On-line Math 21

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.

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.

We'll try it again: now, let's approximate this area using 20 rectangles. The picture is (using the highest points, the circumscribed 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,... ,x_n} to chop the interval [a,b] up into subintervals. Set a = x₀ and set b = x_n 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 i^th subinterval).

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,x_n = b . We usually just refer to the numbers x₀,¼,x_n 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,x_n-x_n-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,x_n of mesh ||P||, and choose (randomly) points x_i^* Î [x_i-1,x_i] . 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.

Solution

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

ó õ 2 1  f(x)dx = 9/2.

Solution

Example 3

ó õ 1 0  x2dx = 1 3 .

Solution

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.

Proof:

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.

Solution

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