If f
is a symbolic expression, then
int(f)
attempts to find another symbolic expression, F
,
so that diff(F)
=
f
. That is, int(f)
returns
the indefinite integral or antiderivative of f
(provided
one exists in closed form). Similar to differentiation,
int(f,v)
uses the symbolic object v
as the variable
of integration, rather than the variable determined by symvar
.
See how int
works by looking at this table.
Mathematical Operation  MATLAB^{®} Command 

$$\int {x}^{n}}dx=\{\begin{array}{ll}\mathrm{log}(x)\hfill & \text{if}n=1\hfill \\ \frac{{x}^{n+1}}{n+1}\hfill & \text{otherwise}\text{.}\hfill \end{array$$ 

$$\underset{0}{\overset{\pi /2}{\int}}\mathrm{sin}(2x)dx=1$$ 

g = cos(at + b) $$\int g(t)dt=\mathrm{sin}(at+b)/a$$ 

$$\int {J}_{1}(z)dz={J}_{0}(z)$$ 

In contrast to differentiation, symbolic integration is a more complicated task. A number of difficulties can arise in computing the integral:
The antiderivative, F
, may not
exist in closed form.
The antiderivative may define an unfamiliar function.
The antiderivative may exist, but the software can't find it.
The software could find the antiderivative on a larger computer, but runs out of time or memory on the available machine.
Nevertheless, in many cases, MATLAB can perform symbolic integration successfully. For example, create the symbolic variables
syms a b theta x y n u z
The following table illustrates integration of expressions containing those variables.
f  int(f) 

syms x n f = x^n;  int(f) ans = piecewise([n == 1, log(x)],... [n ~= 1, x^(n + 1)/(n + 1)]) 
syms y f = y^(1);  int(f) ans = log(y) 
syms x n f = n^x;  int(f) ans = n^x/log(n) 
syms a b theta f = sin(a*theta+b);  int(f) ans = cos(b + a*theta)/a 
syms u f = 1/(1+u^2);  int(f) ans = atan(u) 
syms x f = exp(x^2);  int(f) ans = (pi^(1/2)*erf(x))/2 
In the last example, exp(x^2)
, there is
no formula for the integral involving standard calculus expressions,
such as trigonometric and exponential functions. In this case, MATLAB returns
an answer in terms of the error function erf
.
If MATLAB is unable to find an answer to the integral of
a function f
, it just returns int(f)
.
Definite integration is also possible.
Definite Integral  Command 

$${\int}_{a}^{b}f(x)dx$$ 

$${\int}_{a}^{b}f(v)dv$$ 

Here are some additional examples.
f  a, b  int(f, a, b) 

syms x f = x^7;  a = 0; b = 1;  int(f, a, b) ans = 1/8 
syms x f = 1/x;  a = 1; b = 2;  int(f, a, b) ans = log(2) 
syms x f = log(x)*sqrt(x);  a = 0; b = 1;  int(f, a, b) ans = 4/9 
syms x f = exp(x^2);  a = 0; b = inf;  int(f, a, b) ans = pi^(1/2)/2 
syms z f = besselj(1,z)^2;  a = 0; b = 1;  int(f, a, b) ans = hypergeom([3/2, 3/2],... [2, 5/2, 3], 1)/12 
For the Bessel function (besselj
)
example, it is possible to compute a numerical approximation to the
value of the integral, using the double
function.
The commands
syms z a = int(besselj(1,z)^2,0,1)
return
a = hypergeom([3/2, 3/2], [2, 5/2, 3], 1)/12
and the command
a = double(a)
returns
a = 0.0717
One of the subtleties involved in symbolic integration is the "value" of various parameters. For example, if a is any positive real number, the expression
$${e}^{a{x}^{2}}$$
is the positive, bell shaped curve that tends to 0 as x tends to ±∞. You can create an example of this curve, for a = 1/2, using the following commands:
syms x
a = sym(1/2);
f = exp(a*x^2);
ezplot(f)
However, if you try to calculate the integral
$$\underset{\infty}{\overset{\infty}{\int}}{e}^{a{x}^{2}}}dx$$
without assigning a value to a, MATLAB assumes
that a represents a complex number, and therefore
returns a piecewise answer that depends on the argument of a.
If you are only interested in the case when a is
a positive real number, use assume
to set an assumption
on a
:
syms a assume(a > 0)
Now you can calculate the preceding integral using the commands
syms x f = exp(a*x^2); int(f, x, inf, inf)
This returns
ans = pi^(1/2)/a^(1/2)
To calculate the integral
$$\underset{\infty}{\overset{\infty}{\int}}\frac{1}{{a}^{2}+{x}^{2}}\text{\hspace{0.05em}}dx$$
for complex values of a
, enter
syms a x clear f = 1/(a^2 + x^2); F = int(f, x, inf, inf)
syms
is used with the clear
option
to clear the all assumptions on a
. For more information
about symbolic variables and assumptions on them, see Delete Symbolic Objects and Their Assumptions.
The preceding commands produce the complex output
F = (pi*signIm(1i/a))/a
The function signIm
is defined as:
$$\text{signIm}(z)=\{\begin{array}{ll}1\hfill & \text{if}\mathrm{Im}(z)0,\text{or}\mathrm{Im}(z)=0\text{and}z0\hfill \\ 0\hfill & \text{if}z=0\hfill \\ \text{1}\hfill & \text{otherwise}\text{.}\hfill \end{array}$$
To evaluate F
at a = 1 + i
,
enter
g = subs(F, 1 + i)
g = pi*(1/2  1i/2)
double(g)
ans = 1.5708  1.5708i