# Why is the area between these two graphs positive?

Why is the area between these two functions positive? \begin{align*} f(x) &= (x-1)^3\\ g(x) &= (x-1). \end{align*}

The area between $f(x)$ and $g(x)$ is in two parts.

The first part is in the 4th quadrant with lower limit $0$ and upper limit $1$, $f(x)$ is the upper function.

The second part is in the 1st quadrant with lower limit 1 and upper limit 2, $g(x)$ is the upper function.

When I compute the areas individually I get two equal positive numbers the sum of which is 0.5

Shouldn't the area in the fourth quadrant be negative? Or am I missing something?

I've been trying to understand why area in the fourth quadrant is positive.

Thank you.

-
An area is by definition a positive number... Ok, it could be zero. – Fabian Mar 7 '11 at 22:30

Here is a graph of the two functions with f in pink and g in blue.

By symmetry, the area of each of the two enclosed regions is the same—area, here, meaning the geometric concept which is nonnegative (or perhaps positive), as opposed to signed area. In terms of signed area, one might think of the two pieces as having opposite sign because f is above g for oen piece and g is above f for the other piece. However, typically, a problem asking for the area between curves or the area enclosed by one or more curves is asking for geometric area. The typical method of solution in that instance is to consider each piece separately, integrating (top function) – (bottom function) for each piece, to guarantee a positive (nonnegative) result.

-
There was a typo in the question, f[x] should have been equal to (x-1)^3. Thanks for pointing that out. – user7931 Mar 7 '11 at 22:31
@user7931: Ahh, okay, I've edited my answer. – Isaac Mar 7 '11 at 22:38

"Area between two graphs" is, by definition, positive regardless of where in the plane it lies. This is different from the concept of "net signed area between the graph and the $x$-axis", which is what the regular Riemann integral measures.

This is much like we say that the area of the unit circle is $2\pi$, regardless of where in the plane we "draw" it. (If you were considering the net signed area inside the unit circle centered at the origin, then this would be $0$, since we count the area below the $X$-axis as "negative" and the area over the $X$-axis as "positive", but the net signed area is not the same as the area.)

So here, you are being asked to measure the area, in the usual sense of the word, of the region enclosed by the two graphs. This means integrating the absolute value of their difference (which automatically takes care of converting "net signed area" into regular area), which in turn is done by, as you do, by breaking it up to figure out where $|f-g|$ equals $f-g$ and where it equals $g-f$. But in any case, the "area enclosed" refers to the actual area enclosed, not the "net signed area enclosed".

-

it depends on how you're being asked to measure it, i.e. whether you consider $$\int_0^2(f-g), \int_0^2(g-f), \text{ or } \int_0^2|f-g|$$ or if you want the area totally enclosed (as in another answer) then you only integrate from 1 to 2 as in the graph pictured (as in another answer).

-