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Chapter 6 Integration Section 5 The Fundamental Theorem of Calculus.

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Presentation on theme: "Chapter 6 Integration Section 5 The Fundamental Theorem of Calculus."— Presentation transcript:

1 Chapter 6 Integration Section 5 The Fundamental Theorem of Calculus

2 2 Objectives for Section 6.5 Fundamental Theorem of Calculus ■ The student will be able to evaluate definite integrals. ■ The student will be able to calculate the average value of a function using the definite integral.

3 3 Fundamental Theorem of Calculus If f is a continuous function on the closed interval [a, b], and F is any antiderivative of f, then

4 4 By the fundamental theorem we can evaluate easily and exactly. We simply calculate Evaluating Definite Integrals

5 5 Definite Integral Properties

6 6 Example 1 Make a drawing to confirm your answer. 0 ≤ x ≤ 4 –1 ≤ y ≤ 6

7 7 Example 2 Make a drawing to confirm your answer. 0 ≤ x ≤ 4 –1 ≤ y ≤ 4

8 8 Example 3 0 ≤ x ≤ 4 –2 ≤ y ≤ 10

9 9 Example 4 Let u = 2x, du = 2 dx

10 10 Example 5

11 11 Example 6 This is a combination of the previous three problems

12 12 Example 7 Let u = x 3 + 4, du = 3x 2 dx

13 13 Example 7 (revisited) On the previous slide, we made the back substitution from u back to x. Instead, we could have just evaluated the definite integral in terms of u:

14 14 Numerical Integration on a Graphing Calculator Use some of the examples from previous slides: Example 5: Example 7: 0 ≤ x ≤ 3 –1 ≤ y ≤ 3 –1 ≤ x ≤ 6 –0.2 ≤ y ≤ 0.5

15 15 Example 8 From past records a management service determined that the rate of increase in maintenance cost for an apartment building (in dollars per year) is given by M ´(x) = 90x 2 + 5,000, where M(x) is the total accumulated cost of maintenance for x years. Write a definite integral that will give the total maintenance cost from the end of the second year to the end of the seventh year. Evaluate the integral.

16 16 Example 8 From past records a management service determined that the rate of increase in maintenance cost for an apartment building (in dollars per year) is given by M ´(x) = 90x 2 + 5,000, where M(x) is the total accumulated cost of maintenance for x years. Write a definite integral that will give the total maintenance cost from the end of the second year to the end of the seventh year. Evaluate the integral. Solution:

17 17 Using Definite Integrals for Average Values The average value of a continuous function f over [a, b] is Note this is the area under the curve divided by the width. Hence, the result is the average height or average value.

18 18 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. b)Find the average value of the cost function over the interval [0, 1000]. c)Write a description of the difference between part a) and part b). Example

19 19 a) Find the average cost per unit if 1000 dictionaries are produced Solution: The average cost is Example (continued)

20 20 Example (continued) b) Find the average value of the cost function over the interval [0, 1000] Solution:

21 21 Example (continued) c) Write a description of the difference between part a and part b Solution: If you just do the set-up for printing, it costs $20,000. This is the cost for printing 0 dictionaries. If you print 1,000 dictionaries, it costs $30,000. That is $30 per dictionary (part a). If you print some random number of dictionaries (between 0 and 1000), on average it costs $25,000 (part b). Those two numbers really have not much to do with one another.

22 22 Summary We can find the average value of a function f by We can evaluate a definite integral by the fundamental theorem of calculus:


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