1. 11 The student will learn about: §4.5 Application of Definite Integrals and Area Between Curves. the average value of a function, the average value of a function, and finding the area between two curves.

2. 2 Introduction In this section we will use definite integrals for two important purposes: finding average values of functions and finding areas between curves. Average values are used everywhere. Birth weights of babies are compared with average weights, and retirement benefits are determined by average income. Averages eliminate fluctuations, reducing a collection of numbers to a single “representative” number. Areas between curves are used to find quantities from trade deficits to lives saved by seat belts.

3. 3 Average Value of a Function Intuitively, the average should represent a “leveling off” of the curve to a uniform height, the horizontal line shown on the right. This leveling should use the “hills” to fill in the “valleys,” maintaining the same total area under the curve. Therefore, the area under the horizontal line must equal the area under the curve. Equating the two areas

4. 44 Using Definite Integrals for Average Values Average Value of a Continuous Function f over [a, b]. Note this is the area under the curve divided by the width. Hence, the result is the average height or average value.

5. 55 Application §6.5 # 70. The total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x a. Find the average cost per unit if 1000 dictionaries are produced Continued on next slide. Note that the average cost is 30 What does this mean?

6. 66 Application - continued The total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x b. Find the average value of the cost function over the interval [0, 1000] Continued on next slide. = 20,000 + 5,000 = 25,000 What does this mean?

7. 77 Application - concluded The total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x c. Write a description of the difference between part a and part b From part a the average cost is $30.00 From part b the average value of the cost is $25,000. The average cost per dictionary of making 1,000 dictionaries is $30. The average total cost of making between 0 and 1,000 dictionaries is $25,000.

8. 88 Study of Area Continued In the previous section we studied the area between the graph and the x-axis. The graph was always above the axis but this is not always the case. If the curve is below the x-axis the area is negative. We need only change the sign to positive to get the correct answer.

9. 99 Study of Area Continued With our calculator we can guarantee that the curve will be above the x-axis by graphing y = abs [ f(x) ]. We will use this technique.

10. 10 Area Between Curves The area between two curves can be written as a single integral: Show how to get abs function on the calculator.

11. 11 Study of Area Continued We will now study the area between two curves f (x) and g (x). We will use our knowledge of integrals to find the area under this new curve. f (x) g (x) We will define a new function abs [ (f (x)) – (g (x)) ] and that situation will graph as an entirely different third function.

12. 12 SUMMARY OF AREA PROBLEMS 1. Graph y = abs ( f (x) – g (x) ) in the interval of integration from a to b. In some cases you may need to use minimum to find the interval of integration. That’s it!

13. 13 Find the area bounded by y = 4 – x 2 ; y = 0, 0  x  4. Example 1 - By Calculator

14. 14 Example 2 - By Calculator Find the area bounded by y = 0.5 x 2 + 3; y = 2x, from x = 1 to 3. Form a new function abs [(f (x)) – (g (x))].

15. 15 Example 3 - By Calculator Find the area bounded by y = x 3 + 3x 2 - 1; y = x + 3 from x = -2 to 1. Form a new function abs [(f (x)) – (g (x))].

16. 16 Example 4 - By Calculator Find the area bounded by y = x 2 – 4 and y = 8 – 3x – 2x 2 Find the x-intercepts [minimum] x = - 2.56 and 1.56 Form a new function abs [(f (x)) – (g (x))]. In this problem the limits of integration are not given.

17. 17 SUMMARY OF AREA PROBLEMS 1. Graph y = abs [(f (x)) – (g (x))] in the interval of integration from a to b. In some cases you may need to use minimum to find the interval of integration. That’s it!

18. 18 Summary. We found a methods to calculate the area between two curves. We can find the average value of a function f by:

19. 19 ASSIGNMENT §4.5 on my website. 7, 8, 9, 15, 16, 17.