The Definite Integral

1. The Definite Integral

2. Riemann Sums • Sigma notation enables us to express a large sum in compact form

3. Riemann Sums • LRAM, MRAM,and RRAM are examples of Riemann sums • Sn = This sum, which depends on the partition P and the choice of the numbers ck,is a Riemann sum for f on the interval [a,b]

4. Definite Integral as a Limit of Riemann Sums Let f be a function defined on a closed interval [a,b]. For any partition P of [a,b], let the numbers ck be chosen arbitrarily in the subintervals [xk-1,xk]. If there exists a number I such that no matter how P and the ck’s are chosen, then f is integrable on [a,b] and I is the definite integral of f over [a,b].

5. Definite Integral of a continuous function on [a,b] Let f be continuous on [a,b], and let [a,b] be partitioned into n subintervals of equal length Δx = (b-a)/n. Then the definite integral of f over [a,b] is given by where each ck is chosen arbitrarily in the kth subinterval.

6. Definite integral This is read as “the integral from a to b of f of x dee x” or sometimes as “the integral from a to b of f of x with respect to x.”

7. Using Definite integral notation The function being integrated is f(x) = 3x2 – 2x + 5 over the interval [-1,3]

8. Definition: Area under a curve If y = f(x) is nonnegative and integrable over a closed interval [a,b], then the area under the curve of y = f(x) from a to b is the integral of f from a to b,

9. Nonpositive regions If the graph is nonpositive from a to b then

10. Area of any integrable function = (area above the x-axis) – (area below x-axis)

11. Integral of a Constant If f(x) = c, where c is a constant, on the interval [a,b], then

12. Evaluating Integrals using areas We can use integrals to calculate areas and we can use areas to calculate integrals. Using areas, evaluate the integrals: 1) 2)

13. Evaluating Integrals using areas Evaluate using areas: 3) 4) (a<b)

14. Evaluating integrals using areas Evaluate the discontinuous function: Since the function is discontinuous at x = 0, we must divide the areas into two pieces and find the sum of the areas = -1 + 2 = 1

15. Integrals on a Calculator You can evaluate integrals numerically using the calculator. The book denotes this by using NINT. The calculator function fnInt is what you will use. = fnInt(xsinx,x,-1,2) is approx. 2.04

16. Evaluate Integrals on calculator • Evaluate the following integrals numerically: • = approx. 3.14 • = approx. .89

17. Rules for Definite Integrals • Order of Integration:

18. Rules for Definite Integrals • Zero:

19. Rules for Definite Integrals 3) Constant Multiple:

20. Rules for Definite Integrals 4) Sum and Difference:

21. Rules for Definite Integrals 5) Additivity:

22. Rules for Definite Integrals • Max-Min Inequality: If max f and min f are the maximum and minimum values of f on [a,b] then: min f ∙ (b – a) ≤ ≤ max f ∙ (b – a)

23. Rules for Definite Integrals • Domination: f(x) ≥ g(x) on [a,b] f(x) ≥ 0 on [a,b] ≥ 0

24. Using the rules for integration Suppose: Find each of the following integrals, if possible: • b) c) d) e) f)

25. Using the rules for definite integrals Show that the value of is less than 3/2 The Max-Min Inequality rule says the max f . (b – a) is an upper bound. The maximum value of √(1+cosx) on [0,1] is √2 so the upper bound is: √2(1 – 0) = √2 , which is less than 3/2

26. Average (Mean) Value If f is integrable on [a,b], its average (mean) value on [a,b] is: av(f) = Find the average value of f(x) = 4 – x2 on [0,3] . Does f actually take on this value at some point in the given interval?

27. Applying the Mean Value Av(f) = = 1/3(3) = 1 4 – x2 = 1 when x = ± √3 but only √3 falls in the interval from [0,3], so x = √3 is the place where the function assumes the average.

28. Mean Value Theorem for Definite Integrals If f is continuous on [a,b], then at some point c in [a,b],

29. The Fundamental Theorem of Calculus, Part I If f is continuous on [a,b], then the function F(x) = has a derivative at every point x in [a,b], and

30. Applications of The Fundamental Theorem of Calculus, Part I 1. 2. 3.

31. Applications of The Fundamental Theorem of Calculus, Part I Find dy/dx. y = Since this has an x on both ends of the integral, it must be separated.

32. Applications of The Fundamental Theorem of Calculus, Part I =

33. Applications of The Fundamental Theorem of Calculus, Part I = =

34. The Fundamental Theorem of Calculus, Part 2 If f is continuous at every point of [a,b], and if F is any antiderivative of f on [a,b], then This part of the Fundamental Theorem is also called the Integral Evaluation Theorem.

35. Trapezoidal Rule To approximate , use T = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn) where [a,b] is partitioned into n subintervals of equal length h = (b-a)/n.

36. Using the trapezoidal rule Use the trapezoidal rule with n = 4 to estimate h = (2-1)/4 or ¼, so T = 1/8( 1+2(25/16)+2(36/16)+2(49/16)+4) = 75/32 or about 2.344

37. Simpson’ Rule To approximate , use S = (y0 + 4y1 + 2y2 + 4y3…. 2yn-2 +4yn-1 + yn) where [a,b] is partitioned into an even number n subintervals of equal length h =(b –a)/n.

38. Using Simpson’s Rule Use Simpson’s rule with n = 4 to estimate h = (2 – 1)/4 = ¼, so S = 1/12 (1 + 4(25/16) + 2(36/16) + 4(49/16) + 4) = 7/3