Thanks Dan.  Looks like the original formatting from HtDP didn't carry over from my cut-and-paste.  Still don't really understand how this relates to the question of computing the "area of each trapezoid," though. Cheers, Dave <div class="gmail_quote">On Mon, Feb 9, 2009 at 3:32 PM, <<a href="mailto:danprager@optusnet.com.au">danprager@optusnet.com.au</a>> wrote: <blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"> <div class="Ih2E3d"> > divide the interval into two parts: [a,(a + b/2)] and [(a + b/2),b]; </div>Try: [a,(a+b)/2 and [(a+b)/2, b] (a+b)/2 is the mid-point between a and b: (a + b/2) ain't. -- Dan <div><div></div><div class="Wj3C7c"> > dave yrueta <<a href="mailto:dyrueta@gmail.com">dyrueta@gmail.com</a>> wrote: > > Hi All -- > > Am sending out a message in a virtual bottle for help with HtDP > exercise 23.4.1, reproduced below: > > A general integration function must consume three inputs: a, b, and > the function f. The fourth part, the x axis, is implied. This suggests > the following contract: > > ;; integrate : (number  ->  number) number number  ->  number > ;; to compute the area under the graph of f between a and b > (define (integrate f a b) ...) > Kepler suggested one simple integration method. It consists of three > steps: > > divide the interval into two parts: [a,(a + b/2)] and [(a + b/2),b]; > > compute the area of each trapezoid; and > > add the two areas to get an estimate at the integral. > > Exercise 23.4.1.   Develop the function integrate-kepler. It computes > the area under some the graph of some function f between left and > right using Kepler's rule. > > Specifically, I'm stuck at the "create examples" part of the design > recipe, because I don't understand the "integration method" outlined > by the exercise. > > First, what does it mean to "divide the interval into two parts?" > Suppose I substitute values for variables a and b (a=3 and b=6).  I > end up with a pair of values: [3, 6] and [6, 6].  Are these x,y > coordinates for points on a graph?  If so, how do they relate to the > second step of the integration method, which is to determine the area > for "each trapezoid?" > > I know the area for a trapezoid = a (b1 + b2 /2) where a is the > altitude, and b1 and b2 are the length of the two bases.  My problem > is deriving values for a, b1 and b2 from the integrate-kepler function > arguments. > > Finally, I have no idea how argument f, the line function, fits into > all this.  I don't even know what a function for a line looks like (y > = x + 10?) > > HtDP supplies an illustration to help visualize the problem (http:// > <a href="http://www.htdp.org/2003-09-26/Book/curriculum-Z-H-29.html#node_sec_23.4" target="_blank">www.htdp.org/2003-09-26/Book/curriculum-Z-H-29.html#node_sec_23.4</a>).  I > haven't the faintest idea how to interpret it: > > I'm sure someone even casually acquainted with basic calculus could > solve this problem easily.  Unfortunately, I'm relatively math- > illiterate, so figuring out how all the parts fit together has eluded > me.  I did manage to solve the other mathematical examples which > appeared earlier in the chapter because the mechanics of determining > the solution appeared clear to me.  Not so here. I'm hoping that if > someone can help me formulate some examples, I'd be on my way. > > Thanks! > > Dave Yrueta </div></div>> _________________________________________________ >   For list-related administrative tasks: >   <a href="http://list.cs.brown.edu/mailman/listinfo/plt-scheme" target="_blank">http://list.cs.brown.edu/mailman/listinfo/plt-scheme</a> </blockquote></div>