A = [5, 9; 3, -6];
b = [5;4];
A  b

When you run the file, it displays the following result −

ans =

   1.157895
  -0.087719

In same way, you can solve larger linear systems as given below −

x + 3y -2z = 5

3x + 5y + 6z = 7

2x + 4y + 3z = 8

Expanding and Collecting Equations in MATLAB

The expand and the collect function expands and collects an equation respectively. The following example demonstrates the concepts −

When you work with many symbolic functions, you should declare that your variables are symbolic.

Create a script file and type the following code −

syms x   %symbolic variable x
syms y   %symbolic variable x
% expanding equations
expand((x-5)*(x+9))
expand((x+2)*(x-3)*(x-5)*(x+7))
expand(sin(2*x))
expand(cos(x+y))
 
% collecting equations
collect(x^3 *(x-7))
collect(x^4*(x-3)*(x-5))

When you run the file, it displays the following result −

ans =
   x^2 + 4*x - 45
ans =
   x^4 + x^3 - 43*x^2 + 23*x + 210
ans =
   2*cos(x)*sin(x)
ans =
   cos(x)*cos(y) - sin(x)*sin(y)
ans =
   x^4 - 7*x^3
ans =
   x^6 - 8*x^5 + 15*x^4

Expanding and Collecting Equations in Octave

You need to have symbolic package, which provides expand and the collect function to expand and collect an equation, respectively. The following example demonstrates the concepts −

When you work with many symbolic functions, you should declare that your variables are symbolic but Octave has different approach to define symbolic variables. Notice the use of Sin and Cos, which are also defined in symbolic package.

Create a script file and type the following code −

% first of all load the package, make sure its installed.
pkg load symbolic

% make symbols module available
symbols

% define symbolic variables
x = sym (''x'');
y = sym (''y'');
z = sym (''z'');

% expanding equations
expand((x-5)*(x+9))
expand((x+2)*(x-3)*(x-5)*(x+7))
expand(Sin(2*x))
expand(Cos(x+y))
 
% collecting equations
collect(x^3 *(x-7), z)
collect(x^4*(x-3)*(x-5), z)

When you run the file, it displays the following result −

ans =

-45.0+x^2+(4.0)*x
ans =

210.0+x^4-(43.0)*x^2+x^3+(23.0)*x
ans =

sin((2.0)*x)
ans =

cos(y+x)
ans =

x^(3.0)*(-7.0+x)
ans =

(-3.0+x)*x^(4.0)*(-5.0+x)

Factorization and Simplification of Algebraic Expressions

The factor function factorizes an expression and the simplify function simplifies an expression. The following example demonstrates the concept −

Example

Create a script file and type the following code −

syms x
syms y
factor(x^3 - y^3)
factor([x^2-y^2,x^3+y^3])
simplify((x^4-16)/(x^2-4))

When you run the file, it displays the following result −

ans =
   (x - y)*(x^2 + x*y + y^2)
ans =
   [ (x - y)*(x + y), (x + y)*(x^2 - x*y + y^2)]
ans =
   x^2 + 4

MATLAB – Calculus

MATLAB provides various ways for solving problems of differential and integral calculus, solving differential equations of any degree and calculation of limits. Best of all, you can easily plot the graphs of complex functions and check maxima, minima and other stationery points on a graph by solving the original function, as well as its derivative.

This chapter will deal with problems of calculus. In this chapter, we will discuss pre-calculus concepts i.e., calculating limits of functions and verifying the properties of limits.

In the next chapter Differential, we will compute derivative of an expression and find the local maxima and minima on a graph. We will also discuss solving differential equations.

Finally, in the Integration chapter, we will discuss integral calculus.

Calculating Limits

MATLAB provides the limit function for calculating limits. In its most basic form, the limit function takes expression as an argument and finds the limit of the expression as the independent variable goes to zero.

For example, let us calculate the limit of a function f(x) = (x³ + 5)/(x⁴ + 7), as x tends to zero.

syms x
limit((x^3 + 5)/(x^4 + 7))

MATLAB will execute the above statement and return the following result −

ans =
   5/7

The limit function falls in the realm of symbolic computing; you need to use the syms function to tell MATLAB which symbolic variables you are using. You can also compute limit of a function, as the variable tends to some number other than zero. To calculate lim _x->a(f(x)), we use the limit command with arguments. The first being the expression and the second is the number, that x approaches, here it is a.

For example, let us calculate limit of a function f(x) = (x-3)/(x-1), as x tends to 1.

limit((x - 3)/(x-1),1)

MATLAB will execute the above statement and return the following result −

ans =
   NaN

Let”s take another example,

limit(x^2 + 5, 3)

MATLAB will execute the above statement and return the following result −

ans =
   14

Calculating Limits using Octave

Following is Octave version of the above example using symbolic package, try to execute and compare the result −

pkg load symbolic
symbols

x = sym("x");
subs((x^3+5)/(x^4+7),x,0)

Octave will execute the above statement and return the following result −

ans =
   0.7142857142857142857

Verification of Basic Properties of Limits

Algebraic Limit Theorem provides some basic properties of limits. These are as follows −

Let us consider two functions −

• f(x) = (3x + 5)/(x – 3)
• g(x) = x² + 1.

Let us calculate the limits of the functions as x tends to 5, of both functions and verify the basic properties of limits using these two functions and MATLAB.

Example

Create a script file and type the following code into it −

syms x
f = (3*x + 5)/(x-3);
g = x^2 + 1;
l1 = limit(f, 4)
l2 = limit (g, 4)
lAdd = limit(f + g, 4)
lSub = limit(f - g, 4)
lMult = limit(f*g, 4)
lDiv = limit (f/g, 4)

When you run the file, it displays −

l1 =
   17
  
l2 =
   17
  
lAdd =
   34
 
lSub =
   0
  
lMult =
   289
  
lDiv =
   1

Verification of Basic Properties of Limits using Octave

Following is Octave version of the above example using symbolic package, try to execute and compare the result −

pkg load symbolic
symbols

x = sym("x");
f = (3*x + 5)/(x-3);
g = x^2 + 1;

l1 = subs(f, x, 4)
l2 = subs (g, x, 4)
lAdd = subs (f+g, x, 4)
lSub = subs (f-g, x, 4)
lMult = subs (f*g, x, 4)
lDiv = subs (f/g, x, 4)

Octave will execute the above statement and return the following result −

l1 =
   17.0
l2 =
   17.0
lAdd =
   34.0
lSub =
   0.0
lMult =
   289.0
lDiv =
   1.0

Left and Right Sided Limits

When a function has a discontinuity for some particular value of the variable, the limit does not exist at that point. In other words, limits of a function f(x) has discontinuity at x = a, when the value of limit, as x approaches x from left side, does not equal the value of the limit as x approaches from right side.

This leads to the concept of left-handed and right-handed limits. A left-handed limit is defined as the limit as x -> a, from the left, i.e., x approaches a, for values of x < a. A right-handed limit is defined as the limit as x -> a, from the right, i.e., x approaches a, for values of x > a. When the left-handed limit and right-handed limit are not equal, the limit does not exist.

Let us consider a function −

f(x) = (x – 3)/|x – 3|

We will show that lim_x->3 f(x) does not exist. MATLAB helps us to establish this fact in two ways −

• By plotting the graph of the function and showing the discontinuity.
• By computing the limits and showing that both are different.

The left-handed and right-handed limits are computed by passing the character strings ”left” and ”right” to the limit command as the last argument.

Example

Create a script file and type the following code into it −

f = (x - 3)/abs(x-3);
ezplot(f,[-1,5])
l = limit(f,x,3,''left'')
r = limit(f,x,3,''right'')

When you run the file, MATLAB draws the following plot

After this following output is displayed −

l =
   -1
  
r =
   1

MATLAB – Differential

MATLAB provides the diff command for computing symbolic derivatives. In its simplest form, you pass the function you want to differentiate to diff command as an argument.

For example, let us compute the derivative of the function f(t) = 3t² + 2t^-2

Example

Create a script file and type the following code into it −

syms t
f = 3*t^2 + 2*t^(-2);
diff(f)

When the above code is compiled and executed, it produces the following result −

ans =
6*t - 4/t^3

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

t = sym("t");
f = 3*t^2 + 2*t^(-2);
differentiate(f,t)

Octave executes the code and returns the following result −

ans =
   -(4.0)*t^(-3.0)+(6.0)*t

Verification of Elementary Rules of Differentiation

Let us briefly state various equations or rules for differentiation of functions and verify these rules. For this purpose, we will write f”(x) for a first order derivative and f”(x) for a second order derivative.

Following are the rules for differentiation −

Rule 1

For any functions f and g and any real numbers a and b are the derivative of the function −

h(x) = af(x) + bg(x) with respect to x is given by −

h”(x) = af”(x) + bg”(x)

Rule 2

The sum and subtraction rules state that if f and g are two functions, f” and g” are their derivatives respectively, then,

(f + g)” = f” + g”

(f – g)” = f” – g”

Rule 3

The product rule states that if f and g are two functions, f” and g” are their derivatives respectively, then,

(f.g)” = f”.g + g”.f

Rule 4

The quotient rule states that if f and g are two functions, f” and g” are their derivatives respectively, then,

(f/g)” = (f”.g – g”.f)/g²

Rule 5

The polynomial or elementary power rule states that, if y = f(x) = xⁿ, then f” = n. x^(n-1)

A direct outcome of this rule is that the derivative of any constant is zero, i.e., if y = k, any constant, then

f” = 0

Rule 6

The chain rule states that, derivative of the function of a function h(x) = f(g(x)) with respect to x is,

h”(x)= f”(g(x)).g”(x)

Example

Create a script file and type the following code into it −

syms x
syms t

f = (x + 2)*(x^2 + 3)
der1 = diff(f)
f = (t^2 + 3)*(sqrt(t) + t^3)
der2 = diff(f)
f = (x^2 - 2*x + 1)*(3*x^3 - 5*x^2 + 2)
der3 = diff(f)
f = (2*x^2 + 3*x)/(x^3 + 1)
der4 = diff(f)
f = (x^2 + 1)^17
der5 = diff(f)
f = (t^3 + 3* t^2 + 5*t -9)^(-6)
der6 = diff(f)

When you run the file, MATLAB displays the following result −

f =
   (x^2 + 3)*(x + 2)
 
   der1 =
   2*x*(x + 2) + x^2 + 3
  
f =
   (t^(1/2) + t^3)*(t^2 + 3)
 
   der2 =
   (t^2 + 3)*(3*t^2 + 1/(2*t^(1/2))) + 2*t*(t^(1/2) + t^3)
  
f =
   (x^2 - 2*x + 1)*(3*x^3 - 5*x^2 + 2)
  
der3 =
   (2*x - 2)*(3*x^3 - 5*x^2 + 2) - (- 9*x^2 + 10*x)*(x^2 - 2*x + 1)
 
f =
   (2*x^2 + 3*x)/(x^3 + 1)
  
der4 =
   (4*x + 3)/(x^3 + 1) - (3*x^2*(2*x^2 + 3*x))/(x^3 + 1)^2
  
f =
   (x^2 + 1)^17
  
der5 =
   34*x*(x^2 + 1)^16
  
f =
   1/(t^3 + 3*t^2 + 5*t - 9)^6
  
der6 =
   -(6*(3*t^2 + 6*t + 5))/(t^3 + 3*t^2 + 5*t - 9)^7

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
t = sym("t");
f = (x + 2)*(x^2 + 3) 
der1 = differentiate(f,x) 
f = (t^2 + 3)*(t^(1/2) + t^3) 
der2 = differentiate(f,t) 
f = (x^2 - 2*x + 1)*(3*x^3 - 5*x^2 + 2) 
der3 = differentiate(f,x) 
f = (2*x^2 + 3*x)/(x^3 + 1) 
der4 = differentiate(f,x) 
f = (x^2 + 1)^17 
der5 = differentiate(f,x) 
f = (t^3 + 3* t^2 + 5*t -9)^(-6) 
der6 = differentiate(f,t)

Octave executes the code and returns the following result −

f =

(2.0+x)*(3.0+x^(2.0))
der1 =

3.0+x^(2.0)+(2.0)*(2.0+x)*x
f =

(t^(3.0)+sqrt(t))*(3.0+t^(2.0))
der2 =

(2.0)*(t^(3.0)+sqrt(t))*t+((3.0)*t^(2.0)+(0.5)*t^(-0.5))*(3.0+t^(2.0))
f =

(1.0+x^(2.0)-(2.0)*x)*(2.0-(5.0)*x^(2.0)+(3.0)*x^(3.0))
der3 =

(-2.0+(2.0)*x)*(2.0-(5.0)*x^(2.0)+(3.0)*x^(3.0))+((9.0)*x^(2.0)-(10.0)*x)*(1.0+x^(2.0)-(2.0)*x)
f =

(1.0+x^(3.0))^(-1)*((2.0)*x^(2.0)+(3.0)*x)
der4 =

(1.0+x^(3.0))^(-1)*(3.0+(4.0)*x)-(3.0)*(1.0+x^(3.0))^(-2)*x^(2.0)*((2.0)*x^(2.0)+(3.0)*x)
f =

(1.0+x^(2.0))^(17.0)
der5 =

(34.0)*(1.0+x^(2.0))^(16.0)*x
f =

(-9.0+(3.0)*t^(2.0)+t^(3.0)+(5.0)*t)^(-6.0)
der6 =

-(6.0)*(-9.0+(3.0)*t^(2.0)+t^(3.0)+(5.0)*t)^(-7.0)*(5.0+(3.0)*t^(2.0)+(6.0)*t)

Derivatives of Exponential, Logarithmic and Trigonometric Functions

The following table provides the derivatives of commonly used exponential, logarithmic and trigonometric functions −

Example

Create a script file and type the following code into it −

syms x
y = exp(x)
diff(y)
y = x^9
diff(y)
y = sin(x)
diff(y)
y = tan(x)
diff(y)
y = cos(x)
diff(y)
y = log(x)
diff(y)
y = log10(x)
diff(y)
y = sin(x)^2
diff(y)
y = cos(3*x^2 + 2*x + 1)
diff(y)
y = exp(x)/sin(x)
diff(y)

When you run the file, MATLAB displays the following result −

y =
   exp(x)
   ans =
   exp(x)

y =
   x^9
   ans =
   9*x^8
  
y =
   sin(x)
   ans =
   cos(x)
  
y =
   tan(x)
   ans =
   tan(x)^2 + 1
 
y =
   cos(x)
   ans =
   -sin(x)
  
y =
   log(x)
   ans =
   1/x
  
y =
   log(x)/log(10)
   ans =
   1/(x*log(10))
 
y =
   sin(x)^2
   ans =
   2*cos(x)*sin(x)
 
y =
   cos(3*x^2 + 2*x + 1)
   ans =
   -sin(3*x^2 + 2*x + 1)*(6*x + 2)
  
y =
   exp(x)/sin(x)
   ans =
   exp(x)/sin(x) - (exp(x)*cos(x))/sin(x)^2

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
y = Exp(x)
differentiate(y,x)

y = x^9
differentiate(y,x)

y = Sin(x)
differentiate(y,x)

y = Tan(x)
differentiate(y,x)

y = Cos(x)
differentiate(y,x)

y = Log(x)
differentiate(y,x)

% symbolic packages does not have this support
%y = Log10(x)
%differentiate(y,x)

y = Sin(x)^2
differentiate(y,x)

y = Cos(3*x^2 + 2*x + 1)
differentiate(y,x)

y = Exp(x)/Sin(x)
differentiate(y,x)

Octave executes the code and returns the following result −

y =

exp(x)
ans =

exp(x)
y =

x^(9.0)
ans =

(9.0)*x^(8.0)
y =

sin(x)
ans =

cos(x)
y =

tan(x)
ans =

1+tan(x)^2
y =

cos(x)
ans =

-sin(x)
y =

log(x)
ans =

x^(-1)
y =

sin(x)^(2.0)
ans =

(2.0)*sin(x)*cos(x)
y =

cos(1.0+(2.0)*x+(3.0)*x^(2.0))
ans =

-(2.0+(6.0)*x)*sin(1.0+(2.0)*x+(3.0)*x^(2.0))
y =

sin(x)^(-1)*exp(x)
ans =

sin(x)^(-1)*exp(x)-sin(x)^(-2)*cos(x)*exp(x)

Computing Higher Order Derivatives

To compute higher derivatives of a function f, we use the syntax diff(f,n).

Let us compute the second derivative of the function y = f(x) = x .e^-3x

f = x*exp(-3*x);
diff(f, 2)

MATLAB executes the code and returns the following result −

ans =
9*x*exp(-3*x) - 6*exp(-3*x)

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
f = x*Exp(-3*x);
differentiate(f, x, 2)

Octave executes the code and returns the following result −

ans =

(9.0)*exp(-(3.0)*x)*x-(6.0)*exp(-(3.0)*x)

Example

In this example, let us solve a problem. Given that a function y = f(x) = 3 sin(x) + 7 cos(5x). We will have to find out whether the equation f” + f = -5cos(2x) holds true.

Create a script file and type the following code into it −

syms x
y = 3*sin(x)+7*cos(5*x);  % defining the function
lhs = diff(y,2)+y;        %evaluting the lhs of the equation
rhs = -5*cos(2*x);        %rhs of the equation
if(isequal(lhs,rhs))
   disp(''Yes, the equation holds true'');
else
   disp(''No, the equation does not hold true'');
end
disp(''Value of LHS is: ''), disp(lhs);

When you run the file, it displays the following result −

No, the equation does not hold true
Value of LHS is: 
-168*cos(5*x)

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
y = 3*Sin(x)+7*Cos(5*x);           % defining the function
lhs = differentiate(y, x, 2) + y;  %evaluting the lhs of the equation
rhs = -5*Cos(2*x);                 %rhs of the equation

if(lhs == rhs)
   disp(''Yes, the equation holds true'');
else
   disp(''No, the equation does not hold true'');
end
disp(''Value of LHS is: ''), disp(lhs);

Octave executes the code and returns the following result −

No, the equation does not hold true
Value of LHS is: 
-(168.0)*cos((5.0)*x)

Finding the Maxima and Minima of a Curve

If we are searching for the local maxima and minima for a graph, we are basically looking for the highest or lowest points on the graph of the function at a particular locality, or for a particular range of values of the symbolic variable.

For a function y = f(x) the points on the graph where the graph has zero slope are called stationary points. In other words stationary points are where f”(x) = 0.

To find the stationary points of a function we differentiate, we need to set the derivative equal to zero and solve the equation.

Example

Let us find the stationary points of the function f(x) = 2x³ + 3x² − 12x + 17

Take the following steps −

First let us enter the function and plot its graph.

syms x
y = 2*x^3 + 3*x^2 - 12*x + 17;   % defining the function
ezplot(y)

MATLAB executes the code and returns the following plot −

Here is Octave equivalent code for the above example −

pkg load symbolic
symbols

x = sym(''x'');
y = inline("2*x^3 + 3*x^2 - 12*x + 17");

ezplot(y)
print -deps graph.eps

Our aim is to find some local maxima and minima on the graph, so let us find the local maxima and minima for the interval [-2, 2] on the graph.

syms x
y = 2*x^3 + 3*x^2 - 12*x + 17;   % defining the function
ezplot(y, [-2, 2])

MATLAB executes the code and returns the following plot −

Here is Octave equivalent code for the above example −

pkg load symbolic
symbols

x = sym(''x'');
y = inline("2*x^3 + 3*x^2 - 12*x + 17");

ezplot(y, [-2, 2])
print -deps graph.eps

Next, let us compute the derivative.

g = diff(y)

MATLAB executes the code and returns the following result −

g =
   6*x^2 + 6*x - 12

Here is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
y = 2*x^3 + 3*x^2 - 12*x + 17;
g = differentiate(y,x)

Octave executes the code and returns the following result −

g =
   -12.0+(6.0)*x+(6.0)*x^(2.0)

Let us solve the derivative function, g, to get the values where it becomes zero.

s = solve(g)

MATLAB executes the code and returns the following result −

s =
   1
   -2

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
y = 2*x^3 + 3*x^2 - 12*x + 17;
g = differentiate(y,x)
roots([6, 6, -12])

Octave executes the code and returns the following result −

g =

-12.0+(6.0)*x^(2.0)+(6.0)*x
ans =

  -2
   1

This agrees with our plot. So let us evaluate the function f at the critical points x = 1, -2. We can substitute a value in a symbolic function by using the subs command.

subs(y, 1), subs(y, -2)

MATLAB executes the code and returns the following result −

ans =
   10
ans =
   37

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
y = 2*x^3 + 3*x^2 - 12*x + 17;
g = differentiate(y,x)

roots([6, 6, -12])
subs(y, x, 1), subs(y, x, -2)

ans =
   10.0
ans =
   37.0-4.6734207789940138748E-18*I

Therefore, The minimum and maximum values on the function f(x) = 2x³ + 3x² − 12x + 17, in the interval [-2,2] are 10 and 37.

Solving Differential Equations

MATLAB provides the dsolve command for solving differential equations symbolically.

The most basic form of the dsolve command for finding the solution to a single equation is

dsolve(''eqn'') 

where eqn is a text string used to enter the equation.

It returns a symbolic solution with a set of arbitrary constants that MATLAB labels C1, C2, and so on.

You can also specify initial and boundary conditions for the problem, as comma-delimited list following the equation as −

dsolve(''eqn'',''cond1'', ''cond2'',…)  

For the purpose of using dsolve command, derivatives are indicated with a D. For example, an equation like f”(t) = -2*f + cost(t) is entered as −

”Df = -2*f + cos(t)”

Higher derivatives are indicated by following D by the order of the derivative.

For example the equation f”(x) + 2f”(x) = 5sin3x should be entered as −

”D2y + 2Dy = 5*sin(3*x)”

Let us take up a simple example of a first order differential equation: y” = 5y.

s = dsolve(''Dy = 5*y'')

MATLAB executes the code and returns the following result −

s =
   C2*exp(5*t)

Let us take up another example of a second order differential equation as: y” – y = 0, y(0) = -1, y”(0) = 2.

dsolve(''D2y - y = 0'',''y(0) = -1'',''Dy(0) = 2'')

MATLAB executes the code and returns the following result −

ans =
   exp(t)/2 - (3*exp(-t))/2

MATLAB – Integration

Integration deals with two essentially different types of problems.

• In the first type, derivative of a function is given and we want to find the function. Therefore, we basically reverse the process of differentiation. This reverse process is known as anti-differentiation, or finding the primitive function, or finding an indefinite integral.

• The second type of problems involve adding up a very large number of very small quantities and then taking a limit as the size of the quantities approaches zero, while the number of terms tend to infinity. This process leads to the definition of the definite integral.

Definite integrals are used for finding area, volume, center of gravity, moment of inertia, work done by a force, and in numerous other applications.

Finding Indefinite Integral Using MATLAB

By definition, if the derivative of a function f(x) is f”(x), then we say that an indefinite integral of f”(x) with respect to x is f(x). For example, since the derivative (with respect to x) of x² is 2x, we can say that an indefinite integral of 2x is x².

In symbols −

f”(x²) = 2x, therefore,

∫ 2xdx = x².

Indefinite integral is not unique, because derivative of x² + c, for any value of a constant c, will also be 2x.

This is expressed in symbols as −

∫ 2xdx = x² + c.

Where, c is called an ”arbitrary constant”.

MATLAB provides an int command for calculating integral of an expression. To derive an expression for the indefinite integral of a function, we write −

int(f);

For example, from our previous example −

syms x 
int(2*x)

MATLAB executes the above statement and returns the following result −

ans =
   x^2

Example 1

In this example, let us find the integral of some commonly used expressions. Create a script file and type the following code in it −

syms x n

int(sym(x^n))
f = ''sin(n*t)''
int(sym(f))
syms a t
int(a*cos(pi*t))
int(a^x)

When you run the file, it displays the following result −

ans =
   piecewise([n == -1, log(x)], [n ~= -1, x^(n + 1)/(n + 1)])
f =
sin(n*t)
ans =
   -cos(n*t)/n
   ans =
   (a*sin(pi*t))/pi
   ans =
   a^x/log(a)

Example 2

Create a script file and type the following code in it −

syms x n
int(cos(x))
int(exp(x))
int(log(x))
int(x^-1)
int(x^5*cos(5*x))
pretty(int(x^5*cos(5*x)))
int(x^-5)
int(sec(x)^2)
pretty(int(1 - 10*x + 9 * x^2))
int((3 + 5*x -6*x^2 - 7*x^3)/2*x^2)
pretty(int((3 + 5*x -6*x^2 - 7*x^3)/2*x^2))

Note that the pretty function returns an expression in a more readable format.

When you run the file, it displays the following result −

ans =
   sin(x)
 
ans =
   exp(x)
 
ans =
   x*(log(x) - 1)
 
ans =
   log(x)
 
ans =
(24*cos(5*x))/3125 + (24*x*sin(5*x))/625 - (12*x^2*cos(5*x))/125 + (x^4*cos(5*x))/5 - (4*x^3*sin(5*x))/25 + (x^5*sin(5*x))/5
                                    2             4 
   24 cos(5 x)   24 x sin(5 x)   12 x  cos(5 x)   x  cos(5 x) 
   ----------- + ------------- - -------------- + ------------ 
      3125            625             125              5 
   
        3             5 
 
   4 x  sin(5 x)   x  sin(5 x) 
   ------------- + ----------- 
         25              5
 
ans =
-1/(4*x^4)
 
ans =
tan(x)
        2 
  x (3 x  - 5 x + 1)
 
ans = 
- (7*x^6)/12 - (3*x^5)/5 + (5*x^4)/8 + x^3/2
 
      6      5      4    3 
    7 x    3 x    5 x    x 
  - ---- - ---- + ---- + -- 
     12     5      8     2

Finding Definite Integral Using MATLAB

By definition, definite integral is basically the limit of a sum. We use definite integrals to find areas such as the area between a curve and the x-axis and the area between two curves. Definite integrals can also be used in other situations, where the quantity required can be expressed as the limit of a sum.

The int function can be used for definite integration by passing the limits over which you want to calculate the integral.

To calculate

we write,

int(x, a, b)

For example, to calculate the value of we write −

int(x, 4, 9)

MATLAB executes the above statement and returns the following result −

ans =
   65/2

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
f = x;
c = [1, 0];
integral = polyint(c);

a = polyval(integral, 9) - polyval(integral, 4);
display(''Area: ''), disp(double(a));

Octave executes the code and returns the following result −

Area: 

   32.500

An alternative solution can be given using quad() function provided by Octave as follows −

pkg load symbolic
symbols

f = inline("x");
[a, ierror, nfneval] = quad(f, 4, 9);

display(''Area: ''), disp(double(a));

Octave executes the code and returns the following result −

Area: 
   32.500

Example 1

Let us calculate the area enclosed between the x-axis, and the curve y = x³−2x+5 and the ordinates x = 1 and x = 2.

The required area is given by −

Create a script file and type the following code −

f = x^3 - 2*x +5;
a = int(f, 1, 2)
display(''Area: ''), disp(double(a));

When you run the file, it displays the following result −

a =
23/4
Area: 
   5.7500

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
f = x^3 - 2*x +5;
c = [1, 0, -2, 5];
integral = polyint(c);

a = polyval(integral, 2) - polyval(integral, 1);
display(''Area: ''), disp(double(a));

Octave executes the code and returns the following result −

Area: 

   5.7500

An alternative solution can be given using quad() function provided by Octave as follows −

pkg load symbolic
symbols

x = sym("x");
f = inline("x^3 - 2*x +5");

[a, ierror, nfneval] = quad(f, 1, 2);
display(''Area: ''), disp(double(a));

Octave executes the code and returns the following result −

Area: 
   5.7500

Example 2

Find the area under the curve: f(x) = x² cos(x) for −4 ≤ x ≤ 9.

Create a script file and write the following code −

f = x^2*cos(x);
ezplot(f, [-4,9])
a = int(f, -4, 9)
disp(''Area: ''), disp(double(a));

When you run the file, MATLAB plots the graph −

The output is given below −

a = 
8*cos(4) + 18*cos(9) + 14*sin(4) + 79*sin(9)
 
Area: 
   0.3326

Following is Octave equivalent of the above calculation −

pkg load symbolic
symbols

x = sym("x");
f = inline("x^2*cos(x)");

ezplot(f, [-4,9])
print -deps graph.eps

[a, ierror, nfneval] = quad(f, -4, 9);
display(''Area: ''), disp(double(a));

MATLAB – Polynomials

MATLAB represents polynomials as row vectors containing coefficients ordered by descending powers. For example, the equation P(x) = x⁴ + 7x³ – 5x + 9 could be represented as −

p = [1 7 0 -5 9];

Evaluating Polynomials

The polyval function is used for evaluating a polynomial at a specified value. For example, to evaluate our previous polynomial p, at x = 4, type −

p = [1 7 0  -5 9];
polyval(p,4)

ans = 693

p = [1 7 0  -5 9];
X = [1 2 -3 4; 2 -5 6 3; 3 1 0 2; 5 -7 3 8];
polyvalm(p, X)

ans =
      2307       -1769        -939        4499
      2314       -2376        -249        4695
      2256       -1892        -549        4310
      4570       -4532       -1062        9269

p = [1 7 0  -5 9];
r = roots(p)

r =
   -6.8661 + 0.0000i
   -1.4247 + 0.0000i
   0.6454 + 0.7095i
   0.6454 - 0.7095i

p2 = poly(r)

p2 =

   Columns 1 through 3:

      1.00000 + 0.00000i   7.00000 + 0.00000i   0.00000 + 0.00000i

   Columns 4 and 5:

      -5.00000 - 0.00000i   9.00000 + 0.00000i

p = polyfit(x,y,n)

x = [1 2 3 4 5 6]; y = [5.5 43.1 128 290.7 498.4 978.67];   %data
p = polyfit(x,y,4)   %get the polynomial

% Compute the values of the polyfit estimate over a finer range, 
% and plot the estimate over the real data values for comparison:
x2 = 1:.1:6;          
y2 = polyval(p,x2);
plot(x,y,''o'',x2,y2)
grid on

p =
   4.1056  -47.9607  222.2598 -362.7453  191.1250

laplace(f(t))

syms s t a b w

laplace(a)
laplace(t^2)
laplace(t^9)
laplace(exp(-b*t))
laplace(sin(w*t))
laplace(cos(w*t))

ans =
   1/s^2

ans =
   2/s^3

ans =
   362880/s^10

ans =
   1/(b + s)
  
ans =
   w/(s^2 + w^2)
  
ans =
   s/(s^2 + w^2)

ilaplace(1/s^3)

ans =
   t^2/2

syms s t a b w

ilaplace(1/s^7)
ilaplace(2/(w+s))
ilaplace(s/(s^2+4))
ilaplace(exp(-b*t))
ilaplace(w/(s^2 + w^2))
ilaplace(s/(s^2 + w^2))

ans =
   t^6/720

ans =
   2*exp(-t*w)

ans =
   cos(2*t)

ans =
   ilaplace(exp(-b*t), t, x)

ans =
   sin(t*w)

ans =
   cos(t*w)

syms x 
f = exp(-2*x^2);     %our function
ezplot(f,[-2,2])     % plot of our function
FT = fourier(f)      % Fourier transform

FT =
   (2^(1/2)*pi^(1/2)*exp(-w^2/8))/2

ezplot(FT)

f = ifourier(-2*exp(-abs(w)))

f =
   -2/(pi*(x^2 + 1))

[x,y] = meshgrid(-2:.2:2);
g = x .* exp(-x.^2 - y.^2);
surf(x, y, g)
print -deps graph.eps

f = x^2*cos(x);
ezplot(f, [-4,9])
a = int(f, -4, 9)
disp(''Area: ''), disp(double(a));

a =
8*cos(4) + 18*cos(9) + 14*sin(4) + 79*sin(9)
 
Area: 
   0.3326

pkg load symbolic
symbols

x = sym("x");
f = inline("x^2*cos(x)");

ezplot(f, [-4,9])
print -deps graph.eps

[a, ierror, nfneval] = quad(f, -4, 9);
display(''Area: ''), disp(double(a));

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