**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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