Integration in 1D
We often calculate integrals in physics (electromagnetism, thermodynamics, quantum
mechanics, etc.). Without computers we evaluate integrals analytically. There are some
programs that follow the same steps that you follow to derive the analytical solution. The
Wolfram web page has a 1-D integrator: http://integrals.wolfram.com/index.jsp. On their
web page you can click on compute for a variety of examples, or enter your own
function. The result is a function of the input parameters and variables. Maple and
Mathematica both have analytical integrators and are otherwise similar to Matlab,
allowing you to evaluate the resulting functions and plot the results.
Matlab’s trapz, quad, and quadl
Some functions are too difficult to integrate analytically and for these we need to use the
computer to integrate numerically. A numerical integral goes back to the basic principles
of calculus. The integral is the area under a function and is given by I   f ( x) x in
the limit that x  0. Matlab computes integrals numerically using the trapz function.
The idea is to evaluate f (x) at many x values and then sum over the areas made from
small trapezoids: 12 [ f ( xn )  f ( xn1 )]x.
% Make the function
clear all; close all;
dx = 0.1;
%
x = 0:dx:20;
%
y = x.^2;
%
% Integrate
format long
%
I1 = trapz(x,y)
%
I2 = dx*trapz(y)
%
dx step size
array of x values [0:20]
y is a function of x
print more digits
trapz
trapz again.
The integrals, I1 and I2 should have printed in the Matlab window. To obtain precise
results it is important that x be a small number. Making dx too small, however, will
make enormous arrays that can waste Matlab’s memory. This may also cause the
calculation to take a very long time to finish. Try the above example with dx=1.0 and
dx=0.01. When you calculate an integral, try the calculation with several values of dx to
see that your result hasn’t changed in the significant digits you want to report.
There are several other more sophisticated integration functions in Matlab: quad, and
quadl. For functions with sharp curvatures, these routines will adapt the size of the
interval to produce better estimates of the integral. Users can also specify the accuracy of
the integral by passing it a tolerance. The added complication is that the user has to
provide a reusable function as the first argument in the function calls.
xSquared = @(x) x.^2;
tol = 1e-8;
I3 = quad(xSquared, 0, 20, tol)
I4 = quadl(xSquared, 0, 20, tol)
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Here I’ve created a “handle” to an “anonymous function” by writing xSquared =
@(x) x.^2. The @ (x) means that this function takes the argument x and the equation
that follows is the code that will be run each time @(x) is called. We could also define
the function in the call to quad.
I3 = quad(@(x) x.^2, 0, 20, tol)
In theory we can call any of our reusable functions. In the last assignment, we wrote a
1D function to turn thermistor data into a continuous function. Even though there is no
reason to integrate this function, let’s do it.
Itherms = quad(@thermistor, 1, 3, tol)
The problem I usually have, however, is that my functions have additional parameters.
Remember the myGauss function? It needs sigma as well as x. To integrate
myGauss(x,sigma) I would do the following:
Igauss= quad(@(x) myGauss(x,0.2), -2, 2)
This will call the function myGauss.m (which I created previously) with sigma=0.2.
Technically this is called wrapping my function in an anonymous function.
Trapezoid Rule and Simpson’s Rule
Although “trapz” and “quad” work nicely, take a look at how easy they are to write for
yourself.


f(x)
Trapezoid Rule:
If we take a bunch of trapezoids, the area of each is:
Atrapezoid  1/2[ f (x n 1 )  f (x n )]x .
The integral is then just the sum,
I   Atrapezoid   12 [ f (x n 1)  f (x n )]x .
x
If performed directly, this means that each
n
n+1
f ( xn ) needs to be evaluated twice, once for the trapezoid to its left and again for the
trapezoid to its right. We can eliminate this double computation by rearranging the terms
in the sum:
I  [ 12 f ( x1 ) 
end 1
 f (x ) 
n
1
2
f ( xend )]x .
2
Here’s how to implement the trapezoid sum in Matlab. We’ll first make an array of
coefficients that multiply f(x): [½ 1 1 1 1 1 … ½]. Then we sum.
coef = ones(size(x));
coef(1) = 1/2;
coef(end) = 1/2;
prob = dx*sum(coef.*xSquared(x))
Simpson’s Rule:
Simpson’s Rule is similar to the Trapezoid Rule, except it considers the curvature of the
function. Instead of using trapezoids to estimate the area under the curve, it uses 3
adjacent points to determine a quadratic curve and integrates the area under this
analytically. The result is a sum:
I  [ 13 f ( x1 )  43 f ( x2 )  23 f ( x3 )  ...  43 f ( xend 2 )  43 f ( xend 1 )  13 f ( xend )]x
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Except for the first and last elements, all the even elements have a coefficient of 4/3 and
all the odd elements have a coefficient of 2/3.
coef = 1/3*ones(size(x));
coef(2:2:end-1) = 4/3;
% Even coefficients
coef(3:2:end-1) = 2/3;
% Odd coefficients
prob = dx*sum(coef.*xSquared(x))
Electric field of a rod with constant charge density:
Here is the calculation of the electric field as done previously with a spreadsheet.
Remember, it was a charged rod (charge density  (x ) ) on the x-axis between a and b and
we want the electric field at a point in the plane, ( x0 , y0 ) . First we'll do the constant
charge density case. This time, we can use “pointField.m”, to calculate the field of all
the chunks of charge along the x-axis. Instead of summing directly, however, we’ll use
trapz to perform the integration.
% define constants
lambda = 1.0e-6;
% charge density is 1.0 microC/m
x0 = 3.0;
% position of point (x0, y0) in meters
y0 = 3.0;
dx = 0.1;
xi = 1:dx:2;
% limits of the rod. a=1 to b=2
% electric field at P from a chunk of charge at xi
[E_xi,E_yi] = pointField(x0-xi, y0, lambda*dx);
% integrate along charges on x-axis.
Ex = trapz(E_xi)
Ey = trapz(E_yi)
In the spreadsheet we got: Ex = 349.49 N/C and Ey = 714.43 N/C. Here we get
Ex=349.26 N/C and Ey=714.42 N/C. Which do you think is more accurate?
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Assignment: M7_Integration
1. It is tempting to blindly do a calculation and trust that computers never make mistakes.
Explain why the following example gives imprecise results. Then write code that runs in
less than 5 seconds and integrates correctly using trapz to produce a number that is less
than 10-10.
% This is very imprecise:
theta = [0:0.3:pi]
Iwrong = trapz(theta,cos(theta))
% the answer should be exactly 0.0
2. In Modern Physics or Chemistry you needed to do integrals to find the probability of
finding a particle in a certain region. Consider a particle in a one-dimensional box of
b
2
width L. P(a,b)   (x) dx is the probability of finding the particle between a and b
a
where  is the wavefunction. (You may need to look in a modern physics or chemistry
text to get the wave functions for a particle in a box. Try Google if you don’t have a text
book handy or if you are under 30.) What is the probability of finding the particle
between L/3 and L/2 for the ground state and for the first excited state? To see if these
2
make sense make a plot of  (x) vs. x for L=1. Perform the integral using both trapz
and quad with an accuracy of 6 sig figs. How many bins did you need to use for trapz?
3. . Redo the variable charge density case done previously on the spreadsheet,
 x2
 ( x)  0 2 with 0  2 106 C/m2 and a =1 m and b = 3 m, look at the point
b
(x0, y0 )  (5,3) m. Be sure you get results similar to what you got before. Devise a test to
determine the accuracy of the Matlab calculation. Which is more accurate, Matlab’s
 or Excel’s? Can you suggest possible reasons for the difference(s)?
calculation

4. The CIA was pleased with your analysis of the LabPro data from Baghdad. They now
want to know if you can tell them about any impulses that occurred during the Baghdad
experiments with information recently obtained from a secret agent that the cart in the


tests had a mass of 1000 kg. Impulse can be calculated as J   F (t )dt . First you need
to verify the numerical accuracy of the Impulse-Momentum theorem. Then you should
report your best estimate of the impulse for cia_file1.txt from 1.15 s to 1.45 s and again
from 1.95 s to 2.55 s. (Note: Smoothing has a dramatic affect on the accuracy of
integrating data. Please vary the smoothing window to find the best result.) Can you
draw any conclusions about the impulse that would interest the CIA?
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