Chapter 17 Definite Integrals: Area

Riemann Sums

1. Divide the interval from to into a certain number of equal subintervals , each of length x = ;
2. Draw a vertical line up to the curve at the endpoints of the subintervals;

3. Over each subinterval form a rectangle by going up to the curve at the left endpoint and drawing a horizontal line to the right from that point on the curve;

4. Take the sum of the areas of the thin rectangles.

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n-th (left) Riemann Sum =

Here the summation notation tells us to take all integer values of starting with and ending with , put each value in place of in , and take the sum.

We have , , , and the Riemann sum is

= ,.

Later we will see that the actual area is a little larger, namely 1/3. Notice that rather than multiplying each term by 0.2 and then summing, it is often faster to factor out the term 0.2 , add the remaining terms and then multiply by 0.2 .

In general, if f(x) is an increasing function on the interval (a,b) , then the left endpoint Riemann sum will be a little less than the true area; but if f(x) is a decreasing function on (a,b) , then the left endpoint Riemann sum will be a little greater than the true area.

The larger n is, the closer the Riemann sum will be to the true area. We say that the area is ``the limit of the Riemann sums as n approaches '' and we write it in symbols this way:

In old books, the "integral sign" we use in expressions like is actually the printed form of the letter "S" and "infinitely small'' values of are written . Thus, a good symbol for the limit is

.

In a definite integral it makes no difference what the variable of integration is called. In other words, the area under the function between and means exactly the same thing as the area under the function between and . That is, both and are equal to the same thing because in a definite integral the variable of integration only holds a place, and does not affect the answer.

In the next section, we will discover that there is a relation between the
definite integral and the indefinite integral which you studied in the last chapter.

Computing the Area Under a Curve

for any u between a and b we set

A(u) = area under f(x) between a and u .

If we could find a formula for the function , then we would just substitute to get the area between and that we want.

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= area under between u and u+h which is approximately the area of the rectangle above the interval (u,u+h) of height =

This approximate equality between the area under the curve and the area of the thin rectangle becomes very close to equality when h is small. Dividing through by h , we find that

= (approx) ,

with the approximate equality becoming an equality in the limit as h goes to 0 . That is, the limit of the difference quotient --- the derivative of A(u) --- is equal to f(u) :

.

Once we know that / is , this means that is an antiderivative of . So suppose that we find the antiderivative , using the methods in the last section. Our function must correspond to a certain value of C . Just as before, we need some initial information to determine this value. For the area function A(u) the initial information is: when u = a there is no area, i.e., A(a) = 0 . Thus, = , and hence . We conclude that A(u) = F(u)-F(a) . This means that the total area under the curve between a and b is A(b) = F(b)-F(a) . In other words, the area under f(x) between a and b can be computed by finding an antiderivative of f(x) , substituting b and a in that antiderivative, and subtracting . Using the definite integral notation, we conclude that

Area under f(x) between a and b = = F(x) = , where is an antiderivative of .

Here we used the notation "F(x) " to mean " substitute b and a into F(x) and then subtract .''

Note: You can tell whether an integral is a definite integral or an indefinite integral by whether or not there are limits of integration on the integral sign. The answer to a definite integral problem is usually a specific value, not a function with an unknown constant C as in an indefinite integral problem. However, we will see many definite integrals with variable limits and these evaluate to functions of the limits, but without an arbitrary constant stepping in!

Example 2 : Find the area under:

(a) between and , and

(b) one arch of the sine curve.

(a) = = ( - ) = .

(b) An arch of the sine curve is the region under between
and , so its area is = = -( ) = -(-1)-(-1) = 2 .

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= = - = - =

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Substitution

In integrals evaluated using a u -substitution, as before you must change EVERYTHING from the old variable (or ) to the new variable . In the case of a definite integral, "everything'' includes the limits of integration. You get the new limits of integration (the u -limits of integration) by asking the question: " as x goes from a to b the variable u goes from (what?) to (what?) .'' To answer this question, substitute in place of in the formula that gives in terms of ; this gives the lower -limit of integration. Then substitute in place of in the same formula to get the upper -limit of integration. Here are some examples:

Example 4: Find the area under

(a) between 1 and 2, and

(b) between and 1.

(a) In we make the substitution , so that and . This transforms the integral to

= .

What about the limits of integration? When x is equal to the lower limit of integration 1, we have = 1 ; whereas when x is equal to the upper limit of integration 2, we have = . Thus, our new limits of integration are 1 and 8. So the area is

= = = = 45/28

(b) If we make the substitution , so that = , we see that as x goes from 0 to 1 the variable u goes from 2 to 1. We obtain:

=

=

= - ( )

= ( ) = 0.776

Some Properties of Definite Integrals

1. Like indefinite integrals, they are linear. This means, for example, that any constant inside the integral can be pulled outside (as we did in Example 17.4(a) with the constant 1/7). Also, if you have the definite integral from a to b of f(x)+g(x) , it makes no difference whether you work with the sum all at once, or find the definite integral of f(x) and g(x) separately and then add the results. (The same observation applies to f(x)-g(x) .)

2. If c is a number between a and b , then the area between a and b is equal to the sum of the area between a and c and the area between c and b :

Actually, when you use the information from the next item below, we can get that the above formula is valid even when c is not between a and b .

3. If the lower limit of integration is greater than the upper limit, i.e., if b is on the bottom and a is on the top (so you are traversing the area under the curve from right to left rather than from left to right), then the effect is to put in a negative sign. This is because = . In Example 17.4(b), after making the u -substitution we obtained an integral of the form , which was negative. (We also had a negative constant in front of the integral, so that our final answer was positive.)

4. If the function f(x) is negative, then is negative (assuming that b>a ), i.e., area below the x -axis is counted as negative area. For example, if we compute (see Example 17.2(b)), we obtain as our answer. This makes sense, because Int(sin x , x=Pi..2*Pi) is the "area" of an arch of the sine curve under the x -axis, and so has area -2 .

By property (2) above, we have = 2+(-2) = 0 , i.e., the two equal arches, one above and one below the x -axis, cancel.

Application to Degree-Days

The concept of "degree-days'' arises when one wants to estimate heating costs. Suppose that in the winter you keep a building heated to a constant temperature of F. Let be the outside temperature. Then the cost of heating the building between time t = a and time t = b is roughly proportional to the area between the line y = 65 and the temperature curve y = f(t) from t = a to t = b . For example, if the temperature outside has the constant value f(t) = D , then the area between the two lines y = 65 and y = D is simply (the units are units of temperature times time, i.e., degree-days). In that case all we're saying is that the heating cost is proportional to the temperature difference times the time interval. But when f(t) is not constant we need to use definite integration to find this area.

Example 5: Using the temperature function in Example 8.2 in the section on trig functions, find the number of degree-days in January in an average year in
Seattle.

Our temperature function is . Subtracting this from 65 and simplifying, we obtain . Our time interval is 0 to 30 (recall that we supposed that every month has 30 days). So the number of degree-days is

=
=
= = 821.

Thus, based on this information, every January we can get a rough idea of whether to expect high or low heating bills by comparing the total number of degree-days measured in the particular month with the norm for January, which is 821.

Velocity and Distance Traveled As Area

In Chapter 16 we saw that the velocity is the antiderivative of the acceleration. Another way to say this is that velocity is the area under the acceleration graph . More precisely, the area under between and is equal to the amount the velocity increases during this time interval. In mathematical notation:

.

The graph illustrates this in the special case when a(t) = k is a constant function.

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