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Returning to the Geography project shown in Examples 5.2 and 5.3, suppose we rearrange the list of continents into geographical groupings, with separators between each group. We will also disable Oceans, make Seas invisible, and place a check mark next to Africa (listed under Continents). In addition, we will add check boxes to the main window so that Oceans can be enabled and Seas can be made visible. Finally, we will add the Visual Basic code required by the check boxes, and to toggle the Africa check mark on and off. In order to rearrange the list of continents and introduce separators, the Continents menu items in the Menu Editor must appear as shown below. Note the separators following Europe, South America and Australia.

(5.7)

argument is a pointer to a function that has one doubl e argument and returns a doubl e. In this test

Fig. 5-34

run, we pass it (a pointer to) the cube ( ) function. The other three arguments are the boundaries a and b of the interval [a, b] over which the integration is being performed and the number n of subn rectangles intervals to be used in the sum. The actual Riemann sum is the sum of the areas of the based on these subintervals whose heights are given by the function being integrated: double riemann(double (*) (double), double, double, int); double cube(double); main0

tout tout tout tout CC CC CC CC riemann(cube,0,2,10) CC endl; riemann(cube,0,2,100) CC endl; riemann cube,0,2,1000) CC endl; riemann cube,0,2,10000) CC endl;

(a) BC (b) AB (c) AB (d) AB (e) m/B ( f ) m/1

// Returns [f(a)*h + // where h = (b-a)/n: double riemann(double { double s = 0, h = int i; for (x = a, i = 0; s += (*Pf) bd ; return s*h;

60 120

+ . . . + f(b-h)*h], n)

29, m/2

*Pf ) (double t), double a, double b, int b-a J/n, x;

&Continents ....&Africa ....As&ia ....&Europe ....&North America ....&South America ....A&ustralia ....An&tarctica

y (5.8)

i< n; x += h, i++)

return t*t*t;

5.11. Provide the proofs requested in Fig. 5-35.

In this test run, we are integrating the function y = J over the interval [0,2]. By elementary calculus, the value of this integral is 4.0. The call riemann ( cube, 0 , 2 , 10 ) approximates this integral using 10 subintervals, obtaining 3.24. The call r iemann ( cube, 0 ,2 , 10 0 > approximates the integral using 100 subintervals, obtaining 3.9204. These sums get closer to their limit 4.0 as n increases. With 10,000 subintervals, the Riemann sum is 3.9992. Note that the only significant difference between this ri emann ( ) function and the sum ( ) f u n c tion in Example 6.15 is that the sum is multiplied by the subinterval width h before being returned.

Fig. 5-35

Write a function that returns the numerical derivative of a given functionf at a given point x, using a given tolerance h. Use the formula

5.12. Prove each of the following: (a) If the diagonals of a parallelogram are congruent, the parallelogram is a rectangle.

-f(x-h)

(5.9)

This derivative ( ) function is similar to the sum ( ) function in Example 6.15, except that it implements the formula for the numerical derivative instead. It has three arguments: a pointer to the function5 the x value, and the tolerance h. In this test run, we pass it (pointers to) the cube ( > f u n c tion and the sqr t ( ) function. #include #include ciostream.h> cmath.h> (*) (double), double, double);

The remaining menu items will appear in the same order as in Example 5.3. Figures 5.10(a), (b) and (c) show the Menu Editor settings for the three menu items. Note the status of the check boxes in each figure (for each of the menu items). Fig. 5.11 shows the form design window after the new controls (Check1 and Check2) have been added. Notice that Oceans now appears grayed out, and Seas does not appear on the menu bar at all. We must now add the following two event procedures so that the check box controls will function properly.