Fundamental theorem of calculus

The fundamental theorem: Two flavors

The fundamental theorem of calculus (FTOC) is divided into parts. Often they are referred to as the "first fundamental theorem" and the "second fundamental theorem," or just FTOC-1 and FTOC-2.

Together they relate the concepts of derivative and integral to one another, uniting these concepts under the heading of calculus, and they connect the antiderivative to the concept of area under a curve.

What they say

FTOC-1 says that the process of calculating a definite integral to find the area under a curve, say between x=a and x=b, is nothing more than finding the difference in the antiderivative of the integrand evaluated at points a and b. That's actually quite a remarkable result.

FTOC-2 is a little more abstract, but very important. It lays out the definite integral as a function that can accumulate area under a curve by placing the variable of a function as one of the limits of an integral (usually the upper limit). This concept is important because it allows us to create a whole new class of useful functions that are only defined by the integral — integral-defined functions. One example is the Gaussian distribution function used in statistics and probability, but many others exist.

We'll start with FTOC-1 and in this section we'll use capital letters for functions that are antiderivatives of their lower-case counterparts. So from here on you can assume that F(x) is the antiderivative of f(x), G(x) is the antiderivative of g(x), and so on. Here's the statement of FTOC-1:

Note that some sources swap the numbering of FTOC-1 and 2 from what I use here. It doesn't matter ... it's the concepts that are important.

Fundamental theorem , part 1

If $f(x)$ is a continuous function on the interval $[a, b]$ and $F(x)$ is an antiderivative of $f(x),$ i.e. $F'(x) = f(x),$ then

$$\int_a^b f(x) \, dx = F(b) - F(a)$$

The definite integral is the area under a curve between $x = a$ and $x = b.$ The first fundamental theorem reduces that Riemann sum to the difference between antiderivatives evaluated at $x = a$ and $x = b.$ (That's a pretty remarkable result, if you think about it.)

Proof

We begin by converting the difference F(b) - F(a) into a sum of smaller differences. The figure below shows graphically how this is done. If we plot F(x), we can divide it into segments with endpoints x_o = a, x₁, x₂, ... and so on. I've only gone up to x₅ here, but we could make these segments as narrow as we'd like. We'll call the endpoints a and b, where a = x_o and b = x₅

If we calculate the widths of the segments along the y-axis, we find widths of F(b) - F(x₄), F(x₄) - F(x₃), and so on. Notice (right column) that if we add all of these segments, we get F(b) - F(a) because of all the ± cancellations.

The sum:

$$ \require{cancel} \begin{align} F(\color{#E90F89}{b}) &- \cancel{F(x_4)} \\[3pt] \cancel{F(x_4)} &- \cancel{F(x_3)} \\[3pt] \cancel{F(x_3)} &- \cancel{F(x_2)} \\[3pt] \cancel{F(x_2)} &- \cancel{F(x_1)} \\[-3pt] \cancel{F(x_1)} &- F(\color{#E90F89}{a}) \\[1pt] \underline{-6} \\[1pt] F(\color{#E90F89}{b}) &- F(\color{#E90F89}{a}) \end{align}$$

So we have

$$F(b) - F(a) = \sum_{i = 1}^5 F(x_i) - F(x_{i - 1})$$

where the summation is [F(x₁) - F(x_o)] + [F(x₂) - F(x₁)] + ... + [F(x₅) - F(x₄)]. Now we can in fact make any number of these partitions, so let's just make this small change to reflect that:

$$F(b) - F(a) = \sum_{i = 1}^N F(x_i) - F(x_{i - 1})$$

So far we have restated the right side of the FTOC-1, F(b) - F(a), as a sum of smaller divisions of the antiderivative function.

The next step is to recall the mean value theorem, which says that for every continuous function on an interval [a, b], there exists a number, c, at which the derivative (slope) of the function, f'(c) is equal to the average slope between a and b:

Remember that we really don't care where c is, just that it exists in the interval of interest. We'll rearrange that to

$$f(b) - f(a) = f'(c)(b - a)$$

Now the mean value theorem guarantees the existence of the point c on any interval, including [x_i, x_i-1], so we can rewrite the MVT like this: There must exist some c_i in [x_i, x_i-1] such that F(x_i) - F(x_i-1) = F'(c_i)(x_i - x_i-1). This is just the MVT re-expressed for each of our sub-intervals of [a, b].

Here that is again,

$$F(x_i) - F(x_{i - 1}) = F'(c)(x_i - x_{i - 1})$$

and if we remember that because F(x) is an antiderivative of f(x), then F'(x) = f(x), we get

Now if we replace x_i - x_i-1 with Δx, and sum each side from 1 to N (the number of partitions), we get

$$\sum_{i = 1}^N F(x_i) - F(x_{i - 1}) = \sum_{i = 1}^N f(c_i) \Delta x$$

In the first part of the proof, we showed that the sum on the left is just F(b) - F(a), so we have

$$F(b) - F(a) = \sum_{i = 1}^N f(c_i) \Delta x$$

Finally, what's on the right is just a Riemann sum integral of the area under f(x), where the MVT guarantees that there is some point c somewhere in each partition, no matter what its width, and Δx is just the width of the partition. As the width of those partitions (rectangles) goes to zero (Δx → dx), the interval Δx squeezes down on c and we get the integral of the function:

$$F(b) - F(a) = \int_a^b f(x) `, dx$$

Quod erat demonstrandum

The value of FTOC-1 cannot be understated

It's worth thinking about the first fundamental theorem of calculus one more time. It says the the integral representing the area between a function and the x-axis, an infinite sum of infinitely narrow rectangles, can be reduced to a simple difference of an antiderivative taken at the endpoints of the domain of integration [a, b].

The second part of the FTOC (FTOC-2)

The second part of the fundamental theorem is of the more difficult bits of calculus to wrap your head around, so be sure to give it some time, look at it often and work through the proofs and some examples.

Like many concepts that are difficult at first, the more you look at it and work with it, the easier it gets, so hang in there.

FTOC-2

If a function f is continuous on the interval [a, b], then f has an antiderivative in [a, b]. In fact, the function

$$F(x) = \int_a^x f(t) \, dt$$

qualifies as one.

$$F'(x) = \frac{d}{dx} \left[ \int_a^x f(t) \, dt \right] = f(x)$$

Well, this is a very odd statement. our independent variable, x, is now the upper limit of the integral, and we are meant to treat t as a dummy variable, to be used for integration purposes only. The FTOC-2 says formally that differentiation and integration are inverse operations. Notice in the last line of equations in the box above that one need not actually do the integral to find its derivative. You only need to rewrite f(t) with x inserted for t.

An area-accumulation function

Another way to look at it is that we've invented a new kind of function, G(x), an integral-defined function with its independent variable as one of the limits. It's an area-accumulation function: As x grows, the amount of area under the curve increases.

This graph shows how we can use the integral to accumulate area under the function $f(t) = x^3 - 6x^2 + 12x + 2.$ – just an arbitrary function, nothing special about it.

Slide left and right with your mouse or finger to see how area is accumulated by this function.

Derivative of an integral function – a graphical interpretation

Here's a nice graphical interpretation of why the FTOC-2. Take a function f(t) and graph it. Then it's easy to interpret the integrals between a and x, & a and x+h as areas:

Now if we focus on the area between x and x + h, we can express that area two different ways:

The integral of the area is

$$ \begin{align} A &= \int_a^x \, f(t) \, dt \\[5pt] &= F(x) - F(a), \end{align}$$

where F(t) is an antiderivative of f(t).

Graphically, the picture is

The area is also approximately equal to f(x) · h,

$$F(x + h) - F(x) \approx f(x) \cdot h$$

Now dividing by h gives us an expression on the left that looks like the derivative:

$$\frac{F(x + h) - F(x)}{h} \approx f(x)$$

If we take the limit as h→0, we see that the derivative of the area function is just f(x):

$$\lim_{x\to 0} \frac{A(x + h) - A(x)}{h} = f(x)$$

Area accumulation as a function | An example

The graphs below should help you understand the difference between a function and that function as used to make an integral-defined function. The panel on the left shows f(x) = sin(x²), which does not have an analytic integral (you can't just solve it on paper – it has to be done numerically). You can see that it has regions of positive and negative area, the orange shaded regions.

If you imagine moving our vertical line along the independent variable x, sweeping out area under the curve, that the total area would oscillate as we add negative and positive areas. It's not a stretch to see how the purple curve could be a graph of that area as a function of x. The purple graph is the integral-defined function. It's actually a pretty important function in the field of optics, and it's called the Fresnel (pronounced fruh · nel') function.

Example 1

Consider the integral-defined function $G(x) = \int_0^x \, 2t \, dt,$ and find $G'(x)$

Solution: The simple solution to this problem is that G'(x) is just 2x, found by simply replacing x with t. But let's look in more detail.

In this example, we can easily compare the area defined by the integral with the area calculated geometrically. The area of the green triangle under the linear function f(t) = 2t is (1/2)(x)(2x) = x².

If we integrate (note that the lower limit is zero), then take the derivative of the result, after evaluating the limits, we get:

The result of the definite integral is x², and its derivative is 2x, just what we knew already (but nice to confirm it!).

The lower limit of integration doesn't matter

Now we can show that the lower limit of integration is irrelevant to G'(x) by setting the lower limit of the integral-definition of G(x) to some number, a, instead of zero, where 0 < a < x.

$$G(x) = \int_a^b 2t \, dt$$

The picture now looks like this.

The solution of the definite integral is now x² - a² and its derivative is still 2x. The constant turns out not to matter, so we can conclude that the lower limit of integration in these cases doesn't matter.

The lower limit of integration of an integral-defined function is irrelevant when taking the derivative of the function.

Proof of the FTOC-2

The FTOC-2 posits that:

$$G'(x) = \frac{d}{dx} \left[ \int_a^x f(t) \, dt \right] = f(x)$$

So we need to prove that G'(x), as defined, is equal to f(x). To do so, we define two antiderivatives, G(x) and G(z) according to FTOC-2:

$$G(x) = \int_a^x f(t) \, dt \; \; \color{#E90F89}{\text{&}} \; \; G(z) = \int_a^z f(t) \, dt$$

Now we're going to work toward a merging of the average value of an integral with the definition of a derivative, so the next step is to take the difference between G(z) and G(x), and we'll assume that z > x.

$$G(z) - G(x) = \int_a^z f(t) \, dt - \int_a^x f(t) \, dt$$

We can use two of the properties of definite integrals to flip the limits of integration on the second integral, then combine them into one:

$$= \int_a^z f(t) \, dt + \int_x^a f(t) \, dt = \int_x^z f(t) \, dt$$

Now the average value of that integral is just the sum of all the f(t)'s over the interval, divided by the interval itself, (z - x).

We'll name that average f(c) (with no particular meaning intended for the letter 'c' except that we're heading toward using the mean value theorem)

$$f(c) = \frac{1}{z - x} \int_x^z f(t) \, dt$$

A little rearrangement of the last expression gives us

$$\color{#E90F89}{\longrightarrow} \; \int_x^z f(t) \, dt = (z - x) f(c)$$

Now here's the crux: There's another way to calculate that same average. It's just the change in rise of the antiderivatives over the change in the independent variable t. It is:

$$\frac{G(z) - G(x)}{z - x} = f(c)$$

This looks like a derivative; it's just lacking the limit as x → z to give G'. Recall that we're trying to show that G'(x) = f(x). If we take that limit on both expressions for the average of the integral, we end up "squeezing" f(c) between x and z. After all, the average will always lie between the two extremes. At the limit where x = z, f(c) = f(x), and we've proved our theorem.

$$G'(x) = \lim_{x\to x} \frac{G(z) - G(x)}{z - x} = \lim_{z\to x} f(c)$$

And finally,

A second proof of FTOC-1

Now that we've proved FTOC-2, we can use it in a simpler proof of FTOC-1. Here it is.

$$\text{If} \; F'(x) = f(x), \; \text{then} \; \int_a^b f(x) dx = F(b) - F(a)$$

Now we've proved that G(x) is antiderivative of f(x),

$$G(x) = \int_a^x f(t) \, dt$$

so F(x), postulated to be an antiderivative of f(x), must be equal to G(x) to within an additive constant:

$$F(x) = G(x) + C$$

Then we can simply write:

$$ \begin{align} F(b) - F(a) &= [G(b) + C] - [G(a) + C]\\[5pt] &= G(b) - G(a) \end{align}$$

which we expand to

$$= \int_a^b f(x) dx - \int_a^a f(x) dx = \int_a^b f(x) dx,$$

where the second integral is zero, and we have proved the FTOC-1.

Example 2

Find the integral and its derivative:

Solution: Let's first find the integral in the straightforward way, using the power rule of integration and evaluating the limits:

$$\int_0^x t^2 \, dt = \frac{t^3}{3} \bigg|_0^3 = \frac{x^3}{3}$$

Now the derivative of the integral is:

$$\frac{d}{dx} \frac{x^3}{3} = x^2$$

which is just the integrand of our original integral, with t replaced by x. And that will be the case in all such problems. All together it looks like this:

$$\frac{d}{dx} \left[ \int_0^x t^2 \, dt \right] = x^2$$

Now recall that we showed above that the lower limit of integration doesn't matter. Let's confirm that here by replacing the lower limit of integration with t = a.

$$\int_a^x t^2 \, dt$$

Do the integral in the same way, except now we get the answer above with a constant (-a³/3) added to it:

$$\int_a^x t^2 \, dt = \frac{t^3}{3} \bigg|_a^x = \frac{x^3}{3} - \frac{a^3}{3}$$

Now if we take the derivative, it's the same because the second term is constant. What we find is that the lower limit just doesn't matter in this kind of expression of FTOC-2.

$$\frac{d}{dx} \left[ \frac{x^3}{3} - \frac{a^3}{3} \right] = x^2$$

Putting it all together, the statement is:

$$\frac{d}{dx} \left[ \int_0^x t^2 \, dt \right] = x^2$$

For many FTOC-2 problems, the solution is deceptively obvious:

$$F'(x) = \frac{d}{dx} \left[ \int_a^x f(t) \, dt \right] = f(x)$$

Example 3 — The chain rule

Find the derivative $\frac{d}{dx} \int_0^{x^2} ln(t) \, dt$

Solution: Notice that in this integral-defined function, we actually have a function in a function, or a composite function. Consider the functions g(x) and h(x):

$$g(x) = \int_0^x ln(t) \, dt \; \; \color{#E90F89}{\text{&}} \; \; h(x) = x^2$$

Now we see that our function of interest here is just f(x) = g(h(x)):

$$g(h(x)) = \int_0^{x^2} ln(t) \, dt$$

In order to find the derivative of g(x), we must use the chain rule. We can make this clear by making the substitution u = x². That gives us

$$g'(x) = \frac{d}{du} \left[ \int_0^u ln(t) \, dt \right] \frac{du}{dx}$$

Solving the first derivative with the FTOC-2 and re-substituting x² for u in the second derivative gives

$$ \begin{align} &= ln(u) \cdot \frac{d}{dx} x^2 \\[5pt] &= 2x \, ln(x^2) \end{align}$$

These problems can be tricky, but they're do-able if you remember two things: (1) Using the FTOC-2 on the inner function seems too easy, but it's valid, and (2) You must remember exactly how the chain rule works.

Practice problems

Find G'(x) (the first derivative) of each of the following integral-defined functions.

1.	$$G(x) = \int_a^x \frac{t^4}{4} \, dt$$ Solution $$G'(x) = \frac{x^4}{4}$$
2.	$$G(x) = \int_{-\pi}^x sin(\theta) \, d\theta$$ Solution $$G'(x) = sin(x)$$

3.	$$G(x) = \int_a^x cos(t) \ dt$$ Solution $$G'(x) = cos(x)$$
4.	$$G(x) = \int_{-1}^x \frac{dt}{t}$$ Solution $$G'(x) = \frac{1}{x}$$

Find the derivative of each of the following integral-defined functions (chain rule!).

5.	$$\frac{d}{dx} \int_{-2}^{2x^2} sin(t) \, dt$$ Solution $$4x \cdot sin(x)$$
6.	$$\frac{d}{dx} \int_2^{2x} \frac{1}{\sqrt{t^3 + 1}} \, dt$$ Solution $$\frac{2}{\sqrt{x^3 + 1}}$$

$$\frac{d}{d \theta} \int_0^{tan(\theta)} sec^2 (t) \, dt$$

Solution

$$sec^2(\theta) \cdot sec^2(\theta) = sec^4(\theta)$$

$$\frac{d}{dx} \int_{sin(x)}^{cos(x)} 2 sin(t) \, dt$$

Solution

Because both limits are functions of x, we need to re-experss the integrand as the sum of two integrals:

$$ \begin{align} &= \frac{d}{dx} \int_{sin(x)}^a 2 sin(t) dt + \frac{d}{dx} \int_a^{cos(x)} 2 sin(t) dt\\[5pt] & = \frac{d}{dx} \int_a^{cos(x)} 2 sin(t) dt - \frac{d}{dx} \int_a^{sin(x)} 2 sin(t) dt\\[5pt] & = -2 sin^2(x) - 2 sin(x) cos(x) \end{align}$$

9.	If $f(x) = \int_2^{2x} \frac{1}{\sqrt{t^3 + 1}} \, dt,$ then $f'(1) = ?$ Solution $$ \begin{align} f'(x) &= \frac{2}{\sqrt{x^3 + 1}} \\[5pt] f'(1) &= \frac{2}{\sqrt{2}} = \sqrt{2} \end{align}$$
10.	Find the approximate average rate of change of the function $f(x) = \int_0^x sin(t^2) \, dt$ over the interval [1, 3]. Solution $$ \begin{align} f'(x) &= sin(x^2) \\[5pt] f'(1) &= sin(1) = 0.8415 \\[5pt] f'(3) &= sin(9) = 0.4122 \end{align}$$ $$\text{average} \; = \frac{0.8145 + 0.4122}{3 - 1} = 0.672,$$
11.	On what interval is the graph of $g(x) = \int_0^x sin(2t) \, dt$ both decreasing and concave-upward? Solution
12.	If $g(x) = \int_{\pi/2}^x cos(t) \, dt,$ find the maximum value of g on the closed interval [-π, π]. Solution
13.	If $g(x) = \int_0^x cos(e^{t/2}) \, dt$ for -1 ≤ x ≤ 4, find the instantaneous rate of change of g with respect to x at x - 4. Solution $$ \begin{align} g'(x) &= cos(e^{x/2}) \\[5pt] g'(4) &= cos(e^{4/2}) = cos(e^2) = 0.448 \end{align}$$

A whole new class of useful functions — integral-defined functions

We can use the FTOC-2 to create a bunch of new and useful new functions. One is the Gaussian function, more commonly known as the bell-shaped curve or bell curve, that we use in probability and statistics. It looks like the curve plotted below. A stripped-down version of the equation is:

$$G(x) = \int_0^x \, e^{-t^2} \, dt$$

You can read a lot more about this function in the section on probability distributions. What's important about it for our purpose here is the area under the curve (which is symmetric across the line x=0). The area between the limits -∞ and ∞ should equal one because it represents the total probability of an event happening at all, and we often include other factors to "normalize" it, or to force the total area under the curve to be 1. The ratio of any lesser area, like the one between ±a in the plot below, to that total is equal to the probability of an event occuring.

This integral can't be done analytically (with paper and pencil) – it has to be done by numerical methods, but we can still easily find its first and second derivatives through FTOC-2, and thus plot the function very well.

The Fresnel function

Another important curve defined by an integral function is the Fresnel function (fruh · nel'), graphed here on the right. We also considered it above. This function is very important in certain kinds of optics applications.

The Fresnel cosine function is also used frequently, depending on the situation.

posit

To posit something is to propose it as an explanation.

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