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 Equation 1 • Antidifferentiation formulas don’t tell us how to evaluate integrals such as

4. 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 newvariableu.

5. 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.

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

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

8. INTRODUCTION Equation 3 • Observe that, if F’ = f, then • because, by the Chain Rule,

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

10. 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

11. 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.

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

13. Solution: 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.

14. 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.

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

16. Solution1: E. g. 2—Solution 1 • Thus, the rule gives:

17. Solution2: E. g. 2—Solution 2 • Another possible substitution is • Then, • So, • Alternatively, observe that u2 = 2x + 1. • So, 2udu = 2 dx.

18. Solution2: E. g. 2—Solution 2 • Thus,

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

20. Solution: • Here, we have used a computer to graph both the integrand and its indefinite integral • We take the case C = 0. Figure 5.5.1, p. 335

21. Solution: • Notice that g(x): • Decreases when f(x) is negative • Increases when f(x) is positive • Has its minimum value when f(x) = 0 Figure 5.5.1, p. 335

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

23. 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.

24. Solution: Example 5 • Also, x2 = u – 1; so, x4 = (u – 1)2:

25. DEFINITE INTEGRALS • Evaluate the indefinite integral first and then use the FTC for definite integral. • For instance, using the result of Example 2, we have:

26. SUB. RULE FOR DEF. INTEGRALS Equation 5 • If g’ is continuous on [a, b] and fis continuous on the range of u = g(x), then

27. 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:

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

29. SUB. RULE FOR DEF. INTEGRALS Example 6 • Evaluate using Equation 5. • Using the substitution from Solution 1 of Example 2, we have: u = 2x + 1 and dx = du/2 Figure 5.5.2a, p. 337

30. Solution: Example 6 • 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 Figure 5.5.2b, p. 337

31. Solution: Example 6 • Thus,

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

33. Solution: Example 7 • Thus,

34. SYMMETRY • The next theorem uses the Substitution Rule for Definite Integrals to simplify the calculation of integrals of functions that possess symmetry properties.

35. INTEGS. OF SYMM. FUNCTIONS Theorem 6 • Suppose f is continuous on [–a , a]. • If f is even, [f(–x) = f(x)], then • If f is odd, [f(-x) = -f(x)], then

36. INTEGS. OF SYMM. FUNCTIONS Proof—Equation 7 • We split the integral in two:

37. INTEGS. OF SYMM. FUNCTIONS Proof • In the first integral in the second part, we make the substitution u = –x . • Then, du = –dx, and when x = –a, u = a.

38. INTEGS. OF SYMM. FUNCTIONS Proof • Therefore,

39. INTEGS. OF SYMM. FUNCTIONS Proof—Equation 8 • So, Equation 7 becomes:

40. INTEGS. OF SYMM. FUNCTIONS Proof a • If f is even, then f(–u) = f(u). • So, Equation 8 gives:

41. INTEGS. OF SYMM. FUNCTIONS Proof b • If f is odd, then f(–u) = –f(u). • So, Equation 8 gives:

42. INTEGS. OF SYMM. FUNCTIONS • Theorem 6 is illustrated here. Figure 5.5.3, p. 338

43. INTEGS. OF SYMM. FUNCTIONS • For the case where f is positive and even, part (a) says that the area under y = f(x) from –a to a is twice the area from 0 to a because of symmetry. Figure 5.5.3a, p. 338

44. INTEGS. OF SYMM. FUNCTIONS • Therefore, part (b) says the integral is 0 because the areas cancel. Figure 5.5.3b, p. 338

45. INTEGS. OF SYMM. FUNCTIONS Example 8 • As f(x) = x6 + 1 satisfies f(–x) = f(x), it is even. • So,

46. INTEGS. OF SYMM. FUNCTIONS Example 9 • As f(x) = (tan x)/ (1 + x2 + x4) satisfies f(–x) = –f(x), it is odd. • So,