Suppose we have a function, like this one:

x = chebfun('x',[0 10]); f = round(2*cos(x)); plot(f), ylim(2.5*[-1 1])

The Chebfun command `sum`

returns the definite integral over the prescribed interval, which is just a number:

format long, sum(f)

ans = -1.150444078461245

You can also calculate the definite interval over a subinterval by giving two additional arguments, like this:

sum(f,3,4)

ans = -1.864326901403210

To compute an indefinite integral, use the Chebfun command `cumsum`

. This returns a chebfun defined over the given interval:

g = cumsum(f); plot(g,'m')

Thus another way to compute the integral over a subinterval would be to take the difference of two values of the cumsum:

g(4) - g(3)

ans = -1.864326901403210

As always in calculus, when working with indefinite integrals you must be careful to remember the arbitrary constant that may be added. Thus for example, if you integrate $f$ and then differentiate it, you get $f$ back again:

norm( diff(cumsum(f)) - f )

ans = 0

If you differentiate $f$ and then integrate it, on the other hand, you get something different:

norm( cumsum(diff(f)) - f )

ans = 6.324555320336759

Plotting the two instantly alerts us that we forgot to add back in the value at the left endpoint, namely $f(0) = 2$:

plot(f,'b',cumsum(diff(f)),'r')

Sure enough, adding this number makes the two functions agree:

norm( f(0)+cumsum(diff(f)) - f)

ans = 0