### The fundamental theorem(s) of calculus – 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 turns the definite integral into a function, one with the independent variable as the upper limit of integration, that accumulates area under a curve. 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 such example is the Gaussian distribution function used in statistics and probability, but many exist. Most importantly, the FTOC-2 establishes that differentiation and integration are inverse procedures.

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

The first part of the fundamental theorem says that a definite integral, representing the area under a curve between two points, a & b, in its domain, which we understand to be a Riemann sum taken to the limit of the sum of an infinite number of rectangles of infinitessimal width, is just the simple difference of two antiderivatives, F(b) and F(a):

### 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 xo = 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 = xo 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.

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(xo)] + [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 [xi, xi-1], so we can rewrite the MVT like this: There must exist some ci in [xi, xi-1] such that F(xi) - F(xi-1) = F'(ci)(xi - xi-1). This is just the MVT re-expressed for each of our sub-intervals of [a, b].

Here that is again,

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

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 xi - xi-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, and Δx is just the width of the partition. As the width of those partitions (rectangles) goes to zero (Δx dx), we get the integral of the function:

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

Quod erat demonstrandum

#### Area under a curve

FTOC-1
The area bounded by a smooth curve f(x), the x-axis, and the lines x=a and x=b is just the difference between the antiderivative of f(x) evaluated at points a and b.

where F'(x) = f(x).

#### 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].

### Part the second — 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.

#### Fundamental theorem – part 2

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

Let the function be that antiderivative. Then:

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.

The last line of the box expresses three layers of functions and operations. Inside is the function f(t), a 1:1 function that assigns one y-coordinate to every value of t.

That is inside of an integral function, with independent variable x. That integral function calculates an area below f(t) between limits a and x. Finally, all of that is inside of a derivative.

### 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.

### Derivative of an integral function – a graphical interpretation

Here's a nice graphical interpretation of why the second FTOC works. 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 area is approximately equal to f(x) · h,

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

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

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

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

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

### 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 (orange) 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.

The

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: f(t) = 2t

In this example, we can easily compare the area defined by the integral with the area calculated geometrically. The area of the purple 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:

\begin{align} G'(x) &= \frac{d}{dx} \left[ \int_0^x \, 2t \, dt \right] \\ \\ &= \frac{d}{dx} \, x^2 = 2x = f(x) \end{align}

### The lower limit of integration doesn't matter ...

Now let's do that same problem but this time we'll use a non-zero lower limit of integration, a. We'll take the derivative with respect to x of this integral:

The graph is shown below, and the full integral is worked out. When the limits are evaluated, the value of the integral is x2 - a2.

Now we take the derivative with respect to x and the derivative of the lower limit, just the constant a2, is zero. The lower limit doesn't matter.

\begin{align} G'(x) &= \frac{d}{dx} \, \left[ \int_a^b \, 2t \, dt \right] \\ \\ &= \frac{d}{dx} (x^2 - a^2) = 2x \end{align}

So the FTOC-2 is pretty weird. Notice that when taking the derivative of such an integral, we don't actually need to integrate. We just replace t in the integrand by x, and that's it. Couldn't be simpler.

### Proof of the FTOC-2

The FTOC-2 posits that:

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

$$\text{for all} \; x \in [a, b]$$

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:

$$G(x) = \int_a^x f(t) dt \; \; \text{and} \; \; G(z) = \int_a^z f(t) dt$$

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.

\begin{align} G(z) - G(x) &= \int_a^z f(t) dt - \int_a^x f(t) dt \\ \\ &= \int_a^z f(t) dt + \int_x^a f(t) dt = \int_x^z f(t) dt \end{align}

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')

\begin{align} f(c) &= \frac{1}{z - x} \int_x^z \, f(t) \, dt \\ \\ &\longrightarrow \int_x^z \, f(t) \, dt = (z - x) f(c) \end{align}

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.

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

$$G'(x) = f(x)$$

### A second proof of FTOC-1

With FTOC-2 proved (just above), we can use that result to prove FTOC-1, which says:

\begin{align} If \: F'(x) &= f(x), \\ \\ then \: \int_a^b \, f(x) \, dx &= F(b) - F(a) \end{align}

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] \\ \\ &= G(b) - G(a) \end{align}

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

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

### Example 1

Find the integral and its derivative: $\int_0^x \, t^2 \, dt$

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

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

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:

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

#### The lower limit doesn't matter

Now one thing you might be wondering about is the lower limit of integration, x=0. Let's repeat this problem, except this time with a finite lower limit; let's call it 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:

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

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:

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

### Example 2: You don't even need to integrate

In the simple example to the right, we integrate as usual, evaluating the integral at limits a and x. Notice that the dummy variable t is now gone and the independent variable is x. If we now take the derivative of the result with respect to x, we just get the integrand (t-3) back, but with x substituted by t. The lower limit contributed nothing to the derivative because the integral evaluated there is a constant.

So you see, in these problems, there's no need to integrate at all. It only becomes more complicated when that x in the upper limit is a function of x, so that we'll need some kind of chain rule analogous to the chain rule of differentiation.

\begin{align} G'(x) &= \frac{d}{dx} \, \int_a^x \, t^{-3} \, dt \\ \\ &= \frac{d}{dx} \frac{-1}{2 t^2} \, \bigg|_a^x \\ \\ &= \frac{d}{dx} \left[ \frac{-1}{2x^2} + \frac{-1}{2a^2} \right] \\ \\ G'(x) &= x^{-3} \end{align}

### Example 3 — The chain rule

Consider a problem like this:

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

Now instead of just having an independent variable as the upper limit of integration, we have a function of that variable — it's like a chain rule problem in differentiation. Think of it like this: If

$$F(x) = \int_0^x \, ln(t) \, dt$$

then

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

Now using the chain rule of differentiation, the derivative of F(x2) is the derivative of the outer function F with respect to x2 times the derivative of x2.

Now we can just plug in the solution:

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

With a little practice, you'll recognize that these are just like other chain-rule derivatives you've done.

### Practice problems

(1-4) Find G'(x)

 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 $$G'(x) = \frac{1}{x}$$

(5-8) Find the derivative:

 5 $$\frac{d}{dx} \int_{-2}^{2x^2} \, sin(t) \, dt$$ Solution $$= 4x\cdot sin(x)$$ 6 $$\frac{d}{dx} \int_a^{sin(x)} \, \frac{1}{t} \, dt$$ Solution $$= \frac{cos(x)}{x}$$ 7 $$\frac{d}{d\theta} \int_0^{tan(\theta)} \, sec^2(t) \, dt$$ Solution $$= sec^4(\theta)$$ 8 $$\frac{d}{dx} \int_{sin(x)}^{cos(x)} \, 2 sin(t) \, dt$$ Solution $$= 2\cdot sin^2(x)$$
 9 If   $f(x) = \int_2^{2x} \frac{1}{\sqrt{t^3 + 1}} dt,$   then $f'(1) = ?$ Solution 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 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 $[-\pi, \, \pi]$. 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

### 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:

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.

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