Integrals

The Area and Distance Problems

The Area Problem

We begin by attempting to solve the area problem: Find the area of the region $S$ that lies under the curve $y = f(x)$ from $a$ to $b$.

It isn’t easy, to find the area of a region with curved sides. To do so, we first approximate the region $S$ by rectangles, and then we take the limit of the sum of the areas of the approximating rectangles as we increase the number of rectangles.

We begin approximating each strip by a rectangle that has the same base as the strip and whose height is the same as the right edge of the strip [see Figure 4(b)]. In other words, the heights of these rectangles are the values of the function $f(x)$.

If we let $R_n$ be the sum of the areas of these approximating rectangles, we get

$$ R_n = \sum_{i = 1}^n \Delta x f(x_i) $$

where $\Delta x$ is the constant width of the strips, and $f(x_i)$ is the height of the rectangle at the right-side point of the rectangle for the $i$-th strip. We can also compute the sum of the areas by taking $x_i$ to be the left-side point of the rectangle for the $i$-the strip:

$$ L_n = \sum_{i = 0}^{n - 1} \Delta x f(x_i) $$

We could obtain better estimates by increasing the number of strips, that is, by increasing $n$.

From Figures 8 and 9 it appears that as $n$ increases, both $L_n$ and $R_n$ become better and better approximations to the area of $S$. Therefore, we define the area $A$ to be the limit of the sums of the areas of the approximating rectangles, that is,

$$ A = \lim_{n \to \infty} L_n = \lim_{n \to \infty} R_n $$

It can be proved that the previous limit always exists, since we are assuming that $f$ is continuous. It can also be shown that we get the same value if we use left endpoints

$$ A = \lim_{n \to \infty} L_n = \lim_{n \to \infty} [\Delta x f(x_0) + \Delta x f(x_1) + \cdots + \Delta x f(x_{n - 1})] $$

In fact, instead of using left endpoints or right endpoints, we could take the height of the $i$th rectangle to be the value of f at any number $x_i*$ in the $i$-th subinterval $[x_{i -1}, x_i]$. We call the numbers $[x_1*, x_2*, \cdots, x_n*]$ the sample points. Figure 13 shows approximating rectangles when"the sample points are not chosen to be endpoints. So a more general expression for the area of $S$ is

$$ A = \lim_{n \to \infty} [\Delta x f(x_1^*) + \Delta x f(x_2^*) + \cdots + \Delta x f(x_n^*)] $$

NOTE To approximate the area under the graph of $f$ we can form lower sums (or upper sums) by choosing the sample points $x_i*$ so that $f(x_i*)$ is the minimum (or maximum) value of $f$ on the $i$-th subinterval (see Figure 14).

The Distance Problem

Now let’s consider the distance problem: find the distance traveled by an object during a certain time period if the velocity of the object is known at all times. If the velocity remains constant, then the distance problem is easy to solve by means of the formula

$$ \text{distance} = \text{velocity} \cdot \text{time} $$

Let’s sketch an approximate graph of the velocity function along with rectangles whose heights are the initial velocities for each time interval [see Figure 17(a)]

The area of each rectangle can be interpreted as a distance because the height represents velocity and the width represents time. The sum of the areas of the rectangles in Figure 17(a) is our initial estimate for the total distance traveled.

If we want a more accurate estimate, we could take velocity readings more often, as illustrated in Figure 17(b). You can see that the more velocity readings we take, the closer the sum of the areas of the rectangles gets to the exact area under the velocity curve [see Figure 17(c)]. This suggests that the total distance traveled is equal to the area under the velocity graph.

In general, suppose an object moves with velocity $v = f(t)$, where $a \leq t \leq b$ and $f(t) \geq 0$ (so the object always moves in the positive direction). We take velocity readings at times $t_0 (=a), t_1, \cdots, t_n(=b)$ so that the velocity is approximately constant on each subinterval. If these times are equally spaced, then the time between consecutive readings is $\Delta t = \frac{b - a}{n}$. The total distance traveled during the time interval $[a, b]$ is approximately:

$$ f(t_0) \Delta t + f(t_1) \Delta t + \cdots + f(t_{n - 1}) \Delta t = \sum_{i = 1}^{n} f(t_{i - 1}) \Delta t $$

If we use the velocity at right endpoints instead of left endpoints, our estimate for the total distance becomes

$$ f(t_1) \Delta t + f(t_1) \Delta t + \cdots + f(t_n) \Delta t = \sum_{i = 1}^n f(t_i) \Delta t $$

The more frequently we measure the velocity, the more accurate our estimates become, so it seems plausible that the exact distance d traveled is the limit of such expressions:

$$ d = \lim_{n \to \infty} \sum_{i=1}^n \Delta t f(t_{i - 1}) = \lim_{n \to \infty} \sum_{i=1}^n \Delta t f(t_i) $$

Because the previous equation has the same form as our expressions for the area under a region $S$, it follows that the distance traveled is equal to the area under the graph of the velocity function.

The Definite Integral

Definite Integrals

The precise meaning of the limit for the definite integral is as follows:

For every number $\epsilon > 0$, there is an integer $N$ such that

$$|\int_{a}^b f(x) dx - \sum_{i=1}^{n} f(x_i^*) \Delta x| < \epsilon$$

for every integer $n > N$ and for every choice of $x_i^*$ in $[x_{i - 1}, x_i]$.

This follows from the definition of limits on the infinite.

Note that the definite integral $\int_a^b f(x) dx$ is a number, that is, it does not depend on $x$. In fact we could use any letter in place of $x$ without changin the value of the integral.

Also note that the sum

$$ \sum_{i = 1}^n f(x_i^*) \Delta x $$

is called a Riemann sum after the German mathematician Berhard Riemann. So the Definition of a Definite Integral says that the definite integral of an integrable function ca be approximated to within any desired degree of accuracy by a Riemann sum.

We know that if $f$ happes to be positive, then the Riemann sum can be interpreted as the sum of areas of approximating rectangles (see Figure 1).

If $f$ takes on both positive and negative values, as in Figure 3, then the Riemann sum is the sum of the areas of the rectangles that lie above the $x$-axis and the negatives of the areas of the rectangles that lie below the $x$-axis.

So, a definite integral can be interpreted as a net area, that is, a difference of areas:

$$ \int_a^b f(x) dx = A_1 - A_2 $$

where $A_1$ is the area of the region above the $x$-axis and below the graph of $f$, and $A_2$ is the area of the region below the $x$-axis and above the graph of $f$.

Although we have defined $\int_a^b f(x) dx$ by dividing $[a, b]$ into subintervals of equal width, there are situations in which it is advantageous to work with subintervals of unequal width. If the subinterval widths are $\Delta x_1, \Delta x_2, \cdots, \Delta x_n$ we have to ensure that all these widths approach $0$ in the limiting process.

This happens if the largest width, $\max \Delta x_i$, approaches $0$. Why? For the limit of the sums of rectangle areas to converge to the value of the integral, we need the rectangles to be arbitrarily thin. Imagine some subintervals stay wide, then the sum does not approximate necessarily the real area (the value of the integral). Because in this case, a wide subinterval might enclose great variations on $f(x)$, and so the rectangle defined by this subinterval does not “represent” $f(x)$ well enough. So the following must hold:

$$ \max_i \Delta x_i \to 0 $$

That is, the width of the widest subinterval tends to zero.

So in this case the definition of a definite integral becomes

$$ \int_{a}^b f(x) dx = \lim_{\max \Delta x_i \to 0} \sum_{i = 1}^n f(x_i^*) \Delta x_i $$

We have defined the definite integral for an integrable function, but not all functions are integrable.

Why is $f$ said to be integrable under such conditions?

Let’s start by with the condition of $f$ being continuous. If $f$ is continuous, on very small subintervals (we know the subintervals tend to width zero) then $f$ does not “change abruptly”. Thus any sample point $x_i^*$ on that subinterval will resolve to a height of $f(x_i^*)$, which will be very similar to all the other possible heights for any other sample point on the subinterval.

Thus the rectangles on the Riemann sums will be very similar for any arbitrary sample point. Therefore all the approximations for the area under $f$ tend to the same value.

This makes the limit exist, and thus $f$ is integrable.

What if $f$ is not continuous but has a finite number of jump discontinuities? In this case the discontinuities can create some local error on the Riemann sum. However the total error can be made arbitrarily small by making the subintervals that contain these discontinuities smaller and smaller.

To be more concrete, image an subinterval which contains a jump discontinuity (e.g. there is a jump discontinuity at $x = c$, such that $f(c^-) = 1$ and $f(c^+) = 3$). Then, no matter the sample point you choose, the area will not fully capture the behaviour of $f$ on that subinterval.

Note, however, that the error depends on the width of the subinterval:

$$ \text{error} \approx \text{height differences} \times \text{width} $$

If the width of the subinterval can be made infinitely smaller, then the error will tend to zero. And we do not know that the width of the subintervals do tend to zero, therefore the local error introduced by the discontinuities also do tend to zero, then the limit exists and $f$ is integrable.

But you may ask, why does this not always hold with an infinite number of jump discontinuities? Another way to state Riemann’s definition for integrability is:

A function $f$ on $[a, b]$ is integrable if and only if the set of all its jump discontinuities has measure zero.

What does this mean for us? The fact that $f$ has infinitely many jump discontinuities it is not itself a problem. The problem arises when the measure of these jump discontinuities is not zero, then:

• The difference of the upper and lower Riemann sums do not tend to zero, no matter how small the subintervals.
• Thus the Riemann integral does not exist and $f$ is not integrable.

The previous image illustrates this fenomenon. On the lefthand figure we can see that the function only has two jump discontinuities, whose contribution to the area can be forced to zero by making the subintervals that contain them have width zero.

However, in the figure on the right, the function exhibits an infinite number of dense oscillations, effectively creating infinitely many discontinuities throughout the interval. As a result, the Riemann sums fail to converge, and the function is not integrable.

To simplify the calculation of the integral we often take the sample points to be right endpoints. Then $x_i^* = x_i$ and the definition of an integral is as follows.

Evaluating Definite Integrals

The following four equations give formulas for sums of powers of positive integers which we will use to help us on the evaluation of integrals.

The remaining formulas are simple rules for working with sigma notation:

The Midpoint Rule

We often choose the sample point $x_i^*$ to be the right endpoint of the $i$-the subinterval. But if the purpose is to find an approximation to an integral, it is usually better to choose $x_i^*$ to be the midpoint of the interval, which we denote by $\overline{x}_i$.

This statement is numerically true because using the midpoint does reduce the approximation error when computing Riemann’s sum, whilst using either the right or left endpoint does result in a bigger approximation error.

What would happen if you were to choose the right endpoints? Then

$$ x^{*}_i = x_i $$

Now, if $f$:

• Increases on the $i$-the subinterval, using the right endpoint overestimates the area.
• Decreases on the $i$-the subinterval, using the right endpoint underestimates the area.

As you can see we are systematically introducing bias. However when we take the midpoint we neutralize this bias.

The value of $f$ at $\overline{x}_i$ is generally closer to the mean value of $f$ on the $i$-th subinterval. This is based on the fact that the mean value of a smooth function on a small interval approximates its value on its midpoint. This can be shown by using Taylor’s Polynomial.

Properties of the Definite Integral

When we defined the definite integral $\int_a^b f(x) dx$, we implicitly assumed that $a < b$. But the definition as a limit of Riemann sums makes sense even if $a > b$. Notice that if we interchange $a$ and $b$, then $\Delta x$ changes from $\frac{b - a}{n}$ to $\frac{a - b}{n}$. Therefore

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

as $\Delta x_{ba} = - \Delta x_{ab}$. Also note that if $a = b$, then $\Delta x = 0$ and so:

$$ \int_a^a f(x) dx = 0 $$

We now develop some basic properties of integrals that will help us to evaluate integrals in a simple manner. We assume that $f$ and $g$ are continuous functions.

Property 1 says that the integral ofo a constante function $f(x) = c$ is the constant times the length of the interval. If $c > 0$ and $a < b$, this is to be expected because $c(b - a)$ is the area of the shaded rectangle in Figure 14.

Property 2 says that the integral of a sum is the sum of the integrals. For positive functions it says that the are under $f + g$ is the area under $f$ plus the area under $g$ (see Figure 15).

In general, Property 2 follows from the definition of the value of a definite integral and the fact that the limit of a sum is the sum of the limits:

$$ \int_a^b [f(x) + g(x)] dx = \lim_{n \to \infty} \sum_{i = 1}^n [f(x_i) + g(x_i)] \Delta x $$$$ = \lim_{n \to \infty} \left[\sum_{i = 1}^n f(x_i) \Delta x + \sum_{i = 1}^n g(x_i) \Delta x\right] $$$$ = \lim_{n \to \infty} \sum_{i = 1}^n f(x_i) \Delta x + \lim_{n \to \infty} \sum_{i = 1}^n g(x_i) \Delta x $$$$ = \int_a^b f(x) dx + \int_a^b g(x) dx $$

Property 3 can be proved in a similar manner and says that the integral of a constant times a function is the constant times the integral of the function. Property 4 is proved by writing $f - g = f + (-g)$ and using Properties 2 and 3 with $c = -1$.

The next property tells us how to combine integrals of the same function over adjacent intervals.

This is not easy to prove in general, but for the case where $f(x) \geq 0$ and $a < c < b$, this property can be seen from the geometric interpretation in Figure 16.

Properties 1-5 are true whether $a < b, a = b$ or $a > b$. The following properties, in which we compare sizes of functions and sizes of integrals, are true only if $a \leq b$.

Proof of Property 6

Let $f$ be an integrable function such that $f(x) \geq 0, \forall x \in [a, b]$, then by the definition of a definite integral:

$$ \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i = 1}^n f(x^*_i) \Delta x $$

Because $f(x) \geq 0, \forall x \in [a, b]$, then $f(x^*_i) \geq 0$. Also $a \leq b$, thus $\Delta x \geq 0$. Therefore:

$$ f(x^*_i)\Delta x \geq 0, \forall x^*_i $$

By the properties of real numbers, we know that the sum of non-negative numbers results in a non-negative number, that is:

$$ \lim_{n \to \infty} \sum_{i = 1}^n f(x^*_i) \Delta x \geq 0 $$

Thus

$$ \int_a^b f(x) dx \geq 0 $$

Proof of Property 7

Let $f$ and $g$ be integrable functions such that $f(x) \geq g(x), \forall x \in [a, b]$, then

$$ f(x^*_i) \geq g(x^*_i) $$

By the properties of inequalities, given $\Delta x \geq 0$:

$$ f(x^*_i) \Delta x \geq g(x^*_i) \Delta x $$

where $x^*_i$ are is an arbitrary sample point on the $i$-th subinterval. Therefore, by the properties of limits

$$ \lim_{n \to \infty} \sum_{i = 1}^n f(x^*_i) \Delta x \geq \lim_{n \to \infty} \sum_{j = 1}^n g(x^*_j) \Delta x $$

Thus, by the definition of the definite integral

$$ \int_a^b f(x) dx \geq \int_a^b g(x) dx $$

Property 8 is illustrated by Figure 17 for the case where $f(x) \geq 0$.

Proof of Property 8

Since $m \leq f(x) \leq M$, Property 7 gives

$$ \int_a^b m dx \leq \int_a^b f(x) dx \leq \int_a^b M dx $$

Using Property 1 to evaluate the integrals of the constant functions

$$ m(b - a) \leq \int_a^b f(x) dx \leq M(b - a) $$

Property 8 is useful when all we want is a rough estimate of the size of an integral.