The document discusses the definite integral and how it can be used to find the area under a curve. It defines the definite integral as the limit of Riemann sums, which approximate the area by dividing it into thin rectangles and summing their areas. The midpoint rule for approximating definite integrals with Riemann sums is also explained. Even and odd functions are defined, and formulas given for their integrals over symmetric domains.

1. 5.4 Area and the Definite IntegralGoal: To find the area under a curve

5. As we break our graph into more and more rectangles, we get a better approximation of the area under the curve. We just continue to add the areas of each of the smaller rectangles. Let’s say our graph has n thin strips each of identical width. So we can say that each strip is n wide by f(x) tall.

6. We can think of n as just Δx since it is the just the amount by which the x value changes and it is the same for every rectangle. We can use c1c2,c3… and so on for each of the x values that we are substituting in, in order to calculate f(x). So the area for the first rectangle is Area = f(c1)Δx, for the second Area = f(c2)Δx and so on If we carry this out all the way we can say that: Which means: Or the sum of all the areas of the rectangles

7. Since more rectangles = more accuracy having ∞ rectangles would be best. The limit was first solved without uniform continuity by Cauchy in 1823, it was later proven and put on a solid logical foundation by Georg Friedrick Bernhard Riemann (1823-1866) His work led to the following definition for definite integrals: If f is continuous on [a,b] Exists and is the same number for any choice of ck

8. The Midpoint Rule: If we cannot integrate the function for which we need to find the area under the curve, we can use Riemann sums (the midpoint rule) Basically, we figure out how many rectangles we need. Find the width and then multiply by the sum of the lengths.

9. Use the Midpoint Rule with n = 4 to approximate: First break [0,1] into 4 intervals and find the midpoint of each. 0 - ¼ ¼ - ½ ½ - ¾ ¾ - 1 Δx = ¼ 1/8 3/8 5/8 7/8 are the midpoints for each interval ≈ .4358

10. Use the Midpoint Rule with n = 4 to approximate: First break [2,4] into 4 intervals and find the midpoint of each. 2 – 2.5 2.5-3 3-3.5 3.5-4 Δx = ½ 9/4 11/4 13/4 15/4 are the midpoints for each interval ≈ 5.6419

11. The definite integral in terms of area under the curve: Let f be non-negative and continuous on the closed interval [a,b]. The area of the region bounded by the graph of f, the x-axis and the lines x = a and y = b is denoted by Newton and Leibniz are both credited with achieving this great development in calculus. Newton derived it first, but Leibniz published first.

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15. Average Value of a function or Mean Value Theorem: If f is continuous on [a,b] then the average value of f on [a,b] is

16. Even and Odd Functions If a function is symmetric with respect to the y-axis it is called an even function. so f(-x) = f(x) The value of the function at x and –x is the same If a function is symmetric with respect to the origin it is called an odd function. so f(-x) = -f(x) The value of the function at -x is the opposite value of the function evaluated at f(x).

17. For even functions: For odd functions:

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20. fnInt is found under math num 9fnInt((2x+3)^3,x,1,4)= 1752