1. Chapter 6 Integration Section 4 The Definite Integral

2. 2 Learning Objectives for Section 6.4 The Definite Integral 1. The student will be able to approximate areas by using left and right sums. 2. The student will be able to compute the definite integral as a limit of sums. 3. The student will be able to apply the properties of the definite integral.

3. 3 Introduction We have been studying the indefinite integral or antiderivative of a function. We now introduce the definite integral. This integral will be the area bounded by f (x), the x axis, and the vertical lines x = a and x = b, with notation

4. 4 Estimating One way to approximate the area under a curve is by filling the region with rectangles and calculating the sum of the areas of the rectangles. Take the width of each rectangle to be ∆x = 1. If we use the left endpoints, the heights of the four rectangles are f (1), f (2), f (3) and f (4), respectively. L 4 = f (1) Δx + f (2) Δx + f (3) Δx + f (4) Δx = 2.5 + 4 + 6.5 + 10 = 23 f (x) = 0.5 x 2 + 2 1 2 3 4 5

5. 5 Estimating Area (continued) We can repeat this using the right side of each rectangle to determine the height. The width of each rectangle is again ∆x = 1. The heights of each of the four rectangles are now f (2), f (3), f (4) and f (5), respectively. The sum of the rectangles is then R 4 = 4 + 6.5 + 10 + 14.5 = 35 f (x) = 0.5 x 2 + 2 1 2 3 4 5 The average of L 4 and R 4 would be an even better approximation: Area ≈ (23 + 35)/2 = 29.

6. 6 Estimating Area (continued) The previous average of 29 is very close to the actual area of 28.666…. Our accuracy can be improved if we increase the number or rectangles, and let ∆x get smaller. f (x) = 0.5 x 2 + 2 1 2 3 4 5 The error in our process can be calculated if the function is monotone. That is, if the function is only increasing or only decreasing. Let L n and R n be the approximate areas, using n rectangles of equal width, and the left or right endpoints, respectively.

7. 7 Estimating Area (continued) If the function is increasing, convince yourself by looking at the picture that L n < Area < R n If f is decreasing, the inequalities go the other way. f (x) = 0.5 x 2 + 2 1 2 3 4 5 If you use L n to estimate the area, then Error = |Area – L n | < |R n – L n |. If you use R n to estimate the area, then Error = |Area – R n | < |R n – L n |. Either way you get the same error bound.

8. 8 Theorem 1 It is not hard to show that |R n – L n | = | f (b) – f (a)| ∆x, and that for n equal subintervals, For our previous example: Theorem 1

9. 9 Theorem 2 If f (x) is either increasing or decreasing on [a, b], then its left and right sums approach the same real number I as n → ∞. This number I is the area between the graph of f and the x axis from x = a to x = b.

10. 10 Approximating Area In the previous example, we hadf (x) = 0.5 x 2 + 2 1 2 3 4 5 If we wanted a particular accuracy, say 0.05, we could use the error formula to calculate n, the number of rectangles needed: Solving for n yields n = 960. We would need at least 960 rectangles to guarantee an accuracy of 0.05.

11. 11 Definite Integral as Limit of Sums We now come to a general definition of the definite integral. Let f be a function on interval [a, b]. Partition [a, b] into n subintervals at points a = x 0 < x 1 < x 2 < … < x n–1 < x n = b. The width of the k th subinterval is ∆x k = (x k – x k – 1 ). In each subinterval, choose an arbitrary point c k x k – 1 < c k < x k.

12. 12 Definite Integral as Limit of Sums (continued) Then define S n is called a Riemann sum. Notice that L n and R n are both special cases of a Riemann sum.

13. 13 A Visual Presentation of a Riemann Sum a = x 0 x n = bx 1 x 2 x n – 1... c 1 c 2 c n f (c 1 ) f (c 2 ) Δx The area under the curve is approximated by the Riemann sum

14. 14 Area (Revisited) Let’s revisit our original problem and calculate the Riemann sum using the midpoints for c k. The width of each rectangle is again ∆x = 1. The heights of the four rectangles are now f (1.5), f (2.5), f (3.5) and f (4.5), respectively. The sum of the rectangles is then S 4 = 3.125 + 5.125 + 8.125 + 12.125 = 28.5 f (x) = 0.5 x 2 + 2 1 2 3 4 5 This is quite close to the actual area of 28.666...

15. 15 The Definite Integral Theorem 3. Let f be a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as n → ∞. This limit I of the Riemann sums for f on [a, b] is called the definite integral of f from a to b, denoted The integrand is f (x), the lower limit of integration is a, and the upper limit of integration is b.

16. 16 Negative Values If f (x) is positive for some values of x on [a, b] and negative for others, then the definite integral symbol represents the cumulative sum of the signed areas between the graph of f (x) and the x axis, where areas above are positive and areas below negative. y = f (x) a bA B

17. 17 Examples Calculate the definite integrals by referring to the figure with the indicated areas. Area A = 3.5 Area B = 12 y = f (x) a bA B c

18. 18 Definite Integral Properties

19. 19 Examples Assume we know that A)B) Then

20. 20 Examples (continued) C) D) E)

21. 21 Summary ■ We summed rectangles under a curve using both the left and right ends and the centers and found that as the number of rectangles increased, accuracy of the area under the curve increased. ■ We found error bounds for these sums. ■ We defined the definite integral as the limit of these sums and found that it represented the area between the function and the x axis. ■ We learned how to compute areas under the x axis.