Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.ICN Textbook: Calculus Early Vectors Preliminary Edition By Stewart. Fri, Jan 17 6.2 Area p377: 2,(6),8,9,13,(16),18,(20), Main Idea: Approximate the area with rectangles whose tops touch the curve. Key Words: Area, Riemann Sums, Mesh Size, Midpoint Rule, End Point Rule, Partition. Goal: Continuous curves define area and this area can be computed numerically by any Riemann Sum of sufficiently small mesh size. Previous Assignment 6.1 Summation p368: (28),(30),(48) Page 368 Problem 28 i=4 3-i SUM 2 i=-2 i 3-i 2^(3-i) --------------------------------- -2 5 32 -1 4 16 0 3 8 1 2 4 2 1 2 3 0 1 4 -1 1/2 -------------------- ans 63 1/2 ----------------------------------------------------- Page 368 Problem 30 n Find the value of the sum SUM (2-5i) i=1 n n n n(n+1) -5n^2-n SUM (2-5i) = SUM 2 -5 SUM i = 2n -5 ------- = --------- i=1 i=1 i=1 2 2 ------------------------------------------------------------ Page 368 Problem 48 n 3 Evaluate SUM ---------- i=1 2^(i-1) _ _ i-1 i n 3 n | 1 | 1-(1/2) 6 SUM -------- = 3 SUM |---| = 3 --------- = 6 - ----- i=1 2^(i-1) i=1 |_2_| 1-1/2 2^i ------------------------------------------------------------- End Hand-In-Homework Assignment. Page 368 Problem 38 2 n | n(n+1) | Prove SUM i^3 = |------- | i=1 | 2 | Let P(n) be the following statement: | n(n+1) | 2 1^3+2^3+...+n^3 = |--------| . | 2 | For some n, the statement P may be true; for other values of n, the statement P may be false. We wish to show that P(n) is true for all n. | 1(1+1) | 2 The statement P(1) is: 1^3 = | ------- | which is true. | 2 | | 2(2+1) | 2 The statement P(2) is: 1^3 + 2^3 = | ------- | which is true | 2 | since both sides equal 9 | 3(3+1) | 2 The statement P(3) is: 1^3 + 2^3 + 3^3 = | ------- | which is true. | 2 | since both sides equal 36 Now we assume P(n) is true and prove that P(n+1) is also true. The statement P(n+1) is: | (n+1)(n+2) | 2 1^3+2^3+3^3+...+n^3+(n+1)^3 = |------------| | 2 | We first split off the last term of the left hand side. (1^3+2^3+3^3+...+n^3) + (n+1)^3 Since we are assuming that P(n) is true, the first portion of the sum can be replaced by the formula. | n(n+1) | 2 | ------ | + (n+1)^3 | 2 | We combine these algebraically into a single expression by first factoring out the common factor of (n+1)^2 2 n^2 (n+1) ( --- + n+1) 4 We finally put the whole expression over a common denominator of 4 to have the following. 2 (n+1) (n^2 + 4 n + 4) -------------------------- 4 Recognizing that (n^2 + 4 n + 4) = (n+2)^2, we finally produce the sum of 1^3 + 2^3 + 3^3 + ... + n^3 + (n+1)^3 using the assumption that the formula worked to give us the sum of 1^3 + 2^3 + 3^3 + ... + n^3. | (n+1)(n+2)|2 |-----------| | 2 | Notice that this is the formula evaluated at n --> n+1. The formula has now been shown to also work for n+1. The statement P(n) is true for all integers n. I.e. The formula is correct. Page 368 Problem 44 n 1 | / i \ 3 | Limit SUM --- | |-----| + 1 | n->Infinity i=1 n | \ n / | n 1 | / i \ 3 | n / i \ 3 n SUM --- | |-----| + 1 | = 1/n SUM | --- | +1/n SUM 1 i=1 n | \ n / | i=1 \ n / i=1 | n(n+1) |2 = 1/n^4 |---------| + 1 | 2 | | n(n+1) |2 = |---------| + 1 | 2 n^2 | | 1(1+1/n) |2 = |--------- | + 1 | 2 | The limit of this as n-> infinity is 1/4 + 1 = 5/4. You can view this limit as an area. The area under f(x) = 1+x^3 from zero to 1 should be 5/4. Since the function is increasing and one is taking the right hand endpoint for the height of the rectangular approximations, we are generating an upper sum which should converge downward to the value of the area. The corresponding lower sum is n-1 | / i \ 3 | SUM (1/n) || --- | + 1 | i=0 | n / | u[n_] := Sum[(1/n)( (i/n)^3+1),{i,1,n}] l[n_] := Sum[(1/n)( (i/n)^3+1),{i,0,n-1}] U = Table[{1/n,u[n]},{n,2,100}]; L = Table[{1/n,l[n]},{n,2,100}]; a = ListPlot[U]; b = ListPlot[L]; c = Show[a,b,PlotLabel->"Page 368 Problem 44"]; For an increasing function, the difference between the upper sum and the lower sum can be easily bounded. The difference is <= [f(b)-f(a)]*(meshsize) f[x_] := 1+x^3; h[n_] := Table[ Plot[ f[i/n],{x,(i-1)/n,(i+1)/n}],{i,1,n}]; k[n_] := Table[ Plot[ f[(i+1)/n],{x,(i-1)/n,i/n}],{i,1,n}]; l[n_] := Table[ ParametricPlot[{i/n,y},{y,f[(i-1)/n],f[(i+1)/n]}],{i,1,n}]; Theory of Integration: Given a function y = f(x) we want to find the area under the curve y = f(x) from x=1 to x=b. | | . ' ' . | ' ' . | . ' . | . ' ' . | . ' | ' | . '| | | | | | | | ---------+------------------------------------------------- | a b We partition the interval [a,b] on the x axis by choosing intermediate points a = po < p1 < p2 < p3 < ... < pn = b. The distances between successive points are called the cell widths. It is not required that the points be equally spaced. The maximal cell width is called the "mesh" of the partition. We approximate the curve with rectangles whose sides occur at the points of the partition. | | -----.*-'| ' . | | ' |----'*.-| | .| | |'-*.- | |-----*'-| | | |'*-.- | | . ' | | | | | ' | . '| | | | | | | | | | | | | | | | | | | | ---------+-------|--------|--------|--------|-----|----|--- | a b po p1 p2 p3 p4 p5 The only requirement is that the tops of the rectangles all touch the curve. If the tops of the rectangles are chosen to touch the curve at the highest point, then we get an upper sum. If the tops of the rectangles touch the curve at the lowest point, we get the lower sum. For increasing functions, the upper and lower sums are easy to find because the left hand point is always the minimum of the function and the right hand point is always the maximum. But if the function is not increasing or decreasing, it can be difficult and time consuming to actually find the max and min. The nice thing is that we do not have to. If one uses any height rectangle as long as the top of the rectangle cuts the curve somewhere, the process will give the correct answer. An easy way to get an appropriate height is to simply evaluate the function y = f(x) at some point on the interval. An expression n SUM f(x*) (p -p ) is called i=1 i i+1 i a Riemann Sum. It represents the total area of the approximating rectangles and it approximates the area under the curve. Notice what goes into a Riemann Sum given that one wants to approximate the area of y = f(x) above [a,b]. (1) Choose any partition. p < p < ... < p =b. 0 1 n (2) Choose any x* p <= x* <= p . i i i i+1 Given these choices then the value of the Riemann Sum for this partition and this choice of x* is determined. i In order for the approximation to be close to the true answer, the mesh size should be small. If one graphs the values of the Riemann Sums by mesh size, the appearance of the graph will look like this. Riemann Sum | | .xx| | .xxxx| | .xxxxxxxx| | .xxxxxxxxxxx| | .xxxxxxxxxxxxxxx| Riemann says | .xxxxxxxxxxxxxxxxxxxx| ------------>|... xxxxxxxxxxxxxxxxxxxxxxxx| Limit exists | .xxxxxxxxxxxxxxxxxxxxx| | .xxxxxxxxxxxxxxxxx| | .xxxxxxxxxxxxxx| | .xxxxxxxxxxxx| | .xxxxxxxxxx| | .xxxxxxxx| | .xxxxxx| -------------+--------------------------------------------- | Mesh Size. b-a | In general, one can only say that all of the Riemann sums for a particular partition are boxed in by the upper sums and the lower sums for that particular partition. There is no reason why the upper sums and the lower sums should converge to a single value. But Riemann says that when the function y = f(x) is continuous, then the values of all of the Riemann sums approach a single limiting value as the mesh size decreases to zero. That value is called the integral of the function y = f(x) from x=a to x=b. This value is defined to be the area under the curve. The integral agrees with the standard concept of area for triangles and rectangles. Furthermore it behaves like area should. If one divides a region into disjoint pieces, then the sum of the area of the pieces equals the area of the whole. The concept of area in the traditional sense requires that the curve to lie above the x-axis. Geometry seems to abhor a negative area. The Riemann sums converge just as nicely when y = f(x) is not always positive. However, area below the axis is negative and the result is the combination of the areas with the negative areas canceling positive areas out. The area under the sine curve form x=0 to x=2pi will be zero. In one sense, this is as it should be. Remember that the area under a velocity curve is the distance traveled. There are two ways to measure distance. You can look at the odometer or the gas tank to know how far you went. This gives the total distance traveled. But if you call on your cell phone to report that you ran out of gas 10 miles east of Ames on Route 30, then you want to have subtracted the miles you walked back towards Ames. This is what the area under the velocity curve automatically does when it figures area below the x-axis as negative. If you want the odometer reading, you have to split the curve into positive and negative portions and add the absolute value of the areas. In class problem: (a) Find the area under the curve y = x^2 from x = 0 to x = 3. (b) Find the area under the curve y = x^2 from x = 0 to x = a. (c) Find the area under the curve y = x^2 from x = a to x = b.