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Integration

This example shows how to compute definite integrals using Symbolic Math Toolbox™.

Definite Integrals

To maximize for , first, define the symbolic variables and assume that :

syms a x assume(a >= 0); 

Then, define the function to maximize:

F = int(sin(a*x)*sin(x/a),x,-a,a) 
 F = piecewise([a == 1, 1 - sin(2)/2], [a ~= 1, (2*a*(sin(a^2)*cos(1) - a^2*cos(a^2)*sin(1)))/(a^4 - 1)]) 

Note the special case here for . To make computations easier, use assumeAlso to ignore this possibility (and later check that is not the maximum):

assumeAlso(a ~= 1); F = int(sin(a*x)*sin(x/a),x,-a,a) 
 F = (2*a*(sin(a^2)*cos(1) - a^2*cos(a^2)*sin(1)))/(a^4 - 1) 

Create a plot of to check its shape:

fplot(F,[0 10]) 

Use diff to find the derivative of with respect to :

Fa = diff(F,a) 
 Fa = (2*(sin(a^2)*cos(1) - a^2*cos(a^2)*sin(1)))/(a^4 - 1) + (2*a*(2*a*cos(a^2)*cos(1) - 2*a*cos(a^2)*sin(1) + 2*a^3*sin(a^2)*sin(1)))/(a^4 - 1) - (8*a^4*(sin(a^2)*cos(1) - a^2*cos(a^2)*sin(1)))/(a^4 - 1)^2 

The zeros of are the local extrema of :

hold on fplot(Fa,[0 10]) grid on 

The maximum is between 1 and 2. Use vpasolve to find an approximation of the zero of in this interval:

a_max = vpasolve(Fa,a,[1,2]) 
 a_max = 1.5782881585233198075558845180583 

Use subs to get the maximal value of the integral:

F_max = subs(F,a,a_max) 
 F_max = 0.36730152527504169588661811770092*cos(1) + 1.2020566879911789986062956284113*sin(1) 

The result still contains exact numbers and . Use vpa to replace these by numerical approximations:

vpa(F_max) 
 ans = 1.2099496860938456039155811226054 

Check that the excluded case does not result in a larger value:

vpa(int(sin(x)*sin(x),x,-1,1)) 
 ans = 0.54535128658715915230199006704413 

Multiple Integration

Numerical integration over higher dimensional areas has special functions:

integral2(@(x,y) x.^2-y.^2,0,1,0,1) 
ans = 4.0127e-19 

There are no such special functions for higher-dimensional symbolic integration. Use nested one-dimensional integrals instead:

syms x y int(int(x^2-y^2,y,0,1),x,0,1) 
 ans = 0 

Line Integrals

Define a vector field F in 3D space:

syms x y z F(x,y,z) = [x^2*y*z, x*y, 2*y*z]; 

Next, define a curve:

syms t ux(t) = sin(t); uy(t) = t^2-t; uz(t) = t; 

The line integral of F along the curve u is defined as , where the on the right-hand-side denotes a scalar product. Use this definition to compute the line integral for from 0 to 1:

F_int = int(F(ux,uy,uz)*diff([ux;uy;uz],t),t,0,1) 
 F_int = (19*cos(1))/4 - cos(3)/108 - 12*sin(1) + sin(3)/27 + 395/54 

Get a numerical approximation of this exact result:

vpa(F_int) 
 ans = -0.20200778585035447453044423341349