Single vs. Double Integration for Area Bounded by y=x and y=x^2-2x

[kasse]

If I want to compute the area of the region bounded by the graphs y=x and y=x^2-2x, can I simply compute the integral of (x-(x^2-2x)) dx from x=0 to x=3, or do I have to use double integration?

In any case, why?

 

[christianjb]

For two curves, f(x) and g(x) the area of the element between x and x+dx is simply |f(x)-g(x)|dx. The magnitude sign is there only if you don't want the area to go negative when the curves cross.

So the total area = Int |f(x)-g(x)|dx

 

[neutrino]

"Ordinary" integration will do.

 

[HallsofIvy]

I'm a bit confused by the question. Since you know about "double integration", you must have been studying calculus for some time- and should have learned to find the "area between two curves" quite a while ago!

Here, y= x is above y= x²- 2x for x between 0 and 3 so the area is given by
\int_0^3 (x- (x^2- 2x))dx= \int_0^3 (3x- x^2)dx

Of course, you could use "double integration" with '1' as integrand:
\int_{x=0}^3 \int_{y= x^2- 2x}^x dy dx
but after the first integration, it reduces to the single integral form.