Chapter 5 – The Definite Integral 5.1 Estimating with Finite Sums ex. A train is moving along a track at a steady rate of 75 miles per hour from 7:00 AM to 9:00 AM. What is the total distance traveled by the train? v 퐷 = 푅푇 = 75(2) = 150 miles Suppose we look at the graph of the function representing the trains speed at any given time. 75 . Note the shaded rectangle. What is its area? 75 2 150 . This is not a coincidence. The total distance and the area of the o rectangle are found by multiply the same numerical values. This same connection can be made no matter how fast the train is going or how long or short the time interval was. What if the train had a velocity v that changed or varied t as a function of time? 7 8 9 v The graph to the left is no longer a horizontal line, so the region under the graph (the area) would no longer be a rectangle. Would the area of this region give us the total distance traveled over the time interval? . Newton, Leibniz, and others thought that it would, and that is why they were interested in a calculus for finding areas under curves. o They imagined the time interval being partitioned into many tiny subintervals . Each one so small that the velocity over that interval would literally be constant. This would geometrically be the same as slicing the t a b region into narrow strips, each being a narrow rectangle. Gottfried Wilhelm Leibniz (b. 1646, d. 1716) was a German philosopher, mathematician, and logician who is probably most well known for having invented the differential and integral calculus (independently of Sir Isaac Newton). Ex. A particle starts at x = 0 and moves along the x-axis with velocity v(t) = t2 for time t 0. Where is the particle at t = 3? The graph to the right shows the graph of the velocity function over the interval [0,3] v The region is divided into subintervals of length ¼ and have curved tops that slope upward. To find the area of the strip: o We will use the rectangle whose “top” passes through the midpoint of the interval. o Note the two arrows . The area of the two “triangular” regions are very close to equal, so the area of the rectangle would be a good approximation of the area of the strip. o The smaller the interval (base of the rectangle, the more accurate the area. So here is a table to help us out: t 1 1 1 1 3 3 Subinterval [0, 4 ] [ 4 , 2 ] [ 2 , 4 ] [ 4 ,1] 1 3 5 7 Midpoint (m) 8 8 8 8 2 1 9 25 49 Height (m ) 64 64 64 64 1 2 1 9 25 49 Area ( 4 m ) 256 256 256 256 Continuing this table, we would derive the area of each of the 12 subintervals to be: 1 9 25 49 81 121 169 225 289 361 441 529 2300 256 256 256 256 256 256 256 256 256 256 256 256 256 8.98 We can now conclude that the particle has moved approximate 9 units in 3 seconds. Rectangular Approximation Method (RAM in the book): The above example we used the Midpoint Rectangular Approximation Method (MRAM) o We determine the height using the midpoint of the subinterval. We could use the left endpoint of the subinterval as the height of the rectangle (LRAM) o Suppose we used the same problem with only 6 intervals (the size 1 of the interval would be 2 x ). o This would result in the following sum: 2 1 1 2 1 2 1 3 2 1 2 1 5 2 1 0 ( 2 ) ( 2 ) ( 2 ) 1 ( 2 ) ( 2 ) ( 2 ) 2 ( 2 ) ( 2 ) ( 2 ) 6.875 1 2 3 We could use the right endpoint of the subinterval as the height of the rectangle (RRAM) o This would result in the following sum: 1 2 1 2 1 3 2 1 2 1 5 2 1 2 1 ( 2 ) ( 2 ) 1 ( 2 ) ( 2 ) ( 2 ) 2 ( 2 ) ( 2 ) ( 2 ) (3) ( 2 ) 11.375 Using the MRAM, we would have the following result: 1 2 1 3 2 1 5 2 1 7 2 1 9 2 1 11 2 1 1 2 3 ( 4 ) ( 2 ) ( 4 ) ( 2 ) ( 4 ) ( 2 ) ( 4 ) ( 2 ) ( 4 ) ( 2 ) ( 4 ) ( 2 ) 8.9375 1 Note that the result is not as accurate as when the interval was 4 . Note: You can determine the size of the interval using the following formula: x2 x1 where xn represents the interval [x1, x2] and ni is the # of intervals ni The following table shows how the number of intervals make a difference: n LRAMn MRAMn RRAMn 6 6.875 8.9375 11.375 12 7.90625 8.984375 10.15625 24 8.4453125 8.99609375 9.5703125 48 8.720703125 8.999023438 9.283203125 100 8.86545 8.999775 9.13545 1000 8.9865045 8.99999775 9.0135045 So as the number of intervals increase, LRAM, RRAM, and MRAM get closer and closer to the actual area under the curve. 5.2 Definite Integrals Riemann Sums: In the previous section, we obtained the area as a sum of areas of strips o The terms in the sums were obtained by multiplying selected function values by the lengths of the intervals. o What happens as we make the intervals smaller and smaller and increase the number of intervals? Def: Sigma Notation – enables us to express a large sum in a compact way. o We use the Greek symbol sigma n a a a a ... a a k 1 2 3 n1 n k 1 o The Greek Letter stands for sum in mathematics. The index k tells us where to begin (the number below sigma) and where to end (on top of sigma). If the symbol ∞ appears above the , then it indicates that terms go indefinitely. o We will be looking at Riemann Sums (pronounced “ree-mahn”) after George Friedrich Riemann. LRAM, MRAM, and RRAM are all examples of Riemann sums They are considered Riemann sums because of the way were constructed. Homework: Pg. 269-270 QR #1-10, Ex. 1-6, 15, 16 Pg. 282-283 QR #1-10 5.2 Definite Integrals (con’t) Suppose we have a continuous function f(x) defined on a closed interval [a, b]. Look at the diagram below: a b o Note the graph can have negative values as well as positive values o Now we partition the interval [a, b] into n subintervals by choosing n–1 points, called x1, x2, x3, …, xn-1, between a and b using only the following condition: a < x1 < x2 < … < xn-1 < b o To make things more consistent, we change a to x0 and b to xn. The set P is called the partition of [a, b] and is written P = [x0, x1, x2, …, xn] o This partition P determines n closed subintervals. st nd . The 1 subinterval is [x0, x1], the 2 is [x1, x2], etc… th . The k subinterval would be [xk–1, xk] The length of this subinterval would be xk xk xk 1 . So the partition P divides [a, b] into n subintervals of length x1,x2 ,...,xn . o In each subinterval, we select some number; call it ck, for 1 k n. o On each subinterval, we stand a vertical rectangle that reaches from the x-axis to touch the curve (above or below the x-axis) at (ck, f(ck)) Note: The ck value is an arbitrary point in the interval. c1 c2 It is not a specific a value for the ck cn b interval. 푓(푐) ∙ ∆푥 is the area of the rectangle of the kth subinterval o On each subinterval, we form the product f (ck ) xk . This product can be negative, positive, or zero (depending on f(ck)). Finally, we find the sum of these products: 푎푟푒푎푠 n Sn f (ck ) xk k1 o This partition is a Riemann Sum for f on the interval [a, b]. o As the number of partitions increases (or becomes finer and finer), the sum of the areas of the rectangles would approach the area of the region. o So the lengths of the subintervals must approach zero. We can make sure of this if we have the longest subinterval length (called the norm of the partition and denoted by P ) o This leads us to the definition of an Integral Def: The Definite Integral as a Limit of Riemann Sums Let f be a function defined on the 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 As the length of the n norm approaches 0, then all of the lim f (ck ) xk I P 0 subintervals will k 1 approach 0 no matter how P and the ck’s are chosen, then f is integrable on [a, b] and f is the definite integral of f over [a, b]. o This works only if f is continuous. Theorem: The Existence of Definite Integrals All continuous functions are integrable. That is, if a function f is continuous on an interval [a, b], then its definite integral over [a, b] exists. The 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 ba length x n .