At the end of this lesson you will be able to
GEOMETRIC INTERPRETATION OF DEFINITE INTEGRAL
In Section 4.4 Session 1 notes and on page 342 of your textbook, you saw rectangles associate with the Riemann Sum. In the situations where f (x) is are positive (see the blue rectangles on pages 342 and 343), the product of f (x) Δx is the area of a rectangle. In the yellow-brown rectangles, f (x) is negative and the product f (x) Δx is the negative of the area of the rectangle.
In the definite integral we are taking the limit where the number intervals, and therefore the number of rectangles increase without bound and the width of each rectangle, that is the norm of each partition, gets closer and closer to zero. Therefore, in the situation where f (x) ≥ 0, then the definite integral will be the area bounded by the curve y = f (x), x = a, x = b, and the x-axis. Therefore, when f (x) ≥ 0 , this area is the geometric interpretation of the definite integral. In other words, when f (x) ≥ 0 , this area is the the area under the curve and above the horizontal axis.
RELATIONSHIP WITH DEFINITE INTEGRAL WHEN f (x) < 0
Now when f (x) < 0, the dependent variable values are negative, the product f (x) Δx is negative, and the graph is under the horizontal axis. Therefore, when f (x) < 0, area is the negative of the definite integral.
Now you can go to the Session 25 assignments to continue your assignment.
At the end of this lesson you will be able to
X IS THE INDEPENDENT VARIABLE
The chart on page 368 illustrate the steps that we use to calculate area using x as the independent variable. We are going to expand on those steps so that you can calculate area successfully.
1. Graph the curves carefully on a piece of paper. Use your graphing calculator to help you obtain your graphs. (See the graphs on pages 367, 368, and 369 where the authors drew one vertical rectangle for each area.)
a. Sketch a vertical rectangle on the figure and mark it's width as Δ x or dx.
b. Determine the upper curve and call it the function f and determine the lower curve and call it the function g. These must be the same upper and lower curves throughout the region!
2. Limits of integration. That is, find the limits of integration x = a and x = b. That may require you to find the points where the curves intersect. How do you find these points of intersection?
a. Analytically, that is using algebra, solve y = f (x) and y = g (x) simultaneously for x. That is what your authors were doing in Step 2 of Example 4 on page 368 and of Example 5 at the top of page 370.
b. Graphically, using your graphing calculator. That is what you authors did at the top of page 371. After you have entered both equations into your calculator, here are some options:
- On the TI-86, with your calculator in the graph mode, select MORE, then MATH, select MORE again, and then ISECT above the F3 key. For the TI-83 Plus Silver Edition, you can download from the TI webpage the intersection program.
- Use the ZOOM IN feature repeatedly, called ZIN under the F2 key for the TI-86.
- Find the zeros for f (x) - g (x) using the ROOT function. Enter y = f (x) - g (x) into your graph mode of the calculator. With the TI-86 in the graph mode, it is located in the MATH menu above the F1 key. On the TI-83 Plus, it is located in the 2nd CALC menu as the ZERO function.
- Use the SOLVER, 2nd GRAPH on the TI-86 and in the MATH menu on the TI-83 Plus.
3. Area of the rectangle. From our previous math, we know that area = length * width.
- In Step 1, we already noted the width is dx.
- Now the length. The length is the value at the top of the rectangle, f ( x), minus the value at the bottom, g (x). In other words, the length is f (x) - g (x). Analytically, always simplify f (x) - g (x) before going on.
- The area of the rectangle will be (f (x) - g (x)) dx.
4. Integrate (f (x) - g (x)) dx from a to b. Remember that a and b are the limits of integration that we found in Step 2.
The area for the region between curves will be the definite integral.
Y IS THE INDEPENDENT VARIABLE
When y is the independent variable, the steps are the same except for changes due to a different dependent variable. We will show the changes now.
1. Graph the curves carefully on a piece of paper.
a. Sketch a horizontal rectangle on the figure and mark it's width as Δ y or dy.
b. Determine the right curve and call it the function f and determine the left curve and call it the function g. These must be the same right and left curves throughout the region!
2. Limits of integration. That is, find the limits of integration y = a and y = b. That may require you to find the points where the curves intersect. How do you find these points of intersection?
a. Analytically, that is using algebra, solve x = f (y) and x = g (y) simultaneously for y.
b. Graphically, using your graphing calculator. You will have to switch your variables, instead of x = f (y) use y = f (x) and instead of x = g (y) use y = g (x). Remember that you have switched your variables.
3. Area of the rectangle. From our previous math, we know that area = length * width.
- In Step 1, we already noted the width is dy.
- Now the length. The length is the value at the right of the rectangle, f ( y), minus the value at the left, g (x). In other words, the length is f (y) - g (y). Analytically, always simplify f (y) - g (y) before going on.
- The area of the rectangle will be (f (y) - g (y)) dy.
4. Integrate (f (y) - g (y)) dy from a to b. Remember that a and b are the limits of integration that we found in Step 2.
The area for the region between curves will be the definite integral.
USING THE GRAPHING CALCULATOR
Actually, the graphing calculator cannot evaluate the definite integral. Instead it uses approximation techniques like those in Section 4.7. Do remember that these are approximations of the definite integral.
In the graph mode, first enter f (x) - g (x) for y.
For the TI-86 calculator, with calculator in graph mode select MORE and then MATH. The definite integral will be above the F3 key.
For the TI-83 Plus calculator, select 2nd CALC. Then the definite integral is the 7th option in the CALCULATE menu.
Your answer will be an approximation of the area for the region between curves.
Now you can go to the Session 25 assignments to continue your assignment.