The Rectangle Method

1. The Rectangle Method

2. Introduction • Definite integral (High School material): • A definite integral   a∫bf(x) dxis the integral of a function f(x) with fixed end point a and b: • The integral of a function f(x) is equal to the area under the graph of f(x). • Graphically explained:

3. Introduction (cont.) • Rectangle Method: • The rectangle method (also called the midpoint rule) is the simplest method in Mathematics used to compute an approximation of a definite integral.

4. Rectangle Method: explained • Rectangle Method: • Divide the interval [a .. b] into n pieces; each piece has the same width:

5. Rectangle Method: explained (cont.) The width of each piece of the smaller intervals is equal to: b - a width = ------- n

6. Rectangle Method: explained (cont.) 2. The definite integral (= area under the graph is approximated using a series of rectangles:

7. Rectangle Method: explained (cont.) The area of a rectangle is equal to: We (already) know the width of each rectangle: area of a rectangle = width × height b - a width = ------- n

8. Rectangle Method: explained (cont.) • The height of the rectangles: • The different rectangles has different heights • Heights of each rectangle: • The heights of each rectangle = the function value at the start of the (small) interval

9. Rectangle Method: explained (cont.) • Example: first interval • First (small) interval: [a .. (a + width)] (remember that: width = (b − a)/n) • Height of first (small) interval:

10. Rectangle Method: explained (cont.) • Therefore: height of first rectangle = f(a) • Area of the rectangle = f(a) × width

11. Rectangle Method: explained (cont.) • Example: second interval • the second (small) interval is [(a+width) .. (a+2width)] (remember that: width = (b − a)/n) • Height of first (small) interval:

12. Rectangle Method: explained (cont.) • Therefore: height of first rectangle = f(a+width) • Area of the rectangle = f(a+width) × width

13. Rectangle Method: explained (cont.) • Example: third interval • the third (small) interval is [(a+2width) .. (a+3width)] (remember that: width = (b − a)/n) • Height of first (small) interval:

14. Rectangle Method: explained (cont.) • Therefore: height of first rectangle = f(a+2width) • Area of the rectangle = f(a+2width) × width

15. Rectangle Method: explained (cont.) • We see a pattern emerging: • Height of rectangle 1 = f(a + 0×width) • Height of rectangle 2 = f(a + 1×width) • Height of rectangle 3 = f(a + 2×width) • ... • Height of rectangle n−1 = f(a + (n−2)×width) • Height of rectangle n = f(a + (n−1)×width)

16. Rectangle Method: explained (cont.) • Note: there are n (smaller) interval in total. • Conclusion: Height of rectangle i = f( a + (i-1)*width ) = the function value at the point "a + (i-1)*width" b-a Recall that: width = ----- n

17. Rectangle Method: explained (cont.) • The area of the rectangles: • This figure helps you to visualize the computation:

18. Rectangle Method: explained (cont.) • The width of rectangle i is equal to: b - a width = ------- n

19. Rectangle Method: explained (cont.) • The height of rectangle i is equal to: height = f( a + (i-1)×width )

20. Rectangle Method: explained (cont.) • The area of rectangle i is equal to: area = width × height = width × f( a + (i-1)×width )

21. Rectangle Method: explained (cont.) • The approximation of the definite integral: Approximation = sum of the area of the rectangles = area of rectangle 1 + area of rectangle 2 + ... + area of rectangle n = width × f( a + (1-1)×width ) + width × f( a + (2-1)×width ) + ... + width × f( a + (n-1)×width )

22. The general running sum algorithm • We have seen a running sum computation algorithm previously that adds in simple series of numbers: • Compute the sum: 1 + 2 + 3 + .... + n • Running sum algorithm: sum = 0; // Clear sum for ( i = 1; i <= n ; i++ ) { sum = sum + i; // Add i to sum }

23. The general running sum algorithm (cont.) • The running sum algorithm can be generalized to add a more general series • Example: compute 12 + 22 + 32 + ... + i2 + ... + n2 • The ith term in the sum = i2 • Therefore, the running sum algorithm to compute this sum is: sum = 0; // Clear sum for ( i = 1; i <= n ; i++ ) { sum = sum + i*i; // Add i2 to sum }

24. Algorithm to compute the sum of the area of the rectangles • We can use the running sum algorithm to compute the sum of the area of the rectangles

25. Algorithm to compute the sum of the area of the rectangles (cont.) • Recall: Approximation = sum of the area of the rectangles = width × f( a + (1-1)×width ) + width × f( a + (2-1)×width ) + ... + width × f( a + (i-1)×width ) + ... + width × f( a + (n-1)×width ) Where: width = (b - a) / n

26. Algorithm to compute the sum of the area of the rectangles (cont.) • The ith term of the running sum is equal to: ith term = width × f( a + (i-1)×width )

27. Algorithm to compute the sum of the area of the rectangles (cont.) • Algorithm to compute the sum of the area of the rectangles: Variables: double w; // w contains the width double sum; // sum contains the running sum Algorithm to compute the sum: w = (b - a)/n; // Compute width sum = 0.0; // Clear running sum for ( i = 1; i <= n; i++ ) { sum = sum + w*f( a + (i-1)*w ); }

28. The Rectangle Method in Java • Rough algorithm (pseudo code): input a, b, n; // a = left end point of integral // b = right end point of integral // n = # rectangles used in the approximation Rectangle Method: w = (b - a)/n; // Compute width sum = sum of area of the n rectangles; // Compute area print sum; // Print result

29. The Rectangle Method in Java (cont.) • Algorithm in Java: public class RectangleMethod01 { public static void main(String[] args) { double a, b, w, sum, x_i; int i, n; **** Initialize a, b, n **** /* --------------------------------------------------- The Rectangle Rule Algorithm --------------------------------------------------- */

30. The Rectangle Method in Java (cont.) w = (b-a)/n; // Compute width sum = 0.0; // Clear running sum for ( i = 1; i <= n; i++ ) { x_i = a + (i-1)*w; // Use x_i to simplify formula... sum = sum + ( w * f(x_i) ); // width * height of rectangle i } System.out.println("Approximate integral value = " + sum); } }

31. The Rectangle Method in Java (cont.) • Example 1: compute 0∫1x3 dx(the exact answer = 0.25) public class RectangleMethod01 { public static void main(String[] args) { double a, b, w, sum, x_i; int i, n; a = 0.0; b = 1.0; // 1∫2x3 dx n = 1000; // Use larger value for better approximation /* --------------------------------------------------- The Rectangle Rule Algorithm --------------------------------------------------- */

32. The Rectangle Method in Java (cont.) w = (b-a)/n; // Compute width sum = 0.0; // Clear running sum for ( i = 1; i <= n; i++ ) { x_i = a + (i-1)*w; sum = sum + ( w * (x_i * x_i * x_i) ); // f(x_i) = (x_i)3 } System.out.println("Approximate integral value = " + sum); } }

33. The Rectangle Method in Java (cont.) • Example Program: (Demo above code) • Prog file: http://mathcs.emory.edu/~cheung/Courses/170/Syllabus/07/Progs/RectangleMethod01.java • How to run the program:             • Right click on link and save in a scratch directory • To compile:   javac RectangleMethod01.java • To run:          java RectangleMethod01 • Output: Approximate integral value = 0.2495002499999998Exact answer: 0.25

34. The Rectangle Method in Java (cont.) • Example: compute 1∫2(1/x) dx(the answer = ln(2)) public class RectangleMethod02 { public static void main(String[] args) { double a, b, w, sum, x_i; int i, n; a = 1.0; b = 2.0; // 1∫2(1/x) dx n = 1000; // Use larger value for better approximation /* --------------------------------------------------- The Rectangle Rule Algorithm --------------------------------------------------- */

35. The Rectangle Method in Java (cont.) w = (b-a)/n; // Compute width sum = 0.0; // Clear running sum for ( i = 1; i <= n; i++ ) { x_i = a + (i-1)*w; sum = sum + ( w * (1/x_i) ); // f(x_i) = 1/x_i } System.out.println("Approximate integral value = " + sum); } }

36. The Rectangle Method in Java (cont.) • Example Program: (Demo above code) • Prog file: http://mathcs.emory.edu/~cheung/Courses/170/Syllabus/07/Progs/RectangleMethod02.java • How to run the program:             • Right click on link and save in a scratch directory • To compile:   javac RectangleMethod02.java • To run:          java RectangleMethod02 • Output: Approximate integral value = 0.6933972430599376Exact answer: ln(2) = 0.69314718

37. The effect of the number of rectangles used in the approximation • Difference in the approximations when using different number of rectangles:

38. The effect of the number of rectangles used in the approximation (cont.) • Clearly: • Using more rectangles will give us a more accurate approximation of the area under the graph

39. The effect of the number of rectangles used in the approximation (cont.) • However: • Using more rectangles will make the algorithm add more numbers • I.e., the algorithm must do more work --- it must adds moresmaller values (because the rectangles are smaller and have smaller areas))

40. The effect of the number of rectangles used in the approximation (cont.) • Trade off: • Often, in computer algorithms, a more accurate result can be obtained by a longer running execution of the same algorithm • We call this phenomenon: trade off • You cannot gain somethingwithoutgive up on something else

41. The effect of the number of rectangles used in the approximation (cont.) You can experience the trade off phenomenon by using n = 1000000 in the above algorithm. It will run slower, but give you very accurate results ! Outputs for n = 1000000: • Approximate integral value of 0∫1x3 dx= 0.24999950000025453 • Approximate integral value of 1∫2(1/x) dx= 0.6931474305600139