Calculus/Definite integral

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Definite integral

Suppose we are given a function and would like to determine the area underneath its graph over an interval. We could guess, but how could we figure out the exact area? Below, using a few clever ideas, we actually define such an area and show that by using what is called the definite integral we can indeed determine the exact area underneath a curve.

Contents

Definition of the Definite IntegralEdit

 
Figure 1: Approximation of the area under the curve   from   to  .
 
Figure 2: Rectangle approximating the area under the curve from   to   with sample point   .

The rough idea of defining the area under the graph of   is to approximate this area with a finite number of rectangles. Since we can easily work out the area of the rectangles, we get an estimate of the area under the graph. If we use a larger number of smaller-sized rectangles we expect greater accuracy with respect to the area under the curve and hence a better approximation. Somehow, it seems that we could use our old friend from differentiation, the limit, and "approach" an infinite number of rectangles to get the exact area. Let's look at such an idea more closely.

Suppose we have a function   that is positive on the interval   and we want to find the area   under   between   and   . Let's pick an integer   and divide the interval into   subintervals of equal width (see Figure 1). As the interval   has width  , each subinterval has width   . We denote the endpoints of the subintervals by   . This gives us

 
 
Figure 3: Riemann sums with an increasing number of subdivisions yielding better approximations.

Now for each   pick a sample point   in the interval   and consider the rectangle of height   and width   (see Figure 2). The area of this rectangle is   . By adding up the area of all the rectangles for   we get that the area   is approximated by

 

A more convenient way to write this is with summation notation:

 

For each number   we get a different approximation. As   gets larger the width of the rectangles gets smaller which yields a better approximation (see Figure 3). In the limit of   as   tends to infinity we get the area   .

Definition of the Definite Integral
Suppose   is a continuous function on   and   . Then the definite integral of   between   and   is

 
where   are any sample points in the interval   and   for   .

It is a fact that if   is continuous on   then this limit always exists and does not depend on the choice of the points   . For instance they may be evenly spaced, or distributed ambiguously throughout the interval. The proof of this is technical and is beyond the scope of this section.

Notation

When considering the expression,   (read "the integral from   to   of the   of    "), the function   is called the integrand and the interval   is the interval of integration. Also   is called the lower limit and   the upper limit of integration.
 
Figure 4: The integral gives the signed area under the graph.

One important feature of this definition is that we also allow functions which take negative values. If   for all   then   so   . So the definite integral of   will be strictly negative. More generally if   takes on both positive and negative values then   will be the area under the positive part of the graph of   minus the area above the graph of the negative part of the graph (see Figure 4). For this reason we say that   is the signed area under the graph.

Independence of VariableEdit

It is important to notice that the variable   did not play an important role in the definition of the integral. In fact we can replace it with any other letter, so the following are all equal:

 

Each of these is the signed area under the graph of   between   and   . Such a variable is often referred to as a dummy variable or a bound variable.

Left and Right Handed Riemann SumsEdit

 
Figure 5: Right-handed Riemann sum
 
Figure 6: Left-handed Riemann sum

The following methods are sometimes referred to as L-RAM and R-RAM, RAM standing for "Rectangular Approximation Method."

We could have decided to choose all our sample points   to be on the right hand side of the interval   (see Figure 5). Then   for all   and the approximation that we called   for the area becomes

 

This is called the right-handed Riemann sum, and the integral is the limit

 

Alternatively we could have taken each sample point on the left hand side of the interval. In this case   (see Figure 6) and the approximation becomes

 

Then the integral of   is

 

The key point is that, as long as   is continuous, these two definitions give the same answer for the integral.

ExamplesEdit

Example 1
In this example we will calculate the area under the curve given by the graph of   for   between 0 and 1. First we fix an integer   and divide the interval   into   subintervals of equal width. So each subinterval has width

 

To calculate the integral we will use the right-handed Riemann sum. (We could have used the left-handed sum instead, and this would give the same answer in the end). For the right-handed sum the sample points are

 

Notice that   . Putting this into the formula for the approximation,

 

Now we use the formula

 

to get

 

To calculate the integral of   between   and   we take the limit as   tends to infinity,

 

Example 2
Next we show how to find the integral of the function   between   and   . This time the interval   has width   so

 

Once again we will use the right-handed Riemann sum. So the sample points we choose are

 

Thus

   
 
 
 

We have to calculate each piece on the right hand side of this equation. For the first two,

 
 

For the third sum we have to use a formula

 

to get

 

Putting this together

 

Taking the limit as   tend to infinity gives

   
 
 
 

ExercisesEdit

1. Use left- and right-handed Riemann sums with 5 subdivisions to get lower and upper bounds on the area under the function   from   to   .

Lower bound:  
Upper bound:  

2. Use left- and right-handed Riemann sums with 5 subdivisions to get lower and upper bounds on the area under the function   from   to   .

Lower bound:  
Upper bound:  

Solutions

Basic Properties of the IntegralEdit

From the definition of the integral we can deduce some basic properties. For all the following rules, suppose that   and   are continuous on   .

The Constant RuleEdit

Constant Rule

 

When   is positive, the height of the function   at a point   is   times the height of the function   . So the area under   between   and   is   times the area under   . We can also give a proof using the definition of the integral, using the constant rule for limits,

 

Example

We saw in the previous section that

 

Using the constant rule we can use this to calculate that

  ,
  .

Example

We saw in the previous section that

 

We can use this and the constant rule to calculate that

 

There is a special case of this rule used for integrating constants:

Integrating Constants

If   is constant then  

When   and   this integral is the area of a rectangle of height   and width   which equals   .

Example

 
 
 

The addition and subtraction ruleEdit

Addition and Subtraction Rules of Integration
 

 

As with the constant rule, the addition rule follows from the addition rule for limits:

   
 
 

The subtraction rule can be proved in a similar way.

Example

From above   and   so

 
 

Example

 

ExerciseEdit

3. Use the subtraction rule to find the area between the graphs of   and   between   and  

 

Solution

The Comparison RuleEdit

 
Figure 7: Bounding the area under   on  

Comparison Rule

  • Suppose   for all   . Then
 
  • Suppose   for all   . Then
 
  • Suppose   for all   . Then
 

If   then each of the rectangles in the Riemann sum to calculate the integral of   will be above the   axis, so the area will be non-negative. If   then   and by the first property we get the second property. Finally if   then the area under the graph of   will be greater than the area of rectangle with height   and less than the area of the rectangle with height   (see Figure 7). So

 

Linearity with respect to endpointsEdit

Additivity with respect to endpoints Suppose   . Then

 

Again suppose that   is positive. Then this property should be interpreted as saying that the area under the graph of   between   and   is the area between   and   plus the area between   and   (see Figure 8).

 
Figure 8: Illustration of the property of additivity with respect to endpoints

Extension of Additivity with respect to limits of integration
When   we have that   so

 

Also in defining the integral we assumed that   . But the definition makes sense even when   , in which case   has changed sign. This gives

 

With these definitions,

 
whatever the order of   .

ExerciseEdit

4. Use the results of exercises 1 and 2 and the property of linearity with respect to endpoints to determine upper and lower bounds on   .

Lower bound:  
Upper bound:  

Solution

Even and odd functionsEdit

Recall that a function   is called odd if it satisfies   and is called even if   .

Suppose   is a continuous odd function then for any   ,

 

If   is a continuous even function then for any   ,

 

Suppose   is an odd function and consider first just the integral from   to   . We make the substitution   so   . Notice that if   then   and if   then   . Hence

  .

Now as   is odd,   so the integral becomes

  .

Now we can replace the dummy variable   with any other variable. So we can replace it with the letter   to give

  .

Now we split the integral into two pieces

  .

The proof of the formula for even functions is similar.

5. Prove that if   is a continuous even function then for any   ,
  .

From the property of linearity of the endpoints we have

 

Make the substitution   .   when   and   when   . Then

 

where the last step has used the evenness of   . Since   is just a dummy variable, we can replace it with   . Then

 
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Definite integral