1. §12.5 The Fundamental Theorem of Calculus
The student will learn about:
the definite integral,
the fundamental theorem of calculus,
and using definite integrals to find average values.

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

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

4. Definite Integral Properties
REMINDER !
Definite Integral Properties

5. Example 1
5 · 3 – 5 · 1 =
15 – 5 = 10
Make a drawing to confirm your answer.
0  x  4
- 1  y  6

6. Example 2 4 Make a drawing to confirm your answer. 0  x  4

7. Example 3 1.
9 - 0 = 9
0  x  4
- 2  y  10

8. Example 4
Let u = 2x du = 2 dx 2.

9. Example 5 3.
= ln 2 – ln 1 = ln 2 =

10. Examples 6 This is a combination of the previous three problems 4.
= 9 + (e 6)/2 + ln 3 – 1/3 – (e2)/2 – ln 1 =

11. Examples 7 Let u = x3 + 4 5. du = 3x2 dx (ln 129)/3 – (ln 4)/3 =

12. Examples 7 REVISITED Let u = x3 + 4 5. du = 3x2 dx
At this point instead of substituting for u we can replace the x value in terms of u. If x = 0 then u = 4 and if x = 5, then u = 129.
(ln 129)/3 – (ln 4)/3 =

13. Numerical Integration on a Graphing Calculator
Use some of the examples from above. 3.
0  x  3
- 1  y  3 5.
-1  x  6
- 0.2  y  0.5

14. Example 8
From past records a management services 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 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 seventh year to the end of the seventh year. Evaluate the integral.
30 x 3 + 5,000x
= 10, ,000 – 810 – 15,000
= $29,480

15. 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.

16. a. Find the average cost per unit if 1000 dictionaries are produced
Example
§6.5 # 70. The total cost (in dollars) of printing x dictionaries is C (x) = 20, x
a. Find the average cost per unit if 1000 dictionaries are produced
Note that the average cost is 30
Continued on next slide.

17. Example - continued
The total cost (in dollars) of printing x dictionaries is C (x) = 20, x
b. Find the average value of the cost function over the interval [0, 1000]
20, ,000 = 25,000
Continued on next slide.

18. c. Write a description of the difference between part a and part b
Example - continued
The total cost (in dollars) of printing x dictionaries is C (x) = 20, x
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 us $30. The average total cost of making between 0 and 1,000 dictionaries is $25,000.

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

20. Practice Problems
§6.5; 1, 3, 5, 9, 13, 17, 21, 23, 27, 31, 35, 39, 41, 45, 49, 53, 55, 61, 65, 69, 73, 81, 85.

21. §13.3 Integration by Parts
The student will learn to
integrate by parts.

22. Integration by Parts
Integration by parts is based on the product formula for derivatives:
We rearrange the above equation to get:
And integrating both sides yields:
Which reduces to:

23. Integration by Parts Formula
If we let u = f (x) and v = g (x) in the preceding formula the equation transforms into a more convenient form:
Integration by Parts Formula
The goal is to change the integral from u dv to v du and have the second integral be easier than the first.

24. Integration by Parts Formula
This method is useful when the integral on the left side is difficult and changing it into the integral on the right side makes it easier.
Remember, we can easily check our results by differentiating our answers to get the original integrals.
Let’s look at an example.

25. Example
Our previous method of substitution does not work. Examining the left side of the integration by parts formula yields two possibilities.
Option 1
u dv
Option 2
u dv
Let’s try option 1.

26. Example - continued
We have decided to let u = x and dv = e x dx, (Note: du = dx and v = e x), yielding
Which is easy to integrate
= x e x – e x + C
As mentioned this is easy to check by differentiating.

27. Selecting u and dv
1. The product u dv must equal the original integrand.
2. It must be possible to integrate dv by one of our known methods.
3. The new integral should not be more involved than the original integral
4. For integrals involving x p e ax , try
u = x p and dv = e ax dx
5. For integrals involving x p ln x q , try
u = (ln x) q and dv = x p dx

28. Example 2 Let u = ln x and dv = x 3 dx, yielding
du = 1/x dx and v = x 4/4, and the integral is
Which when integrated is:
Check by differentiating.

29. Summary.
We have developed integration by parts as a useful technique for integrating more complicated functions.
Integration by Parts Formula

30. Practice Problems
§7.3; 1, 3, 5, 7, 11, 15, 19, 21, 23, 25, 29, 33, 37, 41, 43, 47, 49, 57, 63, 65.