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Unformatted text preview: Unit 8 Integration methods Introduction Introduction In this unit you’ll continue your study of integration. In Unit 7 you saw that integration is the reverse of differentiation. You saw that you can use it to solve problems in which you know the values taken by the rate of change of a continuously changing quantity, and you want to work out the values taken by the quantity, or how much the quantity changes over some period. For example, you saw that if you have a formula for the velocity of an object in terms of time, then you can use integration to find a formula for the displacement of the object in terms of time, or work out the change in the displacement of the object over some period of time. In Sections 1 and 2 of this unit, you’ll meet a different way to think about integration. This way of thinking about it involves areas of regions of the plane, rather than rates of change. You’ll see that the link between the two ways of thinking about integration helps you to solve problems both about areas and about rates of change. In Sections 3 to 5 of the unit, you’ll learn some more methods for integrating functions, which you can use in addition to the methods that you learned in Unit 7. These further methods will allow you to integrate a much wider variety of functions than you’ve learned to integrate so far. Many students find some of the material in this unit quite challenging when they first meet it, so don’t worry if that’s the case for you too. As with most mathematics, it will seem more straightforward once you’re more familiar with it. If you find a particular technique difficult, then make sure that you watch the associated tutorial clips, and use the practice quiz and exercise booklet for this unit for more practice. This unit is likely to take you more time to study than most of the other units, so the study planner allows extra time for it. 105 Unit 8 Integration methods 1 Areas, signed areas and definite integrals In this section you’ll look at a type of mathematical problem that you might think has little connection with the mathematics of rates of change that you studied in Units 6 and 7. In Section 2 you’ll see that in fact the two areas of mathematics are closely related. Throughout Sections 1 and 2, we’ll work with continuous functions. Remember that a continuous function is a function whose graph has no breaks in it, over its whole domain. Informally, it’s a function whose graph you can draw without taking your pen off the paper (though you might need an infinitely large piece of paper, as the graph might be infinitely long!). 1.1 Areas and signed areas Sometimes it’s useful to calculate the area of a flat shape with a curved boundary. For example, suppose that an architect is designing a large building shaped as shown in Figure 1(a). The curve of the roof is given by the graph of the function f (x) = 9 − 1 2 36 x (−18 ≤ x ≤ 18), where x and f (x) are measured in metres. This graph is shown in Figure 1(b). Figure 1 (a) The shape of a building (b) the graph that gives the curve of the roof 106 1 Areas, signed areas and definite integrals Suppose that the architect wants to calculate the area shaded in Figure 1(b), so that she can determine the amounts of materials needed to build the end wall of the building. Here’s a way of calculating an approximate value for this area, which you can make as accurate as you wish. You start by dividing the interval [−18, 18], which is the interval of x-values that corresponds to the curve, into a number of subintervals of equal width. For example, in Figure 2 the interval has been divided into twelve subintervals. The right endpoint of the first subinterval is equal to the left endpoint of the next subinterval, and so on. For each subinterval, you approximate the shape of the curve on that subinterval by a horizontal line segment whose height is the value of the function at the left endpoint of the subinterval, as shown in Figure 2. You calculate the areas of the rectangles between these line segments and the x-axis (simply by multiplying their heights by their widths in the usual way), and add up all these areas to give you an approximate value for the required area. The more subintervals you use, the better will be the approximation. Figure 2 A collection of rectangles whose total area is approximately the area in Figure 1(b) In the next example this method is used, with four subintervals, to find an approximate value for the area discussed above. Notice that, in both the example and in Figure 2 above, the rectangle corresponding to the first subinterval has zero height and therefore zero area – this happens because the value of the function at the left endpoint of this subinterval is zero. 107 Unit 8 Integration methods Calculating an approximate value for an area Example 1 Use the method described above, with four subintervals as shown below, to find an approximate value for the area between the graph of 1 2 x and the x-axis. the function f (x) = 9 − 36 Solution We divide the interval [−18, 18] into four subintervals of equal width. The whole interval has width 36, so each subinterval has width 36/4 = 9. The left endpoints of the four subintervals are −18, −18 + 9, −18 + 2 × 9, −9, 9. −18 + 3 × 9, that is −18, 0, So the heights of the rectangles are f (−18), f (−9), f (0), f (9), that is, 9− 1 36 × (−18)2 , 9− 1 36 × (−9)2 , 9− 1 36 × 02 , 9− 1 36 × 92 , which evaluate to 0, 27 4 , 9, 27 4 . So we obtain the following approximate value for the area: 405 27 (0 × 9) + ( 27 4 × 9) + (9 × 9) + ( 4 × 9) = 2 = 202.5. The calculation in Example 1 shows that a (rather crude) approximate value for the area of the cross-section of the roof discussed at the beginning of this subsection is 202.5 m2 . 108 1 Areas, signed areas and definite integrals Note that the units for the area of a region on a graph are, as you’d expect, the units on the vertical axis times the units on the horizontal axis. For example, if the units on both axes are metres, then the units for area are square metres (m2 ). If a graph has no specific units on the axes, then the units for area are simply ‘square units’, and we usually omit them when we state an area. In the next activity you’re asked to find another approximation for the area of the cross-section of the roof discussed earlier, by using the subinterval method with six subintervals instead of four. Activity 1 Calculating an approximate value for an area Use the method described above, with six subintervals as shown below, to find an approximate value for the area between the graph of the function 1 2 x and the x-axis. f (x) = 9 − 36 Table 1 shows the approximate values for the area between the graph of 1 2 the function f (x) = 9 − 36 x and the x-axis that were found in Example 1 and in the solution to Activity 1. It also shows, to three decimal places, some further approximate values that were found in the same way, but using larger numbers of subintervals. The calculations were carried out using a computer. Table 1 Approximate values for the area in Figure 1(b) Number of subintervals 4 6 12 50 100 500 Approximation obtained 202.5 210 214.5 215.914 215.978 215.999 You can see that as the number of subintervals gets larger and larger, the approximation obtained seems to be getting closer and closer to a particular number. We say that this number is the limit of the approximations as the number of subintervals tends to infinity. It’s the 1 2 x exact value of the area between the graph of the function f (x) = 9 − 36 and the x-axis. From Table 1, it looks as if the exact value of this area is 216, or perhaps a number very close to 216. 109 Unit 8 Integration methods In general, if you have a continuous function f whose graph lies on or above the x-axis throughout an interval [a, b], then you can use the method demonstrated above to find, as accurately as you want, the area between the graph of f and the x-axis, from x = a to x = b. Figure 3(a) illustrates an area of this type, and Figure 3(b) illustrates the approximation to the area that’s obtained by using 8 subintervals. Figure 3 (a) The area between a graph and the x-axis, from x = a to x = b (b) A collection of rectangles whose total area is approximately this area Here’s a summary of the method that you’ve met for finding an approximate value for an area of this type. You start by dividing the interval [a, b] into a number, say n, of subintervals of equal width. The width of each subinterval is then (b − a)/n. For each subinterval you calculate the product , b−a left endpoint f , (1) × of subinterval n and you add up all these products. In general, the larger the number n of subintervals, the closer your answer will be to the area between the graph of f and the x-axis, from x = a to x = b. Now suppose that you have a continuous function f whose graph lies on or below the x-axis throughout an interval [a, b], and you want to calculate the area between the graph of f and the x-axis, from x = a to x = b, as illustrated in Figure 4(a). You can use the method above, with a small adjustment at the end, to calculate an approximate value for an area like this. Consider what happens when you apply the method to a graph like the one in Figure 4(a). You start by dividing the interval [a, b] into n subintervals of equal width, then you calculate all the products of form (1) and add them all up. For a graph like this, because the value of f at the left endpoint of each subinterval is negative (or possibly zero), each product of form (1) will also be negative (or zero). In fact, each product of form (1) will be the negative of the area between the line segment that approximates the curve and the x-axis, as illustrated in Figure 4(b). So when you add up all the products of form (1), you’ll obtain an approximate value for the negative of the area between the curve and the x-axis, from x = a to x = b. 110 1 Areas, signed areas and definite integrals Figure 4 (a) The area between another graph and the x-axis, from x = a to x = b (b) A collection of rectangles whose total area is approximately this area That’s not a problem, because you can simply remove the minus sign to obtain the approximate value for the area that you want. However, to help us deal with situations like this, it’s useful to make the following definitions. These definitions will be important throughout the unit. Consider any region on a graph that lies either entirely above or entirely below the x-axis. The signed area of the region is its area with a plus or minus sign according to whether it lies above or below the x-axis, respectively. For example, in Figure 5 the two shaded regions above the x-axis have signed areas +4 and +6, respectively, which you can write simply as 4 and 6, and the shaded region below the x-axis has signed area −3. Figure 5 Regions on a graph If you have a collection of regions on a graph, where each region in the collection lies either entirely above or entirely below the x-axis, then the total signed area of the collection is the sum of the signed areas of the individual regions. For example, the total signed area of the collection of three regions in Figure 5 is 4 + (−3) + 6 = 7. The units for signed area on a graph are, as you’d expect, the same as the units for area on the graph. That is, they’re the units on the vertical axis times the units on the horizontal axis. If there are no specific units on the axes, then we don’t use any specific units for signed area. 111 Unit 8 Integration methods Example 2 Understanding signed areas The areas of some regions on a graph are marked below. In each of parts (a)–(c), use these areas to find the signed area between the graph and the x-axis, from the first value of x to the second value of x. (a) From x = −3 to x = −2. (b) From x = −2 to x = 1. (c) From x = −3 to x = 1. Positive signed area Negative signed area Solution (a) The signed area from x = −3 to x = −2 is −2.2. (b) The signed area from x = −2 to x = 1 is +4.8 = 4.8. (c) The signed area from x = −3 to x = 1 is −2.2 + 4.8 = 2.6. Of course, the signed area value found in Example 2(c) doesn’t correspond to any actual area on the graph, as it’s the sum of a positive signed area and a negative signed area. 112 1 Activity 2 Areas, signed areas and definite integrals Understanding signed areas Consider again the graph in Example 2. In each of parts (a)–(d) below, use the given areas to find the signed area between the graph and the x-axis, from the first value of x to the second value of x. (a) From x = 1 to x = 5. (b) From x = 5 to x = 7. (c) From x = 1 to x = 7. (d) From x = 1 to x = 9. With the definition of signed area that you’ve now seen, the method that you’ve met in this subsection can be described concisely as follows. Strategy: To find an approximate value for the signed area between the graph of a continuous function f and the x-axis, from x = a to x = b Divide the interval between a and b into n subintervals, each of width (b − a)/n. For each subinterval, calculate the product , b−a endpoint of subinterval , f × nearest a n and add up all these products. In general, the larger the number n of subintervals, the closer your answer will be to the required signed area. Notice that the box above uses the phrase ‘endpoint of subinterval nearest a’, rather than ‘left endpoint of subinterval’, which means the same thing. This will be convenient later in this subsection. (Note that, in this phrase, it’s the endpoint of the subinterval, not the subinterval itself, that’s nearest a!) Figures 6 and 7 illustrate the strategy above. If you add up the signed areas of all the rectangles in Figure 6(a), then you’ll obtain an approximate value for the signed area in Figure 6(b). Figure 6 (a) A collection of rectangles whose total signed area is approximately the signed area shown in (b) 113 Unit 8 Integration methods Similarly, if you add up the signed areas of all the rectangles in Figure 7(a), then you’ll obtain an approximate value for the signed area in Figure 7(b). Figure 7 (a) Another instance of a collection of rectangles whose total signed area is approximately the signed area shown in (b) Activity 3 Calculating an approximate value for a signed area Use the method described in the box above, with six subintervals as shown on the left below, to find an approximate value for the signed area between the graph of the function f (x) = 3 − x2 and the x-axis from x = −3 to x = 3, as shown on the right below. In the next activity you can use a computer to explore approximations to signed areas found by using subintervals in the way that you’ve seen. 114 1 Activity 4 Areas, signed areas and definite integrals Calculating more approximate values for signed areas Open the applet Approximations for signed areas. Check that the function is set to f (x) = 3 − x2 , the values of a and b are set to −3 and 3 respectively, and the number of subintervals is set to 6. The resulting approximation for the signed area between the graph of f and the x-axis, from x = a to x = b, should then be the value found in the solution to Activity 3. Now increase the number of subintervals, and observe the effect on the approximation. What do you think is the exact value of the signed area? Experiment by changing the function, the values of a and b, and the number of subintervals, to find approximate values for some other signed areas. So far in this subsection, whenever a signed area from x = a to x = b has been discussed, the value of b has been greater than the value of a. The value of b can also be equal to the value of a, and this gives a signed area of zero, as illustrated in Figure 8. In cases like this, the method that you’ve seen for using subintervals to find approximate values for signed areas gives the answer zero, since the width (b − a)/n of each subinterval is zero. Figure 8 A signed area equal to zero In fact, for reasons that you’ll see later in the unit, it’s useful to extend the definition of signed area a little, to give a meaning to the phrase ‘the signed area between the graph of f and the x-axis from x = a to x = b’ when b is less than a. It’s not immediately obvious what the phrase means in this case, but it turns out that the natural meaning is as follows. This definition will be important throughout the unit. 115 Unit 8 Integration methods Suppose that f is a continuous function whose domain includes the interval [b, a], as illustrated in Figure 9. Then the signed area between the graph of f and the x-axis from x = a to x = b is defined to be the negative of the signed area between the graph of f and the x-axis from x = b to x = a. Figure 9 The graph of a continuous function f whose domain includes the interval [b, a] For example, consider the graph in Figure 10. The signed area from x = 1 to x = 3 is 4, so the signed area from x = 3 to x = 1 is −4. Similarly, the signed area from x = 6 to x = 8 is −5, so the signed area from x = 8 to x = 6 is −(−5) = 5. Figure 10 Areas on a graph Because of this extended definition of signed area, you now have to think a little more carefully when you read or use the phrase ‘signed area’ (and the signed area isn’t zero). If there’s no mention of ‘from x = a to x = b’, then to decide whether the signed area is positive or negative you just need to consider whether it lies above or below the x-axis. However, if the text is of the form ‘the signed area from x = a to x = b’, then to decide whether the signed area is positive or negative you also have to consider whether b is greater than or less than a. This is illustrated in the next example. A helpful way to think about whether b is greater than or less than a is to think about whether x moves forward or backward as it changes from x = a to x = b. 116 1 Example 3 Areas, signed areas and definite integrals Understanding the extended definition of signed area The areas of some regions on a graph are marked below. In each of parts (a)–(d), use the given areas to find the signed area between the graph and the x-axis, from the first value of x to the second value of x. (a) From x = 3 to x = 3. (c) From x = 10 to x = 6. (b) From x = 6 to x = 10. (d) From x = 6 to x = −1. Solution (a) From 3 to 3 is from a number to the same number. The signed area from x = 3 to x = 3 is 0. (b) From 6 to 10 is forward. The signed area from x = 6 to x = 10 is −5.4. (c) From 10 to 6 is backward. The signed area from x = 6 to x = 10 is −5.4, so the signed area from x = 10 to x = 6 is 5.4. (d) From 6 to −1 is backward. The signed area from x = −1 to x = 6 is −2.4 + 4.0 = 1.6, so the signed area from x = 6 to x = −1 is −1.6. In each part of the next activity, start by thinking about whether x moves forward or backward (or stays fixed) as it moves from the first value of x to the second value of x. 117 Unit 8 Integration methods Activity 5 Understanding the extended definition of signed area The graph in Example 3 is repeated below. In each of parts (a)–(g), use the given areas to find the signed area between the graph and the x-axis, from the first value of x to the second value of x. (a) From x = 2 to x = 6. (c) From x = 2 to x = 10. (e) From x = 8 to x = 8. (b) From x = 6 to x = 2. (d) From x = 10 to x = 2. (f) From x = 2 to x = −3. (g) From x = 10 to x = −1. The subinterval method for finding approximate values for signed areas, which is summarised in the box on page 113, applies no matter whether the value of b is greater than, less than, or equal to the value of a. The reason why it works in cases where b is less than a is that in these cases the ‘width’ (b − a)/n of each subinterval is negative. You might like to use the applet Approximations for signed areas to see examples of the method giving approximate values for signed areas in cases where b is less than a. Of course, if you want to work out an area between a curve and the x-axis over an interval [a, b] by using the method that you’ve seen in this subsection, then normally you would work out the signed area from a to b, rather than the signed area from b to a...
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