]> Haese & Harris - Chapter 26 Notes

26: Applications of Integration

A: The Area Problem

Caveat Lector

At first—for this first section, at least—you must forget everything that we did in the previous chapter.

The Problem

Math geeks have this strange fascination with being able to find areas like the one shown below.

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Never mind why—who cares, anyway? The question before you is: how will you do it?

Clearly there are lower and upper bounds. A rectangle of width 3 and height 1 (area 3) fits within the shaded region; a rectangle of width 3 and height 5 (area 15) completely engulfs the shaded region. In fact, a trapezoid with base lengths of 5 and 2, and height 3 (area 10.5), also completely engulfs the shaded area.

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So—using only the simplest geometric formulas possible, how will you find the actual shaded area?

Possible Solutions

Possibility 1: Rectangles

The simplest geometric area is a rectangle…however, using just one won't work (see the work above). Why not divide and conquer? Why not try two rectangles?

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This gives an estimate of 9.375—but it is clearly too large. Perhaps more rectangles will help!

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Now the estimate is 7.40625—still not quite right, though. More rectangles?

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That's an area of 6.632813…how about more?

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50 rectangles gives an area of 6.0918. More, more, more!

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100 rectangles, and an area of 6.04545. Even more?

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1000 rectangles makes 6.004505.

Want to take a guess as to where this is leading?

Possibility 2: Trapezoids

Let's try that with trapezoids and see what happens.

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Two trapezoids makes an area of 7.125.

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Four trapezoids makes an area of 6.28125.

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Eight trapezoids makes an area of 6.070313.

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50 trapezoids; area of 6.0018

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100 trapezoids; area of 6.00045

Again, we seem to be homing in on some value—6, perhaps?

We will discuss (in class) how to use your calculator to create area estimates with rectangles or trapezoids.

Take it to the Limit!

Of course, a finite number of rectangles (or trapezoids) won't give you the answer. For that, we need an infinite number of rectangles.

Alas—infinity is a hard number to work with. However, we could find the limit as the number of (equally spaced) rectangles approaches infinity!

In fact, we define the area between the curve $y = f (x)$ and the x-axis, from $x = a$ to $x = b$ , to be $lim_{n \to ∞} \sum_{x = a}^{b} f (x) Δ x$ ( $f (x)$ is the height of a rectangle, and $Δ x$ is the width of each rectangle).

That limit notation is awfully unpleasant…so much so that we use a simpler notation for the same thing: $\int_{a}^{b} f (x) d x$ .

B: The Fundamental Theorem of Calculus

You may now recall all information from chapter 25…

Integrals are antiderivatives…the area under a curve is an integral…area and antiderivatives are the same thing…area is related to derivatives…area and slope are connected?

Questions swirl; perhaps your view of the universe has been shaken. Good. Very good.

Part 1

Your text has a fantastic motivation for why definite integrals and derivatives are related—I'll not repeat it here.

One-half of The Fundamental Theorem of Calculus: if $f^{'} (x) = F (x)$ , then $\int_{a}^{b} F (x) d x = f (b) - f (a)$ . In words: the integral of a derivative is the original function.

Part 2 (not in the text!)

The relationship between integrals and derivatives goes both ways. The second half of the FTC: if $A (x) = \int_{a}^{x} f (t) d t$ , then $A^{'} (x) = f (x)$ . In words: the derivative of an integral is the original function.

Differentiation and integration are inverse operations—regardless of how you approach them!

Properties of Definite Integrals

We discussed these at the end of the last chapter—they are repeated in your text as a reminder.

Displacement

We know that the derivative of position is velocity— $s^{'} (t) = v (t)$ . As a result, we know $\int_{a}^{b} v (t) d t = s (b) - s (a)$ . Note that the right side of that is the change in position between $t = a$ and $t = b$ : the displacement of the object over the interval $[a, b]$ . Thus, the displacement of an object can be found by evaluating a definite integral on the derivative.

"Why bother?" you ask? Well…there are occasions when we know the velocity function, but not the position function.

Average Value

Question: how do you find average acceleration?

Answer: change in velocity divided by change in time. Or: for the interval $[a, b]$ , the average acceleration is $\frac{v (b) - v (a)}{b - a}$ . We've already determined that this is the slope of the secant line connecting these points on the velocity graph…now, we have another way! The numerator should look familiar…especially if you remember that $v^{'} (t) = a (t)$ .

The average velocity can also be written as $\frac{\int_{a}^{b} a (t) d t}{b - a} = \frac{1}{b - a} \int_{a}^{b} a (t) d t$ .

In general, the average value of $f (x)$ over the interval $[a, b]$ is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ .

C: More on Area

$\int_{a}^{b} f (x) d x = \int_{a}^{b} [f (x) - 0] d x$ is the area between $y = f (x)$ and $y = 0$ .

Why does the second function have to be zero?

Furthermore—why do we have to talk about area in relation to the x-axis…why not in relation to the y-axis?

To the first question: it doesn't.

To the second question: we can do it either way. In particular, if $x = f (y)$ , then the area between the y-axis and the function $x = f (y)$ from $y = a$ to $y = b$ is $\int_{a}^{b} f (y) d y$ .

D: Distance from Velocity

We've already established that $\int_{a}^{b} v (t) d t$ is the displacement (or net change in position) of a moving object. What about the total distance traveled?

For that, you've got to consider the points at which the object changes direction—you've got to find the distance traveled in each interval where the object is moving in one direction, and add those distances. The quick alternative (when using technology): $\int_{a}^{b} | v (t) | d t$ .

E: Problem Solving

Word problems—been there; done that.

There is one item worth mentioning: differential equations. We've already seen these in two ways: indefinite integrals, and indefinite integrals with an initial condition. In both cases, we were given a derivative, and asked to find the original function. An equation that involves a derivative is called a differential equation (ex: $\frac{d y}{d x} = 3 x^{2}$ ). When we take the integral, and come up with a function plus a constant, then we've got a general solution to the differential equation (ex: $y = x^{3} + C$ ). When you're told a bit more about the function (ex: $(1, - 2)$ is on the curve; or $y (1) = - 2$ ), then you've been given an initial condition, and you can find the particular solution to the differential equation (ex: $- 2 = {(1)}^{3} + C ⇒ C = - 3; y = x^{3} - 3$ ).

Page last validated 2010-08-15