# Math Help - definite integration

1. ## definite integration

keep finding that my answers to the area between equation and curve are "correct", only wrong sign. Happens too frequently to be coincidence. Any help?

2. Could you provide a specific example? Area must be a positive quantity - always.

3. So if my answer is negative I simply reverse the sign?

4. Well, it might be more complicated than that. Let me give you an example, and see how you do. Find the area enclosed between the two curves $x^{2}$ and $2-x^{2}$ from $x=-2$ to $x=2.$

Here's a plot (which, incidentally, I would always start out by doing. It gives a great overview of the kind of thing you need to do.)

How would you start?

[EDIT]: Fixed link. Thanks to NOX Andrew.

5. Your link is broken for me. In case it is also broken for hunt798, here is the link to the plot again: http://www.wolframalpha.com/input/?i=Plot[{x^2,2-x^2},{x,-2,2}]

6. Sounds to me like you are mixing up which curve is above the other.

For example, if I were asked to determine the area between the curves $y= x^2$ and $y= x+ 2$, the first thing I would is draw (or at least visualize) the graphs, noting that they intersect when $y= x^2= x+ 2$ so that $x^2- x- 2= (x+1)(x-2)= 0$. That is they intersect at (-1, 1) and (2, 4). I would also recognize that the graph of $y= x^2$ is always below the graph of y= x+ 2 so the area between them is given by
$\int_{-1}^2 ((x+ 2)- x^2)dx= \frac{1}{2}$
Had I mistakenly written the integral as
$\int_{-1}^2 (x^2- (x+2))dx$
(perhaps just because I am used to writing polynomials with highest power first!)
I would get $-\frac{1}{2}$.