INTEGRALS

1. 5 INTEGRALS

2. INTEGRALS 5.5The Substitution Rule • In this section, we will learn: • To substitute a new variable in place of an existing • expression in a function, making integration easier.

3. INTRODUCTION • Due to the Fundamental Theorem of Calculus (FTC), it’s important to be able to find antiderivatives.

4. INTRODUCTION Equation 1 • However, our antidifferentiation formulas don’t tell us how to evaluate integrals such as

5. INTRODUCTION • To find this integral, we use the problem-solving strategy of introducing something extra. • The ‘something extra’ is a new variable. • We change from the variable x to a new variable u.

6. INTRODUCTION • Suppose we let u be the quantity under the root sign in Equation 1, u = 1 + x2. • Then, the differential of u is du = 2xdx.

7. INTRODUCTION • Notice that, if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2x dx would occur in Equation 1.

8. INTRODUCTION Equation 2 • So, formally, without justifying our calculation, we could write:

9. INTRODUCTION • However, now we can check that we have the correct answer by using the Chain Rule to differentiate the final function of Equation 2:

10. INTRODUCTION • In general, this method works whenever we have an integral that we can write in the form ∫f(g(x))g’(x) dx

11. INTRODUCTION Equation 3 • Observe that, if F’ = f, then • ∫F’(g(x))g’(x) dx = F(g(x))+ C • because, by the Chain Rule,

12. INTRODUCTION • If we make the ‘change of variable’ or ‘substitution’ u = g(x), from Equation 3, we have:

13. INTRODUCTION • Writing F’ = f, we get: ∫f(g(x))g’(x) dx = ∫f(u) du • Thus, we have proved the following rule.

14. SUBSTITUTION RULE Equation 4 • If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫f(g(x))g’(x) dx = ∫f(u) du

15. SUBSTITUTION RULE • Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation. • Notice also that, if u = g(x), then du = g’(x) dx. • So,a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.

16. SUBSTITUTION RULE • Thus, the Substitution Rule says: • It is permissible to operate withdx and du after integral signs as if they were differentials.

17. SUBSTITUTION RULE Example 1 • Find ∫x3 cos(x4 + 2) dx • We make the substitution u = x4 + 2. • This is because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral.

18. SUBSTITUTION RULE Example 1 • Thus, using x3dx = du/4 and the Substitution Rule, we have: • Notice that, at the final stage, we had to return to the original variable x.

19. SUBSTITUTION RULE • The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral. • This is accomplished by changing from the original variable x to a new variable u that is a function of x. • Thus, in Example 1, we replaced the integral ∫ x3cos(x4 + 2) dxby the simpler integral ¼ ∫ cos u du.

20. SUBSTITUTION RULE • The main challenge in using the rule is to think of an appropriate substitution. • You should try to choose u to be some function in the integrand whose differential also occurs—except for a constant factor. • This was the case in Example 1.

21. SUBSTITUTION RULE • If that is not possible, try choosing u to be some complicated part of the integrand—perhaps the inner function in a composite function.

22. SUBSTITUTION RULE • Finding the right substitution is a bit of an art. • It’s not unusual to guess wrong. • If your first guess doesn’t work, try another substitution.

23. SUBSTITUTION RULE E. g. 2—Solution 1 • Evaluate • Let u = 2x + 1. • Then, du = 2 dx. • So, dx = du/2.

24. SUBSTITUTION RULE E. g. 2—Solution 1 • Thus, the rule gives:

25. SUBSTITUTION RULE E. g. 2—Solution 2 • Another possible substitution is • Then, • So, • Alternatively, observe that u2 = 2x + 1. • So, 2udu = 2 dx.

26. SUBSTITUTION RULE E. g. 2—Solution 2 • Thus,

27. SUBSTITUTION RULE Example 3 • Find • Let u = 1 – 4x2. • Then, du = -8x dx. • So, x dx = -1/8 du and

28. SUBSTITUTION RULE • The answer to the example could be checked by differentiation. • Instead, let’s check it with a graph.

29. SUBSTITUTION RULE • Here, we have used a computer to graph both the integrand and its indefinite integral • We take the case C = 0.

30. SUBSTITUTION RULE • Notice that g(x): • Decreases when f(x) is negative • Increases when f(x) is positive • Has its minimum value when f(x) = 0

31. SUBSTITUTION RULE • So, it seems reasonable, from the graphical evidence, that g is an antiderivative of f.

32. SUBSTITUTION RULE Example 4 • Calculate ∫e5x dx • If we let u = 5x, then du = 5 dx. • So, dx = 1/5 du. • Therefore,

33. SUBSTITUTION RULE Example 5 • Find • An appropriate substitution becomes more obvious if we factor x5 as x4 . x. • Let u = 1 + x2. • Then, du = 2x dx. • So, x dx = du/2.

34. SUBSTITUTION RULE Example 5 • Also, x2 = u – 1; so, x4 = (u – 1)2:

35. SUBSTITUTION RULE Example 6 • Calculate ∫ tan x dx • First, we write tangent in terms of sine and cosine: • This suggests that we should substitute u = cos x, since then du = – sin x dx,and so sin x dx = – du:

36. SUBSTITUTION RULE Equation 5 • Since –ln |cos x| = ln(|cos x|-1) = ln(1/|cos x|) = ln|sec x|, • the result of the example can also be written as ∫ tan x dx = ln |sec x| + C

37. DEFINITE INTEGRALS • When evaluating a definite integral by substitution, two methods are possible.

38. DEFINITE INTEGRALS • One method is to evaluate the indefinite integral first and then use the FTC. • For instance, using the result of Example 2, we have:

39. DEFINITE INTEGRALS • Another method, which is usually preferable, is to change the limits of integration when the variable is changed. • Thus, we have the substitution rule for definite integrals.

40. SUB. RULE FOR DEF. INTEGRALS Equation 6 • If g’ is continuous on [a, b] and fis continuous on the range of u = g(x), then

41. SUB. RULE FOR DEF. INTEGRALS Proof • Let F be an antiderivative of f. • Then, by Equation 3, F(g(x)) is an antiderivative of f(g(x))g’(x). • So, by Part 2 of the FTC (FTC2), we have:

42. SUB. RULE FOR DEF. INTEGRALS Proof • However, applying the FTC2 a second time, we also have:

43. SUB. RULE FOR DEF. INTEGRALS Example 7 • Evaluate using Equation 6. • Using the substitution from Solution 1 of Example 2, we have: u = 2x + 1 and dx = du/2

44. SUB. RULE FOR DEF. INTEGRALS Example 7 • To find the new limits of integration, we note that: • When x = 0, u = 2(0) + 1 = 1 • When x = 4, u = 2(4) + 1 = 9

45. SUB. RULE FOR DEF. INTEGRALS Example 7 • Thus,

46. SUB. RULE FOR DEF. INTEGRALS Example 7 • Observe that, when using Equation 6, we do not return to the variable x after integrating. • We simply evaluate the expression in u between the appropriate values of u.

47. SUB. RULE FOR DEF. INTEGRALS Example 8 • Evaluate • Let u = 3 - 5x. • Then, du = – 5 dx, so dx =–du/5. • When x = 1, u = – 2, and when x = 2, u = – 7.

48. SUB. RULE FOR DEF. INTEGRALS Example 8 • Thus,

49. SUB. RULE FOR DEF. INTEGRALS Example 9 • Calculate • We let u = ln x because its differential du = dx/xoccurs in the integral. • When x = 1, u = ln 1, and when x = e, u = ln e = 1. • Thus,

50. SUB. RULE FOR DEF. INTEGRALS Example 9 • As the function f(x) = (ln x)/x in the example is positive for x > 1, the integral represents the area of the shaded region in this figure.