The Fundamental Theorem of Calculus - Infinity is Really Big

Calculus 2, Lectures 5A through 6 (Videotaped Fall 2016)

The Fundamental Theorem of Calculus implies that the area under the graph of the speed function gives the distance traveled function.

The Fundamental Theorem of Calculus is often split into two forms in textbooks.

These forms are typically called the “First Fundamental Theorem of Calculus” and the “Second Fundamental Theorem of Calculus”, but they are essentially two sides of the same coin, which we can just call the “Fundamental Theorem of Calculus”, or even just “FTC”, for short.

In my current series of blog posts based on my Calculus 2 lectures at Bethel University from the Fall 2016 semester, I have already been making use of the FTC. This is natural to do for a second semester calculus course, because the FTC is usually introduced in the first semester.

In my second post, “Integration by Substitution (Method of Integration)”, I discuss how the FTC is relevant to the substitution method. The key definite integral equation that is derived there, assuming we have appropriately nice functions $f$ and $g$ , is $\displaystyle\int_{a}^{b}f(g(x))g'(x)\, dx=\displaystyle\int_{g(a)}^{g(b)}f(u)\, du$ . The FTC along with the Chain Rule are the facts that are necessary to prove this.

On the other hand, in my third post, “Integration by Parts (and Linear Motion)”, it is the Product Rule that must be combined with the FTC to obtain the key equation. This is equation is typically written as $\displaystyle\int_{a}^{b}u(x)v'(x)\, dx=u(x)v(x)\Bigr\rvert_{a}^{b}-\displaystyle\int_{a}^{b}u'(x)v(x)\, dx$ , where $u(x)$ and $v(x)$ are suitably nice functions. The notation $u(x)v(x)\Bigr\rvert_{a}^{b}$ is understood to mean $u(b)v(b)-u(a)v(a)$ , just as $F(x)\Bigr\rvert_{a}^{b}$ means $F(b)-F(a)$ .

But what is the FTC (both parts)? And why is it true? Let’s explore!

The First Fundamental Theorem of Calculus

If you ask a studious calculus student what the Fundamental Theorem of Calculus is, what will they answer? In all likelihood, that student will tell you that it gives a method to evaluate definite integrals.

For a function whose graph is always above the horizontal axis, it therefore gives you a way to find the area under its graph and above the axis.

Example 1: Find an Area under the Graph of the Squaring Function using the FTC

For example, consider the definite integral $\displaystyle\int_{1}^{3} x^{2}\, dx$ . This symbol represents the area of the region $\mathcal{R}$ shown below. The region $\mathcal{R}$ is bounded by the graph of $y=f(x)=x^{2}$ , the $x$ -axis, and the vertical lines $x=1$ and $x=3$ .

The Fundamental Theorem of Calculus can be used to find areas of regions under graphs of (positive) functions. In this case, it is an area under the graph of y = x^2 (the squaring function). — The area of the region $\mathcal{R}$ under the graph of $y=x^{2}$ (and above the horizontal axis) between $x=1$ and $x=3$ can be found with the FTC to be equal to $\frac{26}{3}=8.3333\ldots$ . This can be visually confirmed by counting boxes (and partial boxes) under the graph. Note that each box has an area of $0.5\times 1=0.5$ , so you should be able to count just under 17 boxes total.

How is this done? The FTC tells us to find an antiderivative of the integrand function and then compute an appropriate difference.

An antiderivative of $f(x)=x^{2}$ is $F(x)=\frac{1}{3}x^{3}$ . Since the limits of integration in $\displaystyle\int_{1}^{3} x^{2}\, dx$ are $x=1$ and $x=3$ , the FTC tells us that we must compute $F(3)-F(1)$ . The answer is $F(3)-F(1)=9-\frac{1}{3}=\frac{26}{3}=8.3333\ldots$ .

We have indeed used the FTC here. But which version?

The version we just used is typically called the First Fundamental Theorem of Calculus. But what is its precise statement?

Statement of the First Fundamental Theorem of Calculus

Here is one way to state the 1st FTC.

First Fundamental Theorem of Calculus: Suppose that the function $F$ is an antiderivative of the function $f$ on an interval $[a,b]$ , so that $F'(x)=f(x)$ for all $x$ in $[a,b]$ , then $\displaystyle\int_{a}^{b}f(x)\, dx=F(b)-F(a)$ .

This is a pretty amazing theorem. It tells us that finding an antiderivative is most of the way to finding what could be, quite possibly, a very complicated area.

This is a vast (infinite) extension of what you learned about areas of objects in geometry!

On the other hand, there is no guarantee that an antiderivative $F(x)$ will be easy or even possible to find.

Example 2: Find a Complicated Area with the First FTC

For instance, consider the definite integral $\displaystyle\int_{0}^{\sqrt{5\pi}}\left(2x\sin(x^{2})+10\right)\, dx$ . You should check, using the Product Rule and Chain Rule, that $F(x)=10x-\cos(x^{2})$ is an antiderivative of the integrand $f(x)=2x\sin(x^{2})+10$ .

Therefore, by the FTC, we compute

$\displaystyle\int_{0}^{\sqrt{5\pi}}\left(2x\sin(x^{2})+10\right)\, dx=F(x)\Bigr\rvert_{0}^{\sqrt{5\pi}}=F(\sqrt{5\pi})-F(0)$

$=(10\sqrt{5\pi}-\cos(5\pi))-(0-\cos(0))=2+10\sqrt{5\pi}\approx 41.63.$

This is indeed the area of the complicated region $\mathcal{R}$ shown in the figure below.

The Fundamental Theorem of Calculus can be used to find very complicated areas. — The area of the region $\mathcal{R}$ under the graph of $y=2x\sin(x^{2})+10$ (and above the horizontal axis) between $x=0$ and $x=\sqrt{5\pi}\approx 3.96$ can be found to be **exactly** $2+10\sqrt{5\pi}$ with the FTC. **That is pretty amazing!**

Signed Area Interpretation and Computation

More broadly, definite integrals are defined to give what are sometimes called “signed” areas rather than “true” areas.

If $f(x)\geq 0$ for all $x$ in a closed interval $[a,b]$ , then $\displaystyle\int_{a}^{b}f(x)\, dx$ certainly represents a true area. It is the area under the graph of $y=f(x)$ , above the $x-$ axis, and between the vertical lines at $x=a$ and $x=b$ .

But if the graph of $y=f(x)$ goes below the $x-$ axis, then $\displaystyle\int_{a}^{b}f(x)\, dx$ represents something different. It represents the area “under the graph” when it is above the axis minus the area “above the graph” when it is below the axis.

This will be more clear if we consider and example with a graph.

Example 3: Find a Signed Area with the First FTC

Let $f(x)=\cos(x)$ . Then the function $F$ defined by $F(x)=\sin(x)$ is an antiderivative of $f$ (over the whole real number line ${\Bbb R}=(-\infty,\infty)$ ). Then, for example, by the 1st FTC,

$\displaystyle\int_{\pi/3}^{5\pi/3}\cos(x)\, dx=\sin\left(\frac{5\pi}{3}\right)-\sin\left(\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=-\sqrt{3}\approx -1.73.$

We see that the answer is negative. In fact, in the picture below, with the (positive) areas labeled $A_{1}$ , $A_{2}$ , and $A_{3}$ , the definite integral above equals $A_{1}-A_{2}+A_{3}=(A_{1}+A_{3})-A_{2}$ .

In other words, areas that are below the graph when the graph is above the axis contribute positively to the value of the integral, while areas that are above the graph when the graph is below the axis contribute negatively to the integral.

The definite integral $\displaystyle\int_{\pi/3}^{5\pi/3}\cos(x)\, dx$ gives a `signed area’. Assuming the symbols $A_{1}$ , $A_{2}$ , and $A_{3}$ represent positive areas, the definite integral equals $(A_{1}+A_{3})-A_{2}$ . Since $A_{2}>A_{1}+A_{3}$ , the value of the definite integral is negative. In this case, its value is $-\sqrt{3}\approx -1.73$ .

The definite integral $\displaystyle\int_{\pi/3}^{5\pi/3}\cos(x)\, dx$ gives a `signed area’. Assuming the symbols $A_{1}$ , $A_{2}$ , and $A_{3}$ represent positive areas, the definite integral equals $(A_{1}+A_{3})-A_{2}$ . Since $A_{2}>A_{1}+A_{3}$ , the value of the definite integral is negative. In this case, its value is $-\sqrt{3}\approx -1.73$ .

In fact, $A_{1}=\displaystyle\int_{\pi/3}^{\pi/2}\cos(x)\, dx=\sin\left(\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{3}\right)=1-\frac{\sqrt{3}}{2}\approx 0.134$ and $A_{3}=\displaystyle\int_{3\pi/2}^{4\pi/3}\cos(x)\, dx=\sin\left(\frac{4\pi}{3}\right)-\sin\left(\frac{3\pi}{2}\right)=-\frac{\sqrt{3}}{2}+1\approx 0.134$ .

On the other hand,

$A_{2}=\displaystyle\int_{\pi/2}^{3\pi/2}|\cos(x)|\, dx=-\displaystyle\int_{\pi/2}^{3\pi/2}\cos(x)\, dx$ $=-\left(\sin\left(\frac{3\pi}{2}\right)-\sin\left(\frac{\pi}{2}\right)\right)=-(-1-1)=2$

since $|\cos(x)|=-\cos(x)$ when $\frac{\pi}{2}\leq x\leq \frac{3\pi}{2}$ .

Therefore,

$\displaystyle\int_{\pi/3}^{5\pi/3}\cos(x)\, dx=(A_{1}+A_{3})-A_{2}=2-\sqrt{3}-2=-\sqrt{3}.$

The First Fundamental Theorem of Calculus Might Seem Like Magic

When you think about these area and signed area applications, the First Fundamental Theorem of Calculus might seem quite “magical”.

Is there any way to understand it intuitively? Yes, an intuitive (but non-rigorous) approach involving so-called infinitesimals (infinitely small quantities) can be used to gain insight. But we will save that topic for later in this post.

In the next main section, however, we will explore the Second Fundamental Theorem of Calculus. The 2nd FTC is more difficult to understand than the 1st FTC when you first encounter it. However, once the main conceptual ideas about it are understood, it is actually less mysterious. Moreover, the 2nd FTC can actually be used to give a relatively “quick” proof of the 1st FTC.

In Lecture 5A from Fall 2016, I start describing how to gain an understanding the Second Fundamental Theorem of Calculus through examples.

Lecture 5A: the Second Fundamental Theorem of Calculus (and Linear Motion)

I start Lecture 5A by considering a linear motion example with a piecewise linear velocity function.

Calculus 2, Lecture 5A: Velocity and Displacement, 2nd Fundamental Theorem of Calculus

The piecewise linear velocity function has the formula $v(t)=\left\{\begin{array}{cl} -2 & \mbox{if } 0\leq t\leq 4 \\ 2t-10 & \mbox{if } 4<t \end{array}\right\}$ .

This means the velocity is negative if $t<5$ and positive if $t>5$ (see the graph in the thumbnail of the video above). If we think of motion to the right as being a positive velocity and motion to the left as being a negative velocity, then the object starts out by moving to the left before turning around at time $t=5$ and moving to the right thereafter.

Now let $D(t)=\displaystyle\int_{0}^{t}v(\tau)\, d\tau$ . By the signed area interpretation of the definite integral, we can derive the following (see the graph in the thumbnail of the video above):

If $0\leq t\leq 4$ , then $D(t)=-(\mbox{Area of rectangle of base t and height 2})=-2t$ .
If $4<t<5$ , then $D(t)=-(\mbox{Area of rectangle base 4 and height 2})-(\mbox{Area of Trapezoid})$ , so $D(t)=-8-(t-4)\frac{2+(-2t+10)}{2}$ . This simplifies to $D(t)=t^{2}-10t+16$ .
And, if $5<t$ , then $D(t)=-8-1+(\mbox{Area of Triangle})=-9+\frac{1}{2}(t-5)(2t-10)$ . This also simplifies to $D(t)=t^{2}-10t+16$ .

Notice that, no matter what $t$ is, we have $D'(t)=v(t)$ . We have used the signed area interpretation of the definite integral to “construct” an antiderivative $D(t)$ of the integrand function $v(t)$ !!

This is what the Second Fundamental Theorem of Calculus is all about: integrals (signed areas) define antiderivatives.

Statement of the Second Fundamental Theorem of Calculus

Here is the formal statement of the 2nd FTC. It is sometimes called the Antiderivative Construction Theorem, which is very apt.

Second Fundamental Theorem of Calculus: Suppose that the function $f$ is continuous on the closed interval $[a,b]$ . Define a function $F$ on $[a,b]$ by the formula $F(x)=\displaystyle\int_{a}^{x}f(t)\, dt$ . Then $F$ is differentiable on $[a,b]$ and $F'(x)=f(x)$ for all $x$ in $[a,b]$ . In other words, $F$ is an antiderivative of $f$ on $[a,b]$ .

This can also be written concisely as follows. If $f$ is continuous near the number $a$ , then

$\frac{d}{dx}\left(\displaystyle\int_{a}^{x}f(t)\, dt\right)=f(x)$ when $x$ is close to $a$ .

In this equation, it is as if the derivative operator $\frac{d}{dx}$ and the integral operator $\displaystyle \int_{a}^{x}\ldots \, dt$ “undo” each other to leave the original function $f$ .

In fact, this “undoing” property holds with the First Fundamental Theorem of Calculus as well. Its equation can be written as $\displaystyle\int_{a}^{b}F'(x)\, dx=F(b)-F(a)$ . Or, if you prefer, we can rearrange and use different letters and write the equation from the 1st FTC as $F(x)=F(a)+\displaystyle\int_{a}^{x}\frac{d}{dt}(F(t))\, dt$ .

Here we see something similar: the integral operator $\displaystyle \int_{a}^{x}\ldots \, dt$ and the derivative operator $\frac{d}{dt}$ “undo” each other to obtain the function $F$ , although we must add on a “constant” $F(a)$ to get equality.

Example 4: An Example to Help You Believe the Second Fundamental Theorem of Calculus

Here is an example to help you believe that the Second Fundamental Theorem of Calculus is true. Suppose $f(t)=\sqrt{1-t^{2}}$ is the speed of a moving object, in meters per second, as a function of time $t$ , in seconds. For the sake of illustration, let’s do something a bit strange and let $t=-1$ be our starting time. So our overall time interval is $-1\leq t\leq 1$ .

Let $T$ be a time with $0<T<1$ . What we want to see is that the definite integral $\displaystyle\int_{-1}^{T}f(t)\, dt$ , when thought of as a function of $T$ , is an antiderivative of $f(T)$ .

Since the graph of $\mbox{speed}=f(t)=\sqrt{1-t^{2}}$ is the upper half of the unit circle, we can compute the definite integral purely with geometry and trigonometry.

In the graph below, the area of the quarter circle on the left is $\frac{1}{4}\pi\cdot 1^{2}=\frac{\pi}{4}$ .

The region on the right can be split into a sector of a circle (a “piece of pie”) and the right triangle shown. By trigonometry, $\sin(\theta)=\frac{T}{1}$ so that $\theta=\sin^{-1}(T)=\arcsin(T)$ . The area of the sector of the circle is therefore $\frac{1}{2}r^{2}\theta=\frac{1}{2}\arcsin(T)$ . Furthermore, the area of the triangle is $\frac{1}{2}\cdot \mbox{base}\cdot \mbox{height}=\frac{1}{2}T\sqrt{1-T^{2}}$ .

The definite integral $\displaystyle\int_{-1}^{T}\sqrt{1-t^{2}}\, dt$ is the area shaded in red. This area, as a function of $T$ , is an antiderivative of $\sqrt{1-T^{2}}$ . The formula for the area is $\frac{\pi}{4}+\frac{1}{2}T\sqrt{1-T^{2}}+\frac{1}{2}\arcsin(T)$ .

Putting all this together, we see that $F(T)=\displaystyle\int_{-1}^{T}\sqrt{1-t^{2}}\, dt=\frac{\pi}{4}+\frac{1}{2}T\sqrt{1-T^{2}}+\frac{1}{2}\arcsin(T)$ .

Now we must check that $F(T)$ an antiderivative of $\mbox{speed}=f(T)$ . Use the Product Rule and Chain Rule to confirm this. Here goes:

$F'(T)=0+\frac{1}{2}\sqrt{1-T^{2}}+\frac{1}{2}T\cdot \frac{1}{2}(1-T^{2})^{-1/2}\cdot (-2T)+\frac{1}{2\sqrt{1-T^{2}}}$ .

We continue to simplify and get $F'(T)=\frac{1-T^{2}}{2\sqrt{1-T^{2}}}-\frac{T^{2}}{2\sqrt{1-T^{2}}}+\frac{1}{2\sqrt{1-T^{2}}}$ . But this simplifies to $F'(T)=\frac{2(1-T^{2})}{2\sqrt{1-T^{2}}}=\sqrt{1-T^{2}}=f(T)$ , as desired.

Also note that $F(-1)=\frac{\pi}{4}+0+\frac{1}{2}\arcsin(-1)=\frac{\pi}{4}-\frac{\pi}{4}=0$ . In other words, the graph of $F$ also goes through the point $(-1,0)$ .

The case where $-1<T<0$ can be handled geometrically as well, though you would want to think of it as $F(T)=\frac{\pi}{4}-G(T)$ , where $G(T)$ would represent an “unshaded” area to the left of the vertical axis. You should get the same simplified formula for $F(T)$ in the end. Give it a try!

This Antiderivative is the Distance Traveled

In fact, the function output $F(T)$ represents the distance traveled from time $t=-1$ to time $t=T$ . Here is an animation that illustrates all this.

The function $F(T)=\displaystyle\int_{-1}^{T}\sqrt{1-t^{2}}\, dt$ is the distance traveled by the blue dot on the vertical axis from time $t=-1$ to time $t=T$ . Notice that the speed is largest near time $T=0$ and smallest near times $T=\pm 1$ .

The function $F(T)=\displaystyle\int_{-1}^{T}\sqrt{1-t^{2}}\, dt$ is the distance traveled by the blue dot on the vertical axis from time $t=-1$ to time $t=T$ . Notice that the speed is largest near time $T=0$ and smallest near times $T=\pm 1$ .

Again, the purpose of these examples is to help you believe the truth of the Second Fundamental Theorem of Calculus. Let’s consider another example from the lectures before we explore proofs of the First and Second Fundamental Theorems of Calculus.

Lectures 5B and 6: More Integral Examples

The main content of both Lecture 5B and Lecture 6 is to work through more integral examples. Additionally, in the first 13 minutes of Lecture 5B, I review the Second Fundamental Theorem of Calculus and introduce parametric curves, while the last 8 minutes of Lecture 6 are spent extending the 2nd FTC to a problem that also involves the Chain Rule.

Here is Lecture 5B.

Calculus 2, Lecture 5B: Fundamental Theorem of Calculus, Parametric Curves, and Some Trigonometric Integrals

And here is Lecture 6. It is a short lecture because the students took a “mini-exam” that day as well.

Calculus 2, Lecture 6: Trigonometric Integrals, Fundamental Theorem of Calculus

In this blog post, we will just focus on the extended problem where the Chain Rule is needed. It’s a pretty “wild” example: buckle up!

Example 5: A Function Defined by an Integral Whose Derivative Must Be Computed with the Chain Rule as well as the 2nd FTC

The problem is to compute the following very complicated-looking derivative:

$\frac{d}{dx}\left(\displaystyle\int_{7}^{x^{11}}\cos\left(\sqrt{t^{5}+3t^{2}}\right)\, dt\right)$

Before taking this derivative, let’s take the time to graph the integrand function $f(t)=\cos\left(\sqrt{t^{5}+3t^{2}}\right)$ , the appropriate signed area between its graph and the horizontal axis, and the graph of the function $G(x)=\displaystyle\int_{7}^{x^{11}}\cos\left(\sqrt{t^{5}+3t^{2}}\right)\, dt$ .

Suffice it to say that all this should not be attempted without technology.

The Graph of the Integrand

First, so there is no misunderstanding, we plot $f(t)$ on a very “wide” window to show that, while it oscillates like a cosine function, the oscillations do indeed get faster and faster as $t$ increases since $\sqrt{t^{5}+3t^{2}}$ increases faster and faster as $t$ increases.

The graph of $f(t)=\cos\left(\sqrt{t^{5}+3t^{2}}\right)$ over the interval $[0,8]$ .

The graph of $f(t)=\cos\left(\sqrt{t^{5}+3t^{2}}\right)$ over the interval $[0,8]$ .

Now zoom in near $t=7$ , which is the lower limit of integration in the formula for $G(x)=\displaystyle\int_{7}^{x^{11}}\cos\left(\sqrt{t^{5}+3t^{2}}\right)\, dt$ . The graph of $f(t)$ , as well as the shaded signed area representing the value of $G(x)$ , is shown below. There is a small negative signed area contribution to the integral from $t=7$ until $t\approx 7.00364$ that is difficult to see.

The graph of $f(t)=\cos\left(\sqrt{t^{5}+3t^{2}}\right)$ over the interval $[7,7.5]$ . The shaded signed area represents $G(x)=\displaystyle\int_{7}^{x^{11}}\cos\left(\sqrt{t^{5}+3t^{2}}\right)\, dt$ when $x^{11}=7.3$ (so $x=7.3^{1/11}\approx 1.19807$ ).

The Graph of the Integral

And here now is the graph of $G(x)=\displaystyle\int_{7}^{x^{11}}\cos\left(\sqrt{t^{5}+3t^{2}}\right)\, dt$ very close to $x=7^{1/11}\approx 1.19351$ . Note that $G(1.19351)\approx G(7^{1/11})=0$ . Furthermore, note that $G$ has two local maxima before $x=7.3^{1/11}\approx 1.19807$ . This is consistent with the fact that the $G(x)$ is “accumulating” signed areas for the graph of $f(t)$ above, and that the graph of $f(t)$ above changes sign from positive to negative twice on the interval $[0,7.3]$ . Also note that $G\left(7.3^{1/11}\right)\approx G(1.19807)\approx 0.02135$ . This is the accumulated signed area shown in the graph shown above where $x^{11}=7.3$ .

The graph of *$G(x)=\displaystyle\int_{7}^{x^{11}}\cos\left(\sqrt{t^{5}+3t^{2}}\right)\, dt$* over the interval from $x=7^{1/11}\approx 1.19351$ to approximately $7.3^{1/11}\approx 1.19807$ . Its oscillations mirror the fact that the signed area for the graph of $f(t)$ above oscillate between positive an negative contributions to the integral.

When you think about the graph and the formula for $G(x)$ , you should realize that the values of $G(x)$ have to be approximated using numerical (approximate) integration. The function $G(x)$ has no simple formula that does not involve an integral sign!!

In spite of this, we can still use the 2nd FTC and the Chain Rule to find a (relatively) simple formula for $G'(x)$ !!

The Derivative of $G(x)$

The function $G$ is really the composition of two functions. Let $F(x)=\displaystyle\int_{7}^{x}\cos\left(\sqrt{t^{5}+3t^{2}}\right)\, dt$ (note the new upper limit of integration) and $H(x)=x^{11}$ . Then $G(x)=F(H(x))$ .

By the Chain Rule $G'(x)=F'(H(x))\cdot H'(x)$ . But $H'(x)=11x^{10}$ and, by the Second Fundamental Theorem of Calculus, $F'(x)=\cos\left(\sqrt{x^{5}+3x^{2}}\right)$ .

Therefore,

$G'(x)=11x^{10}\cos\left(\sqrt{(x^{11})^{5}+3(x^{11})^{2}}\right)=11x^{10}\cos\left(\sqrt{x^{55}+3x^{22}}\right)$

Isn’t calculus amazing! Who would’ve thunk it?!? Praise God for the details of His creation and His generosity in giving us minds to contemplate and compute such things!

If we graph $G'(x)$ for $7^{1/11}\leq x\leq 7.3^{1/11}$ , we should see that it has five zeros on this interval. These correspond to the five critical points of $G(x)$ on this interval in the graph above (where $G$ has local extreme points). The first critical point for the graph of $G$ is so close to the left endpoint that you can’t really see it (it’s “covered” by the big red dot on the left).

The graph of $G'(x)$ for $7^{1/11}\leq x\leq 7.3^{1/11}$ . It has five zeros on this interval which correspond to the five extreme points of the graph of $G(x)$ on the same interval above.

The graph of $G'(x)$ for $7^{1/11}\leq x\leq 7.3^{1/11}$ . It has five zeros on this interval which correspond to the five extreme points of the graph of $G(x)$ on the same interval above.

Applying these theorems was a lot of fun. But why are the First and Second Fundamental Theorems of Calculus true? We end this blog post by looking at proofs and intuitive justifications.

Proof of the First Fundamental Theorem with the Second Fundamental Theorem

We’ll start by doing something kind of strange. Let’s use the Second Fundamental Theorem of Calculus to prove the First Fundamental Theorem of Calculus. After that, we’ll describe the idea of a proof of the Second Fundamental Theorem of Calculus. Finally, we’ll think about the theorems intuitively using infinitesimals.

Technically speaking, we also need a fact about any two antiderivatives of the same function over an interval: they must be vertical translations of each other (they must “differ by a constant”). This is something that can be proved with the Mean Value Theorem.

Proof of the First Fundamental Theorem of Calculus

Suppose that $F$ is an antiderivative of $f$ on the interval $[a,b]$ . Let $G$ be defined by $G(x)=\displaystyle\int_{a}^{x}f(t)\, dt$ . By the Second Fundamental Theorem of Calculus, we know that $G'(x)=f(x)$ for all $x\in [a,b]$ .

Therefore, $G$ is an antiderivative of $f$ on $[a,b]$ . But this means that there is a constant $C$ such that $G(x)=F(x)+C$ for all $x\in [a,b]$ .

Now the formula for $G$ implies that $G(b)-G(a)=\displaystyle\int_{a}^{b}f(t)\, dt-\displaystyle\int_{a}^{a}f(t)\, dt$ . But $\displaystyle\int_{a}^{a}f(t)\, dt=0$ . Hence, $G(b)-G(a)=\displaystyle\int_{a}^{b}f(t)\, dt$ .

But $G(b)-G(a)=(F(b)+C)-(F(a)+C)=F(b)-F(a)$ as well.

Therefore, $\displaystyle\int_{a}^{b}f(t)\, dt=F(b)-F(a)$ and we are done (this also implies that $\displaystyle\int_{a}^{b}f(x)\, dx=F(b)-F(a)$ since variable names don’t matter). Q.E.D.

That was kind of a “slick” proof. Evidently the “hard” work must be involved in proving the Second Fundamental Theorem of Calculus.

We do not give a rigorous proof of the 2nd FTC, but rather the idea of the proof.

Idea of the Proof of the Second Fundamental Theorem of Calculus

Let $f$ be continuous on the interval $[a,b]$ . Define a function $F$ on $[a,b]$ by the formula $F(x)=\displaystyle\int_{a}^{x}f(t)\, dt$ (we are implicitly assuming this integral “exists” for all $x$ here).

Our goal is to show that $F'(x)=f(x)$ . By the definition of the derivative, we know that $F'(x)=\displaystyle\lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}$ (assuming this limit exists for all $x$ and assuming an appropriate one-sided limit if $x=a$ or $x=b$ ).

By interval-related properties of integrals (draw a picture of appropriate areas under the graph of $f(t)$ to help you understand this), we can write

$\frac{1}{h}(F(x+h)-F(x))=\frac{1}{h}\left(\displaystyle\int_{a}^{x+h}f(t)\, dt-\displaystyle\int_{a}^{x}f(t)\, dt\right)=\frac{1}{h}\displaystyle\int_{x}^{x+h}f(t)\, dt$

But if $h\approx 0$ , then $f(t)\approx f(x)$ for all $t$ between $x$ and $x+h$ . Therefore, $\displaystyle\int_{x}^{x+h}f(t)\, dt\approx f(x)\cdot ((x+h)-h)=h\cdot f(x)$ . Hence, $F'(x)\approx \frac{1}{h}\cdot h\cdot f(x)=f(x)$ when $h\approx 0$ .

In the limit as $h\longrightarrow 0$ , this approximation becomes exact (the continuity of $f$ guarantees this). Therefore, $F'(x)=f(x)$ . Since $x\in [a,b]$ was arbitrary, this is true over the entire interval. Q.E.D.

Intuitive Approaches to Understanding These Theorems

Now let’s approach these theorems intuitively. One way to do so is to make use of “infinitely small quantities”, called infinitesimals. Right away, we should give a caveat: in the standard real number system, there are no such things as infinitesimals. They are “convenient fictions” that help to give intuition and to derive formulas in calculus while avoiding limits.

The 1st FTC

Let’s start with the First Fundamental Theorem of Calculus.

Consider an example. Let $f(t)=t^{2}$ . Moreover, let’s assume this is a speed, in meters per second, which is the derivative of a distance traveled, in meters. In fact, we can write $\frac{dy}{dt}=F'(t)=f(t)=t^{2}$ , where $F(t)$ is the distance traveled, in meters, from time 0 to time $t$ .

Imagine now that $t$ is some specific moment in time. For example, maybe $t=3.2$ as shown in the graph below.

The graph of the speed $\frac{dy}{dt}=f(t)=t^{2}$ with a specific value of $t$ highlighted. Imagine the shaded thin strip has ‘infinitesimal’ width $dt$ . In fact, imagine $dt$ is so small that this thin strip is actually a rectangle with base $dt$ and height $f(t)=t^{2}$ . Its area is then the infinitesimally small product $dy=f(t)dt=t^{2}dt$ . This also gives an infinitesimally small distance traveled (since distance equals rate times time for constant rates).

The graph of the speed $\frac{dy}{dt}=f(t)=t^{2}$ with a specific value of $t$ highlighted. Imagine the shaded thin strip has ‘infinitesimal’ width $dt$ . In fact, imagine $dt$ is so small that this thin strip is actually a rectangle with base $dt$ and height $f(t)=t^{2}$ . Its area is then the infinitesimally small product $dy=f(t)dt=t^{2}dt$ . This also gives an infinitesimally small distance traveled (since distance equals rate times time for constant rates).

Think of The Thin Strip as having an Infinitesimal Area (Infinitesimal Distance Traveled) and Add Up Those Areas By Integrating

In this figure, imagine the shaded thin strip has “infinitesimal” width $dt$ . In fact, imagine $dt$ is so small that this thin strip is actually a rectangle with base $dt$ and height $f(t)=t^{2}$ . Its area is then the infinitesimally small product $dy=f(t)dt=t^{2}dt$ . This also gives an infinitesimally small distance traveled (since distance equals rate times time for constant rates).

For a given “ending time” $b>0$ , the definite integral $\displaystyle\int_{0}^{b}f(t)\, dt=\displaystyle\int_{0}^{b}t^{2}\, dt$ gives the area under this curve between $t=0$ and $t=b$ . In fact, the definite integral can be imagined as “adding” all the infinitesimally small products $dy=f(t)dt=t^{2}dt$ as $t$ ranges from 0 to $b$ .

But, as already mentioned, these infinitesimally small products are also infinitesimally small distances traveled. When we add them up, we have to get the total distance traveled $y=F(b)=F(b)-F(0)$ !

Therefore, $y=F(b)-F(0)=\displaystyle\int_{0}^{b}f(t)\, dt=\displaystyle\int_{0}^{b}F'(t)\, dt$ . This is the First Fundamental Theorem of Calculus (with $a=0$ )!

This can take a long time to sink in. Don’t be too upset if you don’t “get it” at first. Keep coming back to it and thinking about it over and over again. It is worth it!

The equation $y=F(b)-F(a)=\displaystyle\int_{a}^{b}f(t)\, dt=\displaystyle\int_{a}^{b}F'(t)\, dt$ can be summarized this way: to find the total change in a function, you must integrate its rate of change.

The 2nd FTC

We can think about the Second Fundamental Theorem of Calculus with infinitesimals as well.

The key observation is this: the value of $F'(t)=f(t)$ determines how “fast” the infinitesimal areas $dy=F'(t)dt=f(t)dt$ “accumulate”. Assuming positive values, the larger $F'(t)=f(t)$ is, the faster the areas accumulate for infinitesimal changes in the input.

In fact, it is a direct proportionality. If $F'(x_{1})=f(x_{1})$ is twice as big as $F'(x_{0})=f(x_{0})$ , then the infinitesimal area near $t=x_{1}$ are twice as big as the corresponding infinitesimal areas near $t=x_{0}$ (for the same infinitesimal change $dt$ in the input).

Therefore, if $F(x)=\displaystyle\int_{a}^{x}f(t)\, dt$ , the rate of change of this function at $x$ , namely $F'(x)$ , must equal $f(x)$ .

Once again, it takes a long time for this to sink in. It is best thought about in terms of animations.

Revisit the animation of the distance traveled when the speed is $f(t)=\sqrt{1-t^{2}}$ farther above.

Since, for instance, $f(-0.9)\approx 0.436$ and $f(-0.4895)\approx 0.872=2\cdot 0.436$ , infinitesimal areas (distances) near time $t=-0.4895$ are accumulating approximately twice as fast as infinitesimal areas (distances) near time $t=-0.9$ . We could represent this symbolically as $dy\approx 0.872dt$ near $t=-0.4895$ and $dy\approx 0.436dt$ near $t=-0.9$ .

Another way to think of it is this: the graph of the distance traveled function $F(t)$ is approximately twice as steep near $t=-0.4895$ as it is near $t=-0.9$ . In other words, the tangent line has double the slope at $t=-0.4895$ compared to $t=-0.9$ .

Continuing to think about examples like this will help you to get it. Keep at it!