Now using telescoping sums, we can write $$F(b)-F(a)=\sum_{k=1}^n[F(x_k)-F(x_{k-1})]=\sum_{k=1}^nF'(x_k^*)(x_k-x_{k-1}).$$ The rhs is a Riemann sum, call it $S$, for the partition $P$! Then since $F'$ is integrable and $\operatorname{mesh}(P)<\delta$, we must have $$\left|S-\int_a^bF'(x)\,dx\right|=\left|(F(b)-F(a))-\int_a^bF'(x)\,dx\right|<\varepsilon.$$ Since $\varepsilon$ is arbitrary, the theorem holds, i.e. $$F(b)-F(a)=\int_a^bF'(x)\,dx.$$

Warning: The requirement that $F'$ be integrable is needed! There exist functions that are continuous on $[a,b]$ and differentiable on $(a,b)$, but whose derivatives are NOT integrable. An example is the Volterra function.

Yo, let's have some fun. Okay, good, you've agreed, now do those darn problems!