Phys102 Lecture03!08!10Fall Integration

8/14/2019 Phys102 Lecture03!08!10Fall Integration

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Integration procedure:

We will need to define an operatorwhich will

take as input a function limits of integration

possibly, desired accuracy

give as output the result of integratingthe function between the two limits

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From Calculus:

Rectangular rule: Evaluate the function at

the left endpoint of each

interval.Multiply by the width of

the interval.

Equivalent to sum of

rectangular areas.

1

0

( ') ' ( )x n

kL

f x dx x f L k x

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Alternate: Midpoint rule

Midpoint rule. Evaluate the function at the left

endpoint of each interval.

Multiply by the width of the interval.

Equivalent to sum of rectangularareas, but has a better discrete errorbehavior.

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0

( ') ' ( ( 0.5) )

x n

kL

f x dx x f L k x

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A first Implementation: Procedural

#include #include

using namespace std;

double square(double aX) {return aX*aX;

}

double integrate(double aLeftLimit, double aRightLimit,int aNumberOfIntervals, double integrand(double)) {

double dx = (aRightLimit - aLeftLimit)/aNumberOfIntervals;double x = aLeftLimit;double sum = 0;for (int i=0; i loLimit >> upLimit;cout nIntervals;cout

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Implementing code with a desired

accuracy.int main() {double loLimit, upLimit, epsilon, result;int nIntervals;cout loLimit >> upLimit;cout nIntervals >> epsilon;

double oldResult = 1.0e30; // large number, at least iterate onceint maximumIterations = 25;for (int j=0; j

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Using a class: IntegralCalculator.hh#ifndef IntegralCalculator_hh#define IntegralCalculator_hh

class IntegralCalculator {

public:IntegralCalculator(double aIntegrandFunction(double));void setNumberOfIntervals(int aNumberOfIntervals)

{mNumberOfIntervals=aNumberOfIntervals;}void setLeftLimit(int val) {mLeftLimit=val;}void setRightLimit(int val) {mRightLimit=val;}void readInputData();void integrate();void print();

int numberOfIntervals() {return mNumberOfIntervals;}int leftLimit() {return mLeftLimit;}

int rightLimit() {return mRightLimit;}int result() {return mResult;}

private:int mNumberOfIntervals;double mLeftLimit;double mRightLimit;double mResult;double (*mIntegrandFunction) (double);

};#endif

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The source file: IntegralCalculator.cc

#include "IntegralCalculator.hh"

#include #include

using namespace std;

IntegralCalculator::IntegralCalculator(doubleaIntegrandFunction(double)) {

mIntegrandFunction = aIntegrandFunction;}void IntegralCalculator::readInputData() {double loLimit, upLimit;int nIntervals;cout loLimit >> upLimit;cout nIntervals;setLeftLimit(loLimit);setRightLimit(upLimit);setNumberOfIntervals(nIntervals);return;

}

void IntegralCalculator::integrate() {double dx = (mRightLimit -

mLeftLimit)/mNumberOfIntervals;double x = mLeftLimit;double sum = 0;

for (int i=0; i

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The new main() using the class.

#include "IntegralCalculator.hh"

double square(double aX) {

return aX*aX;

}

int main() {

IntegralCalculator IC1(square);IC1.readInputData();

IC1.integrate();IC1.print();

return 0;}

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Compiling and linking with a class

Compiling the class to create amachine readable object file: g++ IntegralCalculator.cc c

Compiling the file containing the mainfunction and linking with thepreviously created object file: g++ integWithClass.cc o integWithClass

IntegralCalculator.o

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For Homework: Numerical Integration

Chapter 13, Section 13.4

Assigment 1 Complete the coding of the IntegratorCalculator classusing the midpoint rule. Verify the convergence should be quadraticin the step length. In particular, plot graphs of the error againstthe inverse of the number of points, n, for the integral of x3 from 0to 1 for both the midpoint rule and the rectangle rule.

Assignment 2 Replace the midpoint rule by the Simpson rule, anddetermine graphically the order of error of this procedure.

Asignment 5 Write a program to take the integral of thenumerically evaluated derivative function you wrote in the previoushomework, use it for the function x4 to show that you recover theoriginal function.

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( ) ( ) 4 ( )

6 2

R ni i

i i

iL

x xx f x dx f x f f x