The man said curves! Assuming he is asking about the area of the region formed by three general curves, he will need to use caluculus.
Exactly how that is done depends on the curves themselves. In the very common situation, a sort of "curvy" triangle, where you have one curve under the other two (between the points where the other two intersect it), then you don't need a double integral. You will need to break the integral into two parts. I'm going to call the curve on the bottom C1, the graph of y= f1(x), and the other two C1 and C2, graphs of y=f2(x), y= f3(x) respectively. Let's say that C2 intersect C1 at x=a, C3 intersects C1 at x= c, and that C2 is below C3 until they intersect at x= b after which C3 is below C2.
Then the area is given by two separate integrals:
[tex]\int_a^b(f2(x)-f1(x))dx+ \int_b^c(f3(x)-f1(x)dx[/tex]