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# Documentation

### Contents

• Functions
• Examples and How To

## Integration

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) or int(x^n,x)

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

int(sin(2*x), 0, pi/2) or int(sin(2*x), x, 0, pi/2)

g  =  cos(at + b)

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

g = cos(a*t + b) int(g) or int(g, t)

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

int(besselj(1, z)) or int(besselj(1, z), 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

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

int(f, a, b)

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

int(f, v, a, b)

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```

### Integration with Real Parameters

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)```

### Integration with Complex Parameters

To calculate the integral

$\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{{a}^{2}+{x}^{2}}\text{ }dx$

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(i/a))/a```

The function signIm is defined as:

To evaluate F at a = 1 + i, enter

`g = subs(F, 1 + i)`
```g =
pi*(1/2 - i/2)```
`double(g)`
```ans =
1.5708 - 1.5708i```