1. INTEGRALS 5

2. If u = g(x), Summary If f is even, f(-x) = f(x), If f is odd, f(-x) = -f(x),

3. The Substitution Rule Definite Integral. INTEGRALS

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

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

6. 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.

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

8. 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: SUB. RULE FOR DEF. INTEGRALS Proof

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

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

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

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

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

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

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

16. Therefore, Proof INTEGS. OF SYMM. FUNCTIONS

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

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

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

20. Theorem 6 is illustrated here. INTEGS. OF SYMM. FUNCTIONS

21. 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. INTEGS. OF SYMM. FUNCTIONS

22. Recall that an integral can be expressed as the area above the x–axis and below y = f(x) minus the area below the axis and above the curve. INTEGS. OF SYMM. FUNCTIONS

23. Therefore, part (b) says the integral is 0 because the areas cancel. INTEGS. OF SYMM. FUNCTIONS

24. As f(x) = x 6 + 1 satisfies f(–x) = f(x), it is even. So, Example 8 INTEGS. OF SYMM. FUNCTIONS

25. As f(x) = (tan x)/ (1 + x 2 + x 4 ) satisfies f( – x) = – f(x), it is odd. So, Example 9 INTEGS. OF SYMM. FUNCTIONS

26. If u = g(x), Summary If f is even, f(-x) = f(x), If f is odd, f(-x) = -f(x),