Substitutions and by Parts, Area Between Curves. Goals: The Method of Substitution Areas Integration by Parts

Transcription

1 Week #7: Substitutions and by Parts, Area Between Curves Goals: The Method of Substitution Areas Integration by Parts 1

2 Week 7 The Indefinite Integral The Fundamental Theorem of Calculus, b a f(x) dx = F (b) F (a), reduces the problem of evaluating an integral to that of finding the anti-derivative F. For that reason the anti-derivative is often called the indefinite integral and written as ( ) f(x) dx = F (x).

3 Week 7 A definite integral has the form It represents b a f(x) dx. Its value is An indefinite integral has the form It represents f(x) dx. Its value is

4 Week 7 Problem. Find ( ) t 3 t dt. Note: When you are asked to find an indefinite integral, as in the previous question, it is important to add the +C to indicate that there is more than one anti-derivative, and that all of them differ from each other by a constant.

5 Week 7 Finding antiderivatives is surprisingly difficult. You would think that if we know how to find derivatives then we should know how to find antiderivatives. In fact, however, the latter is much more difficult, as illustrated next.

6 Week 7 Problem. Calculate 1 x dx

7 Week 7 Guess and Check To prepare us for a later more systematic integration technique, the Method of Substitution, let us first explore an informal method often called guess and check. Problem. Find cos(5x) dx.

8 Week 7 Problem. Find cos(x 2 ) dx.

9 Problem. Now what if we were given the problem x cos(x 2 ) dx? Week 7 So why does guess and check work for x cos(x 2 ) dx but not for cos(x 2 ) dx?

10 Week 7 Problem. Which of the following anti-differentiations can be done by the guessand-check method, do you think? 1. x 2 e x3 dx 2. x 2 e x2 dx 3. xe x2 dx A. 1 B. 2 C. 3 D. 1 and 3 E. 1 and 2

11 Problem. Calculate cos(x) e sin(x) dx. Week 7 Next we will turn these ideas into a formal procedure called Substitution.

12 Substitution Method - Introduction - 1 Substitution Rule The Method of Substitution is a formalization of our loose guess and check method. Consider the previous example: e sin(x) cos(x) dx. Here is the procedure: (1) Let u = sin x. (a guess) (2) Differentiate u:

13 Substitution Method - Introduction - 2 (3) Rewrite the integral entirely in terms of u and du: If this step cannot be carried out, it means that you should go back and try a different choice of u!

14 (4) Solve this anti-derivative: Substitution Method - Introduction - 3 Again, if this step cannot be carried out, it means that you should go back and try a different choice of u!

15 Substitution Method - Introduction - 4 (5) Rewrite u in terms of x: Therefore, e sin(x) cos(x) dx =

16 Problem. Calculate 3x 1 + x 2 dx. Substitution Method - Examples - 1

17 Substitution Method - Examples - 2 Final answer: 3x 1 + x 2 dx =

18 Problem. Calculate x 2 dx. Substitution Method - Examples - 3

19 Substitution Method - Examples - 4 Final answer: x 2 dx =

20 Substitution Method - Examples - 5 Once you do many examples using this method (practice, practice practice!), you will become so familiar with the procedure that you will not need to follow it step by step. It will become so automatic that your answers will be one-liners.

21 Substitution Method - Strategy and Pitfalls - 1 Problem. What substitution(s) might be helpful to help evaluate the indefinite integral x cos(1 + x 2 ) dx? (1.) u = cos(1 + x 2 ) (2.) u = x 2 (3.) u = 1 + x 2 (4.) u = x A. Option 1 B. Options 2 and 3 C. Options 2, 3 and 4 D. All options

22 Substitution Method - Strategy and Pitfalls - 2 Try out some of the substitution options for x cos(1 + x 2 ) dx.

23 Problem. Calculate tan x dx. Substitution Method - Definite Integrals - 1

24 Substitution Method - Definite Integrals - 2 Problem. Find π 4 0 tan x dx.

25 Substitution Method - Definite Integrals - 3 Special procedure for using substitution on a definite integral: In the last example, we evaluated F (cos }{{} x) at x = π 4 and x = 0 and subtracted. But when x = π 4 then u = cos(π 4 ) = 1 2 and when x = 0 then u = cos(0) = 1. u

26 Substitution Method - Definite Integrals - 4 Problem. Evaluate π 4 make your substitution. 0 tan x dx, but update the limits of integration when you

27 Substitution Method - Further Examples of Definite Integrals - 1 When we use the method of substitution on a definite integral, we can omit subbing back to the original variable, provided we change the limits of integration when we change variables. Problem. π 2 /4 π 2 /9 cos( t) t dt

28 Areas Between Curves We are now ready to use integration in some simple applications. Areas Between Curves - 1 Problem. Find the area enclosed between the graphs of y = x and y = x 2.

29 Areas Between Curves - 2

30 Problem. Find the area enclosed between the graphs of Areas - Vertical vs Horizontal Slices - 1 y = x, x = 6 y 2, and with y 0 using vertical slices.

31 Areas - Vertical vs Horizontal Slices - 2 Problem. Find the area enclosed between the graphs of y = x, x = 6 y 2, and with y 0 using horizontal slices.

32 Areas - Separating Regions for Sign - 1 Problem. Find the area enclosed between the graphs of y = x 3 2x and y = x 2.

33 Areas - Separating Regions for Sign - 2 y = x 3 2x and y = x 2.

34 Integration by Parts - Introduction - 1 Integration by Parts To allow us to apply integration to a wider variety of problems, we need additional techniques for finding anti-derivatives. Integration by Parts is the first of these. The main idea of Integration by Parts is to use the Product Rule in reverse. d ( ) f(x) g(x) = dx If we integrate both sides, we obtain

35 Integration by Parts - Introduction - 2 If we let u = f(x) and v = g(x) then we can develop the rule:

36 Integration by Parts - Introduction - 3 The main goal of Integration by Parts is to end up with a simpler integral, v du, than the given integral, u dv. Integration by Parts Formula u dv = uv v du. }{{}}{{} given simpler

37 Integration by Parts - Introduction - 4 u dv = uv Problem. Consider the integral xe x dx. v du Divide the integrand into u and dv parts, and apply integration by parts.

38 Integration by Parts - Introduction - 5 After the first step, if the new integral is much more complicated than the original one, go back and set u and dv equal to something else so that you end up with an easier integral to solve. You can avoid this by mentally checking what v du will be before actually choosing what u and dv are.

39 Problem. Calculate x sin(4x) dx. Integration by Parts - Examples - 1

40 Integration by Parts - Examples - 2 Could we have foreseen that our choice of u and dv would work? What would have happened if we had reversed our selections of u and dv?

41 Integration by Parts - Examples - 3 Integration by Parts is very easy once you learn to make smart choices for u and dv. Practice a lot. Problem. Calculate x 3 e x2 dx.

42 Problem. Calculate ln x dx. Integration by Parts - Definite Integrals - 1

43 Integration by Parts - Definite Integrals - 2 Problem. Find the area under the graph of ln x between x = 1 and x = 2.

44 Integration by Parts - Definite Integrals - 3 Problem. Calculate arcsin(x) dx.

45 Problem. If you want to simplify the integral by parts then you should choose: A. u = x 5 1 and dv = dx 1 + x x x 3 Integration by Parts - Strategy - 1 dx using integration B. u = x 4 and dv = x 1 + x 3 dx C. u = x 3 and dv = x x 3 dx D. u = x 2 and dv = x x 3 dx E. u = x x 3 and dv = x3 dx

46 Integration by Parts - Strategy - 2 General integration advice: Look for a substitution in your integral first - they are the simplest method to use, and usually the most obvious. Only try integration by parts if substitution fails. With all methods, you may need to experiment with your choice of u, dv, or your substitution.