One of my last assignments for my first semester of calculus is a list of area, volume, and work problems using integrals, I worked one out, but can't find the answer anywhere, and just want to make sure I'm correct, here's my work:
Find the area of the region bounded by the curves $y=x^2$ and $y=x^4$
$x^4=x^2$
$x^4-x^2=0$
$x^2(x^2-1)=0$
$x=1, -1$
$\int_1^1(x^4-x^2)dx$
$\left[\frac{x^5}{5}-\frac{x^3}{3}\right]$ from $-1$ to $1$
$\left[\frac{(1)^5}{5}-\frac{(1)^3}{3}\right] – \left[\frac{(-1)^5}{5}-\frac{(-1)^3}{3}\right]$
$\left[\frac{1}{5}-\frac{1}{3}\right] – \left[\frac{-1}{5}+\frac{1^3}{3}\right]$
$\left[\frac{3-5}{15}-\frac{-3+5}{15}\right]$
$\left[\frac{-2}{15}+\frac{-2}{15}\right] = \frac{-4}{15}$
I double checked the definite integral online and it got $\frac{-4}{15}$ as well, so I guess that means I"m correct I'm just confused as to how an area can be negative unless I'm missing something obvious.
My prof also left a note that says "*Make sure you have the entire region", where "entire" is underlined, is there more to this problem than what I've done?
Best Answer
I think you do indeed have the entire region. The only problem is that your integral should have been
$$\int_{-1}^1(x^2-x^4)dx$$
instead, because $x^2≥x^4$ in the region where $-1<x<1$.