· To use the Fundamental Theorem of Calculus to determine areas of regions bounded by functions, vertical lines, and the x-axis. Supported by learning objectives 1, 2, and 3.
| I.1 |
| I.2 |
| I.3 |
Objective
I.1 Back to Top
To use the Fundamental Theorem of Calculus to determine the area bounded by a function above, the x -axis below, and the vertical lines 
and 
.
To represent problems and understandings through a variety of communication modes (COM).
Instructional Notes
In section H.3, it was noted that definite integration is related to finding area under a curve. In this section we want to prove that is the case, and then use our result to determine objective areas bounded by a function above, the x -axis below, and the vertical lines 
and 
.
Suppose that 
is a continuous function on an open interval I and that 
for all points x in I . Then for any interval 
in I , 
is equal to the area of the region under the graph of 
and above the x -axis between the vertical lines 
and 
. This result is significant in that to find the area we need only find an anti-derivative 
of 
and then evaluate 
.
Proof: Let 
be any point in the interval I that is to the left of a . Let 
be a function that gives the area under the graph of 
and above the x -axis between the vertical lines drawn at 
and any point x in I that is to the right of 
as shown in the figure below.
Then 
is the area under the graph of 
and above the x -axis between the vertical lines drawn at 
and any point 
( h being positive) in I that is to the right of x as shown in the figure below.
If 
is the area below 
and above the x -axis from 
to 
(figure 3), and if 
is the area below 
and above the x -axis from 
to x (figure 2), then 
is the area between x and 
as shown in the figure 4 below.
Examine the shaded area in figure 4 more closely. The distance along the bottom is 
or h . The length of the left side is the same as the y value of the function at x , namely 
. The length of the right side is the same as the y value of the function at 
, namely 
.
Suppose we could create a rectangle whose width was h (the same as the width of the shaded region) and whose length, L , was such that the area of the rectangle was exactly the same as the shaded area in figure 4, namely 
. Since the area of the rectangle would be 
, we would have 
. Thus, the length of the rectangle would be 
.
When figures 5 and 6 are superimposed, it can be seen that if the area of the rectangle is the same as the shaded area, then the upper edge of the rectangle will cross the curved top of the shaded area (see figure 7).
Notice that as h becomes smaller, the length, L , of the rectangle, namely 
, comes closer to the length of the left edge of the shaded region, namely 
(see figure 8).
Thus, 
. Since the length of the rectangle is 
, we have: 
.
However, 
is just 
, by the definition of the derivative. Thus, 
. The derivative of the area function is the same as the function under which the area was contained.
Integrating statement 1 yields 
‚ . Since 
is the anti-derivative of 
, integration ‚ results in 
. Thus, 
.
What is the expression 
? Recall that 
was our area function giving the area under the curve from 
to any point x . Thus 
is the area under the curve from 
to b . Similarly, 
is the area under the curve from 
to a . Thus, 
is the area under the curve between a and b .
Although the proof is long, its visual aspects make it quite understandable. Having proven the theorem, students can now apply it to find area. Stress the importance of the conditions of the theorem – the function must be continuous and must not dip below the x -axis. Illustrate the theorem with two or three examples. Show students what happens if the conditions of the theorem are ignored.
Example: Find the area bounded by the curve 
, the x -axis, and the lines 
and 
.
Solution: Insist that the students draw a sketch of the graph and have them shade in the region whose area they are finding. They should not blindly apply the theorem without the sketch, because they must be sure that the function is continuous and lies entirely above the x- axis. If they have trouble with the sketch, have them use a table of values, although by this point in the course students should know the graphs of the basic functions.
Note the use of 
to denote area units. If the axes had been marked off in cm, then we could report the area as 
.
Example: Find the area bounded by the curve 
, the x -axis, and the lines 
and 
.
Solution: Draw a sketch of the situation. Notice that the function is continuous and does not dip below the x -axis on this interval. 
It is common to leave the answer in this format. 
and, thus, the area can be approximated by a decimal value.
Example: Find the area bounded by the curve 
, the x -axis, and the lines 
and 
.
Solution: Notice that the function is continuous and does not dip below the x -axis on this interval. Thus, the area is given by 
. If this integral is evaluated by u -substitution, one obtains 
.
Thus 
Note that the function has an infinite discontinuity at 
. If one was asked to find the area under the curve and above the x -axis between 
and 
, the appropriate response would be that the Fundamental Theorem of Calculus does not apply because the function is discontinuous in that interval.
If one blindly applies the theorem, you would obtain 
This answer is nonsensical. Firstly, area cannot be a negative value. Secondly, the vertical asymptote line at 
would likely cause the area between 
and 
to become unbounded.
Example: Find the area bounded by the line 
, the x -axis, and the vertical lines 
and 
.
Solution: If students try to apply the Fundamental Theorem of Calculus without drawing a sketch, they will obtain 
. Clearly, something is inconsistent as area should not be 0. If a sketch is drawn, one realizes that in the interval between 
and 
, the function dips below the x -axis and thus the Fundamental Theorem of Calculus does not apply.
If the theorem is inappropriately applied to areas below the x -axis, these areas will turn out to be negative values. In this example, since the area above the x -axis (a positive value) is congruent to the area below the x -axis (a negative value), the two values offset one another resulting in the answer of 0 that was obtained.
This shows the importance of first sketching the graph of the situation so students can be sure that the Fundamental Theorem of Calculus applies.
Examples/Activities
The first few questions should give students the opportunity to find the area of a region using the Fundamental Theorem of Calculus and then to verify that this area is correct by using formulas from elementary geometry.
1. Use the Fundamental Theorem of Calculus to find the area bounded by the line 
, the x -axis, and the lines 
and 
. Confirm your result by finding the same area using the formula for the area of a rectangle.
2. Use the Fundamental Theorem of Calculus to find the area bounded by the line 
, the x -axis, and the lines 
and 
. Confirm your result by finding the same area using the formula for the area of a triangle.
3. Use the Fundamental Theorem of Calculus to find the area bounded by the line 
, the x -axis, and the lines 
and 
. Confirm your result by finding the same area using the formula for the area of a trapezoid ( 
).
4. Find the area bounded by the x -axis below, 
above, and the given pair of vertical lines. Draw a sketch of the situation (COM).
(a) 
, 
, 
(b) 
, 
, 
(c) 
, 
, 
(d) 
, 
, 
(e) 
, 
, 
(f) 
, 
, 
(g) 
, 
, 
(h) 
, 
, 
(i) 
, 
, 
(j) 
, 
, 
(k) 
, 
, 
(l) 
, 
, 
(m) 
, 
, 
Students who have studied physics realize that the area under a velocity/time graph gives the distance that the object has travelled. For example, if a cyclist travels at a constant speed of 20 km/h, we know that the cyclist will have travelled 
km between the first and fourth hour of travelling. This is the area shown in the figure below.
This area can also be expressed as 
by using the Fundamental Theorem of Calculus.
If an object travels at a velocity of 
from time 
to time 
where 
, then the area under the velocity/time graph, in other words the distance travelled by the object, will be given by the definite integral 
. Use this result to find the distance travelled by a cyclist from time 
to time 
if the velocity of the cyclist is given by the function 
as shown.
1. 
2. 
3. 
4. 
If you earn $10/hour and you work for 4 hours, you will have earned 
. Graphically, we can represent your earnings as the area above the t -axis and under the curve 
(wage at time t is $10/h) bounded by the lines 
and 
. Your wage could be calculated using the Fundamental Theorem of Calculus as 
.
Suppose your wage/hour after working t hours was $ t /hour. Thus after working 2 hours, you would be making $2/h; after working 20 hours, you would be making $20/h. Then the graph of your wage/hour versus time, t , would look as follows:
To find the amount of money that you would have made in total, after 4 hours of work, you need to determine the area in the figure above. Clearly, you can do that without the Fundamental Theorem of Calculus ( 
). If you did apply the Fundamental Theorem of Calculus, however, you would find the area using 
.
In general, if your hourly earnings were given by the function 
, then the total amount you would have earned after working b hours would be given by the integral 
, as this integral would determine the area above the t -axis and under the function 
bounded by the vertical lines 
and 
. Use this result to find how much money you would earn in total after 8 hours if your wages were given by the function 
given below. Then divide your answer by 8 to determine your average wage/hour.
1. 
2. 
3. 
4. 
5. 
Suggestions/Extensions
The graphing calculator can show and calculate areas under curves. To find the area bounded by the curve 
, the x -axis, and the lines 
, and 
as we did previously in this section, proceed as follows.
1. Draw the graph of the function using a window that shows the region whose area is to be determined.
2. With the graph of the function on the screen, press 2nd GRAPH to access the CALC screen.
Press 7.
To “lower limit?”, respond with 1 and press ENTER. To “upper limit?”, respond with 3 and press ENTER.
The area is shaded and determined.
To try and find the area under a curve such as 
, above the x -axis and to the right of the line 
, proceed as follows.
Find the area from 
to 
, where 
as follows:
Thus, the area from 
to 
is 
. As b gets larger, find the area by taking 
, that is 
. Clearly, this limit is 
. Thus, there is an infinite area in this region.
Use the technique above to show that the area below the curve 
, above the x -axis, and to the right of 
is not infinite but actually 
.
Prove that the area bounded by the arch 
and the x -axis is two-thirds the base times the height.
Objective
I.2 Back to Top
To determine the area bounded by the x -axis above, a function below, and the vertical lines 
and 
.
To represent problems and understandings through a variety of communication modes (COM).
Instructional Notes
Before explaining to students, present them with the challenge of this objective. It is highly likely that someone in the class will think of reflecting 
about the x -axis as is explained below.
In section I.1 we proved the Fundamental Theorem of Calculus which permitted us to find areas bounded by a function above, the x -axis below, and the vertical lines 
and 
. To find the area bounded by the x -axis above, a function below, and the vertical lines 
and 
, we need only reflect the function above the x -axis first and then apply the Fundamental Theorem of Calculus from I.1.
If the graph of 
is below the x -axis, then the graph of 
will be above the x -axis.
Reflection about the x -axis will not change the area. Since 
is above the x -axis, we can apply the Fundamental Theorem of Calculus and state that the required area is 
.
Thus if 
is a continuous function on an open interval I , and if 
for all points x in I , then for any interval 
in I , 
is equal to the area of the region above the graph of 
and below the x -axis between the vertical lines 
and 
.
Example: Find the area bounded by the x -axis, the curve 
, and the vertical lines 
and 
.
Solution: As in I.1, require students to draw a sketch of the situation (COM).
Since the graph of 
lies below the x -axis between 
and 
, we will find the required area using 
. Thus, we have:
Example: Find the area trapped by the function 
and the x -axis.
Solution: A cubic polynomial function could intersect the x -axis at most three times. To find the x -intercepts, it will be necessary to solve 
, since y- coordinates are 0 along the x -axis. If 
, then 
, and factoring further, we have 
. Thus the x -intercepts are 
, 
, and 
. We should sketch the curve from 
to 
.
Because in region I, 
is above the x -axis, its area can be determined by 
. Evaluating, we have:
In region II, 
lies below the x -axis. Thus, the area can be determined by 
or 
. Evaluating, we have:
Thus, the total area is 
.
Examples/Activities
Use definite integration to find the area bounded by the function 
below, the x -axis above, and the given pair of vertical lines. Draw a sketch of the situation (COM). All answers must be positive. Do not merely switch the sign of the answer at the end, but set up the integrals properly so that the answer will be positive.
1. 
, 
, 
2. 
, 
, 
3. 
, 
, 
4. 
, 
, 
5. 
, 
, 
6. 
, 
, 
7. 
, 
, 
8. 
, 
, 
9. 
, 
, 
10. 
, 
, 
Find the area between the graph of 
, the x -axis, and the two vertical lines. Draw a sketch of the situation (COM). In each of these questions, the function will be both above and below the x -axis. You will need a good sketch of the situation. Begin by finding the x -intercept(s) of the function.
1. 
, 
, 
2. 
, 
, 
3. 
, 
, 
(Hint: There are three regions.)
4. 
, 
, 
5. 
, 
, 
Find the area trapped by the function and the x -axis.
1. 
2. 
Suggestions/Extensions
Consider the region bounded by the curve 
, the x -axis, and the vertical lines 
and 
. Find the equation of the vertical line that would divide the region in half. (Answer: 
)
Determine the area of the region shaded in the figure below. The function is 
.
Objective
I.3 Back to Top
To determine the area of the region bounded by the continuous functions 
above, 
below, and the vertical lines 
and 
.
To represent problems and understandings through a variety of communication modes (COM).
Instructional Notes
As in I.2, present the students with the challenge in this objective and ask them how they would determine the required area. Quite likely, they will be able to describe the required procedure.
Case One: 
Find the area shaded below.
The integral 
gives the shaded area below.
The integral 
gives the shaded area below.
Thus, the area we wish to determine in figure 1 is the difference of the areas shown in figures 2 and 3. We can express this as 
. Since the bounds of integration are the same, we can write the difference of the two integrals as the one integral: 
.
Case Two: 
Find the area shown below.
The integral 
gives the shaded area below.
The integral 
gives the shaded area below. Note that 
lies below the x -axis, hence the negative sign.
Thus, the area we wish to determine in figure 1 is the sum of the areas shown in figures 2 and 3. We can express this as 
. Since the bounds of integration are the same, we can write the sum of the two integrals as the one integral: 
. Note that this is the same result obtained in Case One.
Case Three: 
Find the area shown below.
The integral 
gives the shaded area below.
The integral 
gives the shaded area below.
Thus, the area we wish to determine in figure 1 is the difference of the areas shown in figures 2 and 3. We can express this as 
. Since the bounds of integration are the same, we can simplify and write the difference of the two integrals as the one integral: 
. Note that this is the same result obtained in Cases One and Two.
Over any interval 
in which 
, both being continuous functions, the area bounded by the lines 
and 
is always given by 
.
Illustrate the result with a few examples. Be sure to stress the importance of drawing a sketch (COM).
Example: Find the area bounded by the functions 
, 
, and the vertical lines 
and 
.
Solution: A sketch of the required region is essential. A short table of values can be used to guide the sketch, as you can tell that one function is quadratic, the other linear. You need only concern yourself with a sketch over the interval 
.
Since 
throughout the interval 
, the area can be determined using 
.
Example: Find the area bounded by the curves 
and 
.
Solution: First find where the curves intersect one another. At their point(s) of intersection, the y -coordinates of both functions will be the same. Thus, we set 
and solve for x . If 
, then 
, and 
. Factoring, we have 
. Solving, 
, 
, and 
. Sketch the graph of both functions in the interval 
.
Note that, in the example above, both 
and 
are odd functions (have origin symmetry). Thus, we could have just found the area of region II and doubled that answer.
Region I is formed by 
above and 
below in the interval 
. Thus, the area of region I is 
. Region II is formed by 
above and 
below in the interval 
. Thus, the area of region II is 
. The total area enclosed is 
.
Examples/Activities
Find the area bounded by the two curves and the given pair of lines. Draw a sketch of the situation.
1. 
, 
, 
, 
2. 
, 
, 
, 
3. 
, 
, 
, 
4. 
, 
, 
, 
(leave 
in your answer)
5. 
, 
, 
, 
6. 
, 
, 
, 
7. 
, 
, 
, 
(leave your answer in terms of e )
8. 
, 
, 
, 
9. 
, 
, 
, 
10. 
, 
, 
, 
Find the area trapped by the following curves. Draw a careful sketch of the situation (COM).
1. 
, 
2. 
, 
(there are two regions to consider)
3. 
, 
4. 
, 
5. 
, 
, in the interval 
6. 
, 
7. 
, 
(there are two regions to consider)
A contractor with a mathematical bent wants to pour a concrete driveway in the shape of a sine wave. The equation of the top curve is 
while the equation of the bottom curve is 
. The horizontal length (left to right) of the driveway is 
metres.
Find the area of the driveway above.
A second contractor, seeing the curves of the driveway above, thought that a lot of concrete would be wasted, and suggested the driveway should be rectangular in shape, 2 metres wide by 
metres long. Is the area of this driveway different from the one above?
Is the width of the curved driveway really 2 m?
Suggestions/Extensions
Find the area bounded by the curves 
and 
. A careful sketch is essential. There are two regions to consider. (Answer 
)