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Study Guides > Business Calculus

Reading: The Definite Integral

The previous chapters dealt with Differential Calculus. We started with the "simple" geometrical idea of the slope of a tangent line to a curve, developed it into a combination of theory about derivatives and their properties, techniques for calculating derivatives, and applications of derivatives. This chapter deals with Integral Calculus and starts with the "simple" geometric idea of area. This idea will be developed into another combination of theory, techniques, and applications.

PreCalculus Idea—The Area of a Rectangle

If you look on the inside cover of nearly any traditional math book, you’ll find a bunch of area and volume formulas – the area of a square, the area of a trapezoid, the volume of a right circular cone, and so on. Some of these formulas are pretty complicated. But you still won’t find a formula for the area of a jigsaw puzzle piece or the volume of an egg. There are lots of things for which there is no formula. Yet we might still want to find their areas. One reason areas are so useful is that they can represent quantities other than simple geometric shapes. If the units for each side of the rectangle are meters, then the area will have the units meters× meters = square meters = m2.   But if the units of the base of a rectangle are hours and the units of the height are miles/hour, then the units of the area of the rectangle are hours × miles/hour = miles, a measure of distance (figure 1a). Similarly, if the base units are centimeters and the height units are grams (figure 1b), then the area units are gram-centimeters, a measure of work.
Fig_4_1_1 Figure 1
The basic shape we will use is the rectangle; the area of a rectangle is base × height. The only other area formulas I’ll expect you to know are for triangles ([latex] A = \frac{1}{2}bh [/latex]) and for circles ([latex] A = \pi r^2 [/latex]).

The Definite Integral

Distance from Velocity

Example

Suppose a car travels on a straight road at a constant speed of 40 miles per hour for two hours. See the graph of its velocity in figure 2. How far has it gone?
Figure 2 Figure 2
Solution
We all remember distance = rate × time, so this one is easy. The car has gone 40 miles per hour × 2 hours = 80 miles.

Example

But now suppose that a car travels so that its speed increases steadily from 0 to 40 miles per hour, for two hours. (Just be grateful you weren’t stuck behind this car on the highway.) See the graph of its velocity in figure 3. How far has this car gone?
Figure 3 Figure 3
The trouble with our old reliable distance = rate × time relationship is that it only works if the rate is constant. If the rate is changing, there isn’t a good way to use this formula. But look at Fig. 1 again. Notice that distance = rate × time also describes the area between the velocity graph and the t-axis, between t = 0 and t = 2 hours. The rate is the height of the rectangle, the time is the length of the rectangle, and the distance is the area of the rectangle. This is the way we can extend our simple formula to handle more complicated velocities: And this is the way we can answer the second example.
Solution
The distance the car travels is the area between its velocity graph, the t-axis, t = 0 and t = 2. This region is a triangle, so its area is ½bh = ½(2 hours)(40 miles per hour) = 40 miles. So the car travels 40 miles during its annoying trip. In our distance/velocity examples, the function represented a rate of travel (miles per hour), and the area represented the total distance traveled. This principle works more generally. For functions representing other rates such as the production of a factory (bicycles per day), or the flow of water in a river (gallons per minute) or traffic over a bridge (cars per minute), or the spread of a disease (newly sick people per week), the area will still represent the total amount of something. 
Figure 4 Figure 4

Example

Figure 5 shows the flow rate (cubic feet per second) of water in the Skykomish river at the town of Goldbar in Washington state. (For comparison, the flow over Niagara Falls is about 2.12x105 cf/s.)
Figure 5 Figure 5
The area of the shaded region represents the total volume (cubic feet) of water flowing past the town during the month of October. We can approximate this area to approximate the total water by thinking of the shaded region as a rectangle with a triangle on top. Total water = total area ≈ area of rectangle + area of the “triangle” ≈ (2000 cubic feet/sec)(30 days) + = (2750 cubic feet/sec)(30 days) Note that we need to convert the units to make sense of our result: Total water ≈ (2750 cubic feet/sec)(30 days) = (2750 cubic feet/sec)(2,592,000 sec) ≈ 7.128 x 109 cubic feet. About 7 billion cubic feet of water flowed past Goldbar in October.

Approximating with Rectangles

How do we approximate the area if the rate curve is, well, curvy? We could use rectangles and triangles, like we did in the last example. But it turns out to be more useful (and easier) to simply use rectangles. The more rectangles we use, the better our approximation is.
Figure 6 Figure 6
Suppose we want to calculate the area between the graph of a positive function f and the interval [a, b] on the x–axis (figure 6). The Riemann Sum method is to build several rectangles with bases on the interval [a, b] and sides that reach up to the graph of f (figure 7). Then the areas of the rectangles can be calculated and added together to get a number called a Riemann Sum of f on [a, b]. The area of the region formed by the rectangles is an approximation of the area we want.
Figure 7 Figure 7

Example

Approximate the area in figure 8a between the graph of f and the interval [2, 5] on the x–axis by summing the areas of the rectangles in figure 8b.
Figure 8 Figure 8
Solution
The total area of rectangles is (2)(3) + (1)(5) = 11 square units.

Example

Figure 9 Figure 9
Let A be the region bounded by the graph of f(x) = 1/x, the x–axis, and vertical lines at x = 1 and x = 5. We can’t find the area exactly (with what we know now), but we can approximate it using rectangles. When we make our rectangles, we have a lot of choices. We could pick any (non-overlapping) rectangles whose bottoms lie within the interval on the x-axis, and whose tops intersect with the curve somewhere. But it’s easiest to choose rectangles that—(a) have all the same width, and (b) take their heights from the function at one edge. Figures 9 and 10 below show two ways to use four rectangles to approximate this area. In figure 9, we used left-endpoints; the height of each rectangle comes from the function value at its left edge. In figure 10, we used right-hand endpoints.
Left-Hand Endpoints
Figure 10 Figure 10
The area is approximately the sum of the areas of the rectangles. Each rectangle gets its height from the function [latex] f(x) = \frac{1}{x} [/latex] You can find the area of each rectangle using area = height × width. So the total area of the rectangles, the left-hand estimate of the area under the curve, is [latex] f(1)(1) + f(2)(1) + f(3)(1) + f(4)(1) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{12} \cong 2.08 [/latex] Notice that because this function is decreasing, all the left endpoint rectangles stick out above the region we want – using left-hand endpoints will overestimate the area.
Right-Hand Endpoints
The right-hand estimate of the area is [latex] f(2)(1) + f(3)(1) + f(5)(1) + f(5)(1) = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{77}{60} \cong 1.28 [/latex] All the right-hand rectangles lie completely under the curve, so this estimate will be an underestimate. We can see that the true area is actually in between these two estimates. So we could take their average: Average: [latex] \frac {25/12 + 77/60}{2} = \frac{101}{60} \cong 1.68 [/latex] In general, the average of the left-hand and right-hand estimates will be closer to the real area than either individual estimate. My estimate of the area under the curve is about 1.68. (The actual area is about 1.61.) If we wanted a better answer, we could use even more, even narrower rectangles. But there’s a limit to how much work we want to do by hand. In practice, it’s probably best to choose a manageable number of rectangles. We’ll have better methods to get more accurate answers before long. These sums of areas of rectangles are called Riemann sums. You may see a shorthand notation used when people talk about sums.   We won’t use it much in this book, but you should know what it means.

Riemann Sum

A Riemann sum for a function f(x) over an interval [a, b] is a sum of areas of rectangles that approximates the area under the curve. Start by dividing the interval [a, b] into n subintervals; each subinterval will be the base of one rectangle. We usually make all the rectangles the same width Δx. The height of each rectangle comes from the function evaluated at some point in its sub interval. Then the Riemann sum is: [latex-display] f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + \dots + f(x_n)\Delta x [/latex-display]

Sigma Notation

The upper-case Greek letter Sigma Σ is used to stand for Sum. Sigma notation is a way to compactly represent a sum of many similar terms, such as a Riemann sum. Using the Sigma notation, the Riemann sum can be written [latex] \sum_{i=1}^{n} f(x_i)\Delta x [/latex]. This is read aloud as “the sum as i = 1 to n of f of x sub i Delta x.” The i is a counter, like you might have seen in a programming class. 

Definition of the Definite Integral

Because the area under the curve is so important, it has a special vocabulary and notation.
  • The definite integral of a positive function f(x) over an interval [a, b] is the area between f, the x-axis, x = a and x = b.
  • The definite integral of a positive function f(x) from a to b is the area under the curve between a and b.
  • If f(t) represents a positive rate (in y-units per t-units), then the definite integral of f from a to b is the total y-units that accumulate between t = a and t = b

Notation for the Definite Integral

The definite integral of f from a to b is written [latex-display] \int_a^b f(x)dx [/latex-display] The [latex] \int [/latex] symbol is called an integral sign; it’s an elongated letter S, standing for sum. (The [latex] \int [/latex] is actually the Σ from the Riemann sum, written in Roman letters instead of Greek letters.) The dx on the end must be included; you can think of [latex] \int [/latex] and dx as left and right parentheses. The dx tells what the variable is—in this example, the variable is x. (The dx is actually the [latex] \Delta x [/latex] from the Riemann sum, written in Roman letters instead of Greek letters.) The function f is called the integrand. The a and b are called the limits of integration.

Verb Forms

We integrate, or find the definite integral of a function. This process is called integration.

Formal Algebraic Definition

[latex-display] \int_{a}^{b} f(x)dx = {\lim_{n \to \infty}}_{\Delta x \to 0} \sum_{i = 1}^{n} f(x_i)\Delta x [/latex-display]

Practical Definition

The definite integral can be approximated with a Riemann sum (dividing the area into rectangles where the height of each rectangle comes from the function, computing the area of each rectangle, and adding them up). The more rectangles you use, the narrower the rectangles are, the better your approximation will be.

Looking Ahead

We will have methods for computing exact values of some definite integrals from formulas soon. In many cases, including when the function is given to you as a table or graph, you will still need to approximate the definite integral with rectangles.

Example

Fig. 11 shows y = r(t), the number of telephone calls made per hours (a rate!) on a Tuesday. Approximately how many calls were made between 9 pm and 11 pm? Express this as a definite integral and approximate with a Riemann sum.
Figure 11 Figure 11
Solution
We know that the accumulated calls will be the area under this rate graph over that two-hour period, the definite integral of this rate from t = 9 to t = 11. The total number of calls will be [latex] \int_{9}^{11} r(t)dt [/latex]. The top here is a curve, so we can’t get an exact answer. But we can approximate the area using rectangles. I’ll choose to use 4 rectangles, and I’ll choose left-endpoints:
Figure 12 Figure 12
[latex-display] \int_{9}^{11} r(t)dt \cong 100(.5) + 150(.5) + 180(.5) + 195(.5) = 312.5 [/latex-display] The units are calls per hour × hours = calls.   My estimate is that about 312 calls were made between 9 pm and 11 pm. Is this an under-estimate or an over-estimate?

Example

Describe the area between the graph of [latex] f(x) = \frac{1}{x} [/latex], the x–axis, and the vertical lines at x = 1 and x = 5 as a definite integral.
Solution
This is the same area we estimated to be about 1.68 before. Now we can use the notation of the definite integral to describe it. Our estimate of [latex] f(x) = \int_{1}^{5} \frac{1}{x} dx [/latex] was 1.68. The true value of [latex] f(x) = \int_{1}^{5} \frac{1}{x} dx [/latex] is about 1.61.

Example

Using the idea of area, determine the value of [latex] f(x) = \int_{1}^{3} (1 + x)dx [/latex].
Solution
[latex] f(x) = \int_{1}^{3} (1 + x)dx [/latex] represents the area between the graph of f(x) = 1+x, the x–axis, and the vertical lines at 1 and 3 (figure 13).
Figure 13 Figure 13
Since this area can be broken into a rectangle and a triangle, we can find the area exactly. The area equals 4 + ½ (2)(2) = 6 square units.

Example

The table shows rates of population growth for Berrytown for several years. Use this table to estimate the total population growth from 1970 to 2000:
Year (t) 1970 1980 1990 2000
Rate of population growth R(t) (thousands of people per year) 1.5 1.9 2.2 2.4
Solution
The definite integral of this rate will give the total change in population over the thirty-year period. We only have a few pieces of information, so we can only estimate. Even though I haven’t made a graph, we’re still approximating the area under the rate curve, using rectangles. How wide are the rectangles? I have information every 10 years, so the rectangles have a width of 10 years. How many rectangles? Be careful here – this is a thirty-year span, so there are three rectangles. Using left-hand endpoints: (1.5)(10) + (1.9)(10) + (2.2)(10) = 56; Using right-hand endpoints: (1.9)(10) + (2.2)(10) + (2.4)(10) = 65; Taking the average of these two: [latex] \frac{56 + 65}{2} = 60.5 [/latex] My best estimate of the total population growth from 1970 to 2000 is 60.5 thousand people.

Signed Area

You may have noticed that until this point, we’ve insisted that the integrand (the function we’re integrating) be positive. That’s because we’ve been talking about area, which is always positive. If the “height” (from the function) is a negative number, then multiplying it by the width doesn’t give us actual area, it gives us the area with a negative sign. But it turns out to be useful to think about the possibility of negative area. We’ll expand our idea of a definite integral now to include integrands that might not always be positive. The “heights” of the rectangles, the values from the function, now might not always be positive.

The Definite Integral and Signed Area

  • The definite integral of a function f(x) over an interval [a, b] is the signed area between f, the x-axis, x = a and x = b.
  • The definite integral of a function f(x) from a to b is the signed area under the curve between a and b.
If the function is positive, the signed area is positive, as before (and we can call it area.) If the function dips below the x-axis, the areas of the regions below the x-axis come in with a negative sign. In this case, we cannot call it simply “area.” These negative areas take away from the definite integral. [latex] \int_{a}^{b} f(x)dx [/latex] = (Area above x-axis) – (Area below x-axis). If f(t) represents a positive rate (in y-units per t-units), then the definite integral of f from a to b is the total y-units that accumulate between t = a and t = b. If f(t) represents any rate (in y-units per t-units), then the definite integral of f from a to b is the net y-units that accumulate between t = a and t = b.

Example

Find the definite integral of of f(x) = –2 on the interval [1,4].
Figure 14 Figure 14
Solution
[latex] \int_{1}^{4} -2dx [/latex] is the signed area of the region shown to the right. The region lies below the x-axis, so the area (6) comes in with a negative sign. So the definite integral is [latex] \int_{1}^{4} -2dx = -6 [/latex]. Negative rates indicate that the amount is decreasing. For example, if f(t) is the velocity of a car in the positive direction along a straight line at time t (miles/hour) , then negative values of f indicate that the car is traveling in the negative direction, backwards. The definite integral of f is the change in position of the car during the time interval. If the velocity is positive, positive distance accumulates. If the velocity is negative, distance in the negative direction accumulates. This is true of any rate. For example, if f(t) is the rate of population change (people/year) for a town, then negative values of f would indicate that the population of the town was getting smaller, and the definite integral (now a negative number) would be the change in the population, a decrease, during the time interval.

Example

In 1980 there were 12,000 ducks nesting around a lake, and the rate of population change (in ducks per year) is shown in figure 15. Write a definite integral to represent the total change in the duck population from 1980 to 1990, and estimate the population in 1990.
Figure 15 Figure 15
Solution
The change in population = [latex] \int_{1980}^{1990} f(t)dt = -\text{area between} f \text{and axis} [/latex] ≈ −{200 ducks/year} × {10 years} = − 2000 ducks. Then {1990 duck population} = {1980 population} + {change from 1980 to 1990} = {12,000} + { –2000} = 10,000 ducks.

Example

A bug starts at the location x = 12 on the x–axis at 1 pm walks along the axis with the velocity v(x) shown in figure 16. How far does the bug travel between 1 pm and 3 pm, and where is the bug at 3 pm?
Figure 16 Figure 16
Solution
Note that the velocity is positive from 1 until 2:30, then becomes negative. So the bug moves in the positive direction from 1 until 2:30, then turns around and moves back toward where it started. The area under the velocity curve from 1 to 2:30 shows the total distance traveled by the bug in the positive direction; the bug moved 12.5 feet in the positive direction. The area between the velocity curve and the x-axis, between 2:30 and 3, shows the total distance traveled by the bug in the negative direction, back toward home; the bug traveled 2.5 feet in the negative direction. The definite integral of the velocity curve, [latex] \int_{1}^{3} v(t)dt [/latex], shows the net change in distance: [latex-display] \int_{1}^{3} v(t)dt = 12.5 - 2.5 = 10 [/latex-display] The bug ended up 10 feet further in the positive direction than he started. At 3 pm, the bug is at x = 22.

Example

Use Figure 17 to calculate [latex] \int_{0}^{2}f(x)dx [/latex], [latex] \int_{2}^{4}f(x)dx [/latex], [latex] \int_{4}^{5}f(x)dx [/latex], and [latex] \int_{0}^{5}f(x)dx [/latex].
Figure 17 Figure 17
Solution
[latex] \int_{0}^{2}f(x)dx = 2 [/latex], [latex] \int_{2}^{4}f(x)dx = -5 [/latex], [latex] \int_{4}^{5}f(x)dx = 2 [/latex], and [latex] \int_{0}^{5}f(x)dx = (\text{area above}) - (\text{area below}) = (2+2)-(5) = -1 [/latex].