the rectangle method. introduction definite integral (high school material): a definite integral a...
TRANSCRIPT
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The Rectangle Method
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Introduction
• Definite integral (High School material):
• A definite integral a∫b f(x) dx is 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:
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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.
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Rectangle Method: explained
• Rectangle Method:
1. Divide the interval [a .. b] into n pieces; each piece has the same width:
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Rectangle Method: explained (cont.)
The width of each piece of the smaller intervals is equal to:
b - a width = ------- n
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Rectangle Method: explained (cont.)
2. The definite integral (= area under the graph is approximated using a series of rectangles:
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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
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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
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Rectangle Method: explained (cont.)
• Example: first interval
• First (small) interval: [a .. (a + width)] (remember that: width = (b − a)/n)
• Height of first (small) interval:
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Rectangle Method: explained (cont.)
• Therefore: height of first rectangle = f(a)
• Area of the rectangle = f(a) × width
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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:
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Rectangle Method: explained (cont.)
• Therefore: height of first rectangle = f(a+width)
• Area of the rectangle = f(a+width) × width
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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:
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Rectangle Method: explained (cont.)
• Therefore: height of first rectangle = f(a+2width)
• Area of the rectangle = f(a+2width) × width
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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)
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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
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Rectangle Method: explained (cont.)
• The area of the rectangles:
• This figure helps you to visualize the computation:
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Rectangle Method: explained (cont.)
• The width of rectangle i is equal to:
b - a width = ------- n
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Rectangle Method: explained (cont.)
• The height of rectangle i is equal to:
height = f( a + (i-1)×width )
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Rectangle Method: explained (cont.)
• The area of rectangle i is equal to:
area = width × height = width × f( a + (i-1)×width )
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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 )
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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 }
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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 }
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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
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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
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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 )
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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 ); }
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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
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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 --------------------------------------------------- */
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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); } }
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The Rectangle Method in Java (cont.)
• Example 1: compute 0∫1 x3 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 --------------------------------------------------- */
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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); } }
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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 = 0.2495002499999998 Exact answer: 0.25
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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 --------------------------------------------------- */
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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); } }
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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 = 0.6933972430599376 Exact answer: ln(2) = 0.69314718
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The effect of the number of rectangles used in the approximation
• Difference in the approximations when using different number of rectangles:
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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
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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 more smaller values (because the rectangles are smaller and have smaller areas))
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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 something without give up on something else
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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:
1. Approximate integral value of 0∫1 x3 dx
= 0.24999950000025453
2. Approximate integral value of 1∫2 (1/x) dx
= 0.6931474305600139