Recall that the area below the graph of the function f (x) from a to b is the definite
integral
f (x)dx 

where area counts as negative when f (x) < 0. If the function f (x) takes on both
positive and negative values in the interval [a, b], and we want to compute the total area
counting all areas as positive, we need to refine our method. The correct thing to do is to
break the integral up into several integrals corresponding to the parts of the interval on
which the function is positive and those on which it is negative.
For example, let us calculate the area between the graph of f (x) = sin(x) and the xaxis
from 0 to 2Π. If we were simply to compute the integral
sin(x)dx 

we would obtain 0, because the areas above and below the xaxis exactly cancel each
other out weighted with opposite signs. Instead, we must take the integral of the absolute
value of f, splitting it into two separate integrals in order to evaluate it:
 sin(x) dx  =   sin(x) dx +  sin(x) dx 

 =  sin(x)dx +  sin(x)dx 

 =  cos(x)_{0}^{Π} + cos(x)_{Π}^{2Π} 

 =  (1 + 1) + (1 + 1) 

 =  4 

Alternately, we could have noted from the symmetry of the graph of sin(x) that it is
enough to calculate the area below the graph from 0 to Π and double it.
Integrals also enable us to calculate the area between the graphs of two functions (up to
this point, the second function has always been f (x) = 0, with graph equal to the x
axis). For this, we note that the area between the graphs of two functions f and g is
the difference of the area between the graph of f and the xaxis and the area between
the graph of g and the xaxis. Hence the area between the graphs of f and g from a to b is given by:
f (x)dx  g(x)dx = f (x)  g(x)dx 

where the area is counted as positive when f (x) > g(x) and as negative when f (x) < g(x).