Integral Expressions

1. Feb 26, 2009

Mentallic

$$\int_{-1}^{1} x(x-1)(x+1)dx$$

The problem is that the question doesn't say anything about area, and we have been notified in class that unless otherwise stated, don't express the answer in sq. units.
However, I also have a problem that, if this were to find the area on a graph, I would notice the symmetry about the origin and y-axis and thus express the integral as:

$$-2 \int_{0}^{1} x(x-1)(x+1)dx$$

Now, the first integral (if not area) is equal to 0, while the second is not. Which would be the correct answer?

2. Feb 26, 2009

CompuChip

They do not give the same answer, because they are not equivalent.
You said that there is a symmetry around the origin and y-axis, but note that it is actually an anti-symmetry. That is, if f(x) = x(x - 1)(x + 1), then f(-x) = -f(x) and not f(-x) = f(x). You can note this by sketching the graph, or noticing that (x - 1)(x + 1) is symmetric and x is anti-symmetric, so their product is anti-symmetric.
So you are arguing
$$\int_{-1}^{1} f(x)dx=\int_{-1}^{0} f(x) dx+\int_{0}^{1} f(x) dx$$
If you substitute x to -x in the first integral they don't add up, they cancel out.

3. Feb 26, 2009

Mentallic

Thats on behalf of my poor attempt at putting my thoughts into words.
By symmetry, I meant that they are odd functions, as you stated.

I see you are heading into the direction of making it equal zero. If the question were asked slightly different. Say, rather than 'evaluate' the question asked 'find the area of', I would instead convert the integral so as not to get a value of zero.

Basically my question is, if the question doesn't mention area, am I safe to assume this integral equals zero? or would I have to find it as if it were finding area/volume etc. or would I have to find both?

p.s. my teacher is really picky on the details and will try oppose any 100% marks as I've noticed from a previous assessment task.