The definition of the integral

Integration has its theoretical basis in the idea of finding areas of regions in the plane. That is, to find, for example, the area bounded by the graphs
y = x2, y = 0, and x = 3.
The process begins by finding a way to approximate that area. We know the area of a rectangle, so we use that. But, approximating that region by a rectangle is rather a poor approximation, so we instead approximate the pieces of the region obtained by slicing it by the lines, say, x = 1 and x = 2 , that is, we approximate the area by three rectangles, not one. That's not a very good approximation either, but it's a start.

Now, what should be the heights of the rectangles in each piece? The tallest point on the curve y = x2 ? That's the way it's pictured. Or, perhaps the lowest point? Somewhere in between? You can choose any one. Obviously, choosing the tallest point gives a rectangle with more area than the original region, while choosing the lowest gives less area. If you are very lucky, choosing rectangles in between will give exactly the right area. Let's do some estimates.

Using the highest point on each piece (the right-edge of each region gives the height as f(x) , where x is at the right edge), we get:
Area » 1×1+4×1+9×1 = 14,
while using the lowest point gives:
Area » 0×1+1×1+4×1 = 5.

The real area is somewhere in between.

We'll try it again: now, let's approximate this area using 20 rectangles. The picture is (using the highest points, the circumscribed rectangles):

The approximate area of this is 9.68625. I got that using Maple, which has a special package for these calculations. Using the lowest points in each subinterval, that is, using the inscribed rectangles, gives an area of 8.33625 for 20 rectangles.

How do we do this calculation?

The reality is that, for most curves, it doesn't matter whether you use the right edge, or the left, or anywhere in between. You also don't have to take the subintervals to all be the same length, just so they get smaller, to get a better and better approximation of the real area.

We will be spending quite a bit of time dealing with integration as a summation process, and it will get awkward writing down all these sums. We need a better notation. In addition, we need some specific formulas for certain sums, which you may not have seen before. Look through this topic now:

Summation notation and special formulas

5.1  The definition of the Riemann Integral

Let's look at the process in general, for a function f(x) over an interval [a,b] . Choose any old set of points {x0,x1,... ,xn} to chop the interval [a,b] up into subintervals. Set a = x0 and set b = xn so that we refer to them all the same way. For each of those subintervals, find a point
xi* Î [xi-1,xi]
(this is the ith subinterval).

Then, f(xi*) is the height of the curve at that point, so that height gives the height of a rectangle whose top crosses the curve. Well, if f(xi*) < 0 , it's below the axis, so there is a minus sign there. The area of the ith rectangle is then
f(xi*)Dxi,
where where Dxi: = xi-xi-1 is the width of the ith subinterval. This ``area'' is negative when f(xi*) < 0 , but even then the absolute value is the area of the rectangle.

Then, the ``area under the curve'' is approximated by the sum
n
å
i = 1 
f(xi*)Dxi,

The reason the ``area under the curve'' is in quotation marks is that we will still use the idea of this limit of sums, even if the function is negative somewhere, and so the ``area under the curve'' doesn't quite make sense. However, if you count the areas of those regions where f(x) < 0 as negative area, it does give the ``area under the curve'' as this limit of sums.

Definition 1 A partition P of an interval [a,b] is a decomposition of the interval into smaller pieces, as above, a = x0,x1,¼,xn = b . We usually just refer to the numbers x0,¼,xn as the partition P of [a,b] .

The mesh ||P|| of a partition P is the length of the longest section (or subinterval) of the partition, ||P||: = max{x1-x0,x2-x1,¼,xn-xn-1} .

Now comes the formal definition of an integral.

Definition 2 f is some function, defined for x Î [a,b] (you actually can leave out ``a few'' points without causing any problems). For each number d > 0 , choose a partition P = x0,¼,xn of mesh ||P||, and choose (randomly) points xi* Î [xi-1,xi] . The Riemann sum of the function f corresponding to that partition and set of sampling numbers is the sum:
n
å
i = 1 
f(xi*)Dxi.
The number you get from that process depend on the partition and the sampling numbers, as well as the function itself. However, if as ||P||® 0 (as the mesh gets finer without bound), the limit:

lim
||P||® 0 
n
å
i = 1 
f(xi*)Dxi: = ó
õ
b

a 
f(x)dx
exists, independently of the partitions (as long as the mesh is getting smaller), and of the sampling numbers, then we say that the function f is integrable over the interval [a,b] , and the integral is the limit value, written as
ó
õ
b

a 
f(x)dx.

Integration involves a limit of sums. But, the limit is ``sloppy'' in this case in that so many things are not determined: the subintervals themselves, the point you pick in each subinterval, the number of subintervals. Still, rather remarkably, ``most'' functions are integrable in this sense, even ones that are not continuous.

Example 1 If
f(x) = ì
í
î
1,
if 0 £ x £ 1
1/2,
if 1 < x £ 2
,
then
ó
õ
2

0 
f(x)dx = 3/2.

Example 2 If f(x) = 3x , then

ó
õ
2

1 
f(x)dx = 9/2.

Example 3

ó
õ
1

0 
x2dx = 1
3
.

Exercise 1 Show that

ó
õ
2

-1 
x3dx = 15
4
.

Hint: Use the formulas for the sum of the first n cubes in the review section, and, as before, presume that the limit exists, so that you can use uniform subintervals and circumscribed rectangles.

Answer:

Proposition 1 Rules


  1. ó
    õ
    b

    a 
    c dx = c(b-a).


  2. ó
    õ
    b

    a 
    f(x)dx = - ó
    õ
    a

    b 
    f(x)dx.
    Actually, this rule is a definition. We hadn't had a definition for the right-hand side (at least when a < b ); this helps for some later formulas.

  3. ó
    õ
    a

    a 
    f(x)dx = 0.
    As before, this is really a definition.

  4. ó
    õ
    b

    a 
    cf(x)dx = c ó
    õ
    b

    a 
    f(x)dx.


  5. ó
    õ
    b

    a 
    f(x)±g(x)dx = ó
    õ
    b

    a 
    f(x)dx± ó
    õ
    b

    a 
    g(x)dx.

  6. If f(x) ³ 0 for all x Î [a,b] , a < b , then
    ó
    õ
    b

    a 
    f(x)dx ³ 0.

  7. If f(x) ³ g(x) for all x Î [a,b] , a < b , then
    ó
    õ
    b

    a 
    f(x)dx ³ ó
    õ
    b

    a 
    g(x)dx.

  8. For all a , b , and c ,
    ó
    õ
    c

    a 
    f(x)dx+ ó
    õ
    b

    c 
    f(x)dx = ó
    õ
    b

    a 
    f(x)dx.

  9. If m £ f(x) £ M for all x Î [a,b] , a < b , then
    m(b-a) £ ó
    õ
    b

    a 
    f(x)dx £ M(b-a).


  10. ê
    ê
    ó
    õ
    b

    a 
    f(x)dx ê
    ê
    £ ó
    õ
    b

    a 
    | f(x)| dx.

Theorem 1 If f is continuous on [a,b] , then

ó
õ
b

a 
f(x)dx
exists.

Use the rules and theorems above to sove the following exercises.

Exercise 2 Find
ó
õ
3

-2 
| x| dx =

Hint: Split the integral up into easier pieces, and evaluate the integral of each piece by high-school geometry.

Exercise 3 Show that
ó
õ
p/4

0 
sin3xdx £ ó
õ
p/4

0 
sin2xdx.

Answer:

Example 4 Show that
ó
õ
2p

0 
sin2xdx = p.
The same is true for cos2x ,
ó
õ
2p

0 
cos2xdx = p.

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  • Copyright 2000 David L. Johnson