The purpose of this set of supplemental notes is to provide you with more detailed examples of this topic. As you read this set of supplemental notes, make sure that you understand where every step is coming from.

EXAMPLE 1:

Find the area between the line f (x) = 3 - x and the x-axis on the interval [0, 3].

SOLUTION:

STEP 1: Graph the function. (See figure 1)

STEP 2: Set up the integral.

Notice that the area that we want to determine is the area from x = 0 to x = 3. This area is above the x-axis, so the area will positive. You must determine where the graph is above or below the x-axis. The reason for this is that we are finding the

figure 1

area under the curve and if part of the curve lies below the x-axis, then that part of the area will be negative. Therefore, we will have to compensate for it. We will talk more about this in the next example. Recall that the area under the curve is determined by the following formula.

The upper curve in this example is y = 3 - x and the lower curve is the x-axis (y = 0). The integral for this area is the following definite integral.

STEP 3: Evaluate the definite integral.

EXAMPLE 2:

Find the area between the line f (x) = 3 - x and the x-axis on the interval [0, 5].

SOLUTION: We are considering the same function, f(x) = 3 - x, but the interval that we will be integrating over is now [0, 5]. The definite integral for this area is the following.

What is wrong with this solution? Is this the total area? If it is, why is the value smaller than the area that was found when the interval was [0, 3]? Well, to answer these questions, this is not the total area. Part of the area lies below the x-axis. Hence, the area is negative. To take care of this problem, we must divide the interval into two pieces: [0, 3] and [3, 5]. For the interval, [3, 5], the upper curve is the x-axis (y = 0), and the lower curve is y = 3 -x. So, the true area will be the following.

Does this answer seem more reasonable? Is there an algebraic way to calculate this area? (HINT: you have two triangles)

EXAMPLE 3:

Determine the area under the function, f(x) = - x ³+ 6x and the x-axis on the interval [-2, 2].

SOLUTION:

STEP 1: Graph the function. (See figure 2)

STEP 2: Set up the integrals and evaluate.

Notice that on the interval [-2, 0], the graph of the function lies below the x-axis, and on the interval [0, 2], the graph lies above the x-axis. Therefore, we will have to break the area into two integrals. The solution for this area is the following.

figure 2

Would there have been an easier way to calculate this area? Does this graph have any symmetry? Yes, it has symmetry about the origin. Therefore, I could have found the area under the curve for the interval [0, 2] and multiplied the result by 2. Here is how this integral would have been worked.

Notice that it is a lot less work when you use symmetry to simplify the integral. Only use symmetry when calculating the area under a curve.

EXAMPLE 4:

Find the area between the curve f (x) = cos p x on the interval [0, 2].

SOLUTION:

STEP 1: Graph the function. (See figure 3)

STEP 2: Set up the integrals and evaluate.

Notice that the area we have to find is in three pieces. The intervals [0, .5] and [1.5, 2] are above the x-axis, and the interval [.5, 1.5] is below. Therefore, we will need to have three integrals. Also notice that symmetry cannot be used in this problem.

figure 3

The method for determining the area between two or more curves is an important application of integral calculus. It lets us determine the area of non-standard shapes by evaluating the definite integral. There are three main steps to this process. They are:

1.

Graph the two or more equations.

2.

Determine the points of intersection.

3.

Set up and evaluate the definite integral.

In the following examples, I will hopefully illustrate this process.

EXAMPLE 5:

Find the area between the curves f (x) = 4 - x ² and g (x) = x ² - 4.

SOLUTION:

STEP 1: Graph the functions. (See figure 4)

The reason for graphing the two equations is to be able to determine which function is on top and which one is on the bottom. Sometimes, you can also determine the points of intersection. From this graph, it is cleat that f (x) is the upper function, g (x) is the lower function, and that the points of intersection are x = -2 and x = 2.

figure 4

STEP 2: Determine the points of intersection.

If you did not determine the points of intersection from the graph, solve for them algebraically or with your calculator. To find them algebraically, set each equation equal to each other.

4 - x ² = x ² - 4 ® -2x ² = -8 ® x ² = 4 ® x = -2 or x = 2

STEP 3: Set up and evaluate the integral.

Recall from early in the notes, when we were finding the area between the curve and the x-axis, we had to determine the upper and the lower curve. Then the area was defined to be the following integral.

So the definite integral would be the following.

Now, let us evaluate the integral.

If you look at the graph of the two functions carefully, you should have noticed that we could have used some symmetry when setting up the integral. The region is symmetric with respect to both the x- and the y-axis. If we had used the y-axis symmetry, the resulting integral would have had bounds of 0 and 2, and we would have had to take 2 times the area to find the total area. Here is this integral.

If we had used both symmetries, the resulting integral would still have bounds of 0 and 2, but the upper function would have been f (x) and the lower function would be y = 0 (the x-axis). To find the total area, we would have to take this area times 4. Here is this integral.

EXAMPLE 6:

Find the area between the curves f (x) = 0.5sec ² x and g (x) = -4sin ² x over the interval [-p /3, p /3].

SOLUTION:

STEP 1: Graph the functions.

Notice that f (x) is the upper function, and g (x) is the lower function. Since the interval that we will be integrating over is provided, we can skip step 2.

STEP 3: Set up and evaluate the integral.

figure 5

(I will let you do the algebra to complete this problem.)

Now, let us tackle a problem in which we integrate with respect to y.

EXAMPLE 7:

Find the area between the curves x = y ³ and x = y ² that is contained in the first quadrant.

SOLUTION: STEP 1: Graph the functions. (See figure 6)

Since both equations are x in terms of y, we will integrate with respect to y. When integrate with respect to x, we have to determine the upper function and the lower function. Now that we are integrating with respect to y, we must determine what function is the farthest from the y-axis. The function that is the farthest from the y-axis is x = y ². So that will be our upper curve. The lower curve will be the curve that is nearest to the y-axis. In this case, it is the function x = y ³.

STEP 2: Find the points of intersection.

Set the two equations equal to each other.

y ² = y ³ ® y ² - y ³ = 0 ® y ² (1 - y) = 0 ® y = 0 or y = 1

STEP 3: Set up and evaluate the integral.

EXAMPLE 8:

Find the area of the region bounded by the curves y = x, y = 1, and y = x ²/4 that lies in the first quadrant.

SOLUTION:

STEP 1: Graph the functions. (See figure 7)

Now, let us get extremely personal with this graph. You will notice that if we decide to integrate with respect to x, then we will have to divide the area into two pieces. One will be on the interval [0, 1], and the other on the interval [1, 2]. If we decide to integrate with respect to y, then we will

figure 7

only have one area to contend with. So let us do it with respect to y first. The upper function will be y = x ²/4, and the lower function will be y = x.

STEP 2: Determine the points of intersection.

Notice that the one common intersection point of the two functions is the point (0, 0). So we will integrate from y = 0 to y = 1.

STEP 3: Set up and evaluate the integral.

To integrate with respect to y, we must solve y = x ²/4 for x in terms of y.

Now, let us integrate with respect to x. Remember that we will have to have two integrals. For the interval [0, 1] the upper function is y = x and the lower function is y = x ²/4. On the interval [1, 2] the upper function is y = 1 and the lower function is y = x ²/4.

EXAMPLE 9:

Find the area enclosed by the curves y = 2x ² and y = x ⁴ - 2x ².

SOLUTION:

STEP 1: Graph the functions. (See figure 8)

Notice that this area is symmetric, so we can use symmetry to simplify the process of determining the area between the curves. Notice that y = 2x ² is the upper curve.

STEP 2: Determine the intersection points.

figure 8

We must determine the points of intersection. To do this, set the two equations equal to each other and solve for x. You should bind out that x = -2, x = 0, and x = 2 are the points of intersection. Since we are using symmetry, we will be integrating over the interval [0, 2] and taking that area times 2 for the total area.

STEP 3: Set up and evaluate the integral.