# Area of region enclosed by lines and curves?

• Apr 25th 2011, 10:55 AM
Attenel
Area of region enclosed by lines and curves?
Please help with this math problem; I’ve been working on it for weeks (literally). How do I solve it?

Find the area of the regions enclosed by the lives and curves.

14. y = x^4 – 4x^2 + 4 and y = x^2

I've begun to work it out somewhat:
∫_(-1)^(-2) [x^2-(x^k-4x^2+4) ]ⅆx + ∫_1^2 [x_^4-4x^2+4-(x^2 ) ]ⅆx

The rest I've made effort to work is too long to post; I got 0 as an answer which I'm sure isn't correct.
• Apr 25th 2011, 11:51 AM
TheEmptySet
Attachment 21478

I could not get my Latex to compile so here it is. It is also in the .pdf

Code:

\documentclass{amsart} \usepackage{amsmath}% \usepackage{amsfonts}% \usepackage{amssymb}% \usepackage{graphicx} \begin{document} \begin{equation*} \int_{-2}^{-1} {x}^{2}-({x}^{4}-4{x}^{2}+4)dx + \int_{-1}^{1} ({x}^{4}-4{x}^{2}+4)-{x}^{2}dx + \int_{1}^{2} {x}^{2}-({x}^{4}-4{x}^{2}+4)dx \end{equation*} Or by using symmetry this gives (Both functions are even) \begin{equation*} 2 \left[ \int_{0}^{1} ({x}^{4}-4{x}^{2}+4)-{x}^{2}dx + \int_{1}^{2} {x}^{2}-({x}^{4}-4{x}^{2}+4)dx \right]\end{equation*} From here just simplify and use the power rule. \begin{figure}[h]         \centering                 \includegraphics[width=0.90\textwidth]{plot.jpg} \end{figure} \end{document}
• Apr 25th 2011, 11:51 AM
SpringFan25
Edit: didn't see earlier post
y=x^4 -4x^2
which factorises to
y=x^2(x - 2)(x + 2)

Find the intersection points:
x^2(x - 2)(x + 2)=x^2

[first solution: x=0]

(x - 2)(x + 2)=1

this is a quadratic. solve to obtain x=+/- sqrt{5}

http://www.wolframalpha.com/input/?i...x^2%2C+y%3Dx^2

This means the limits for your integration should be:
(-sqrt{5}, 0)

(0, sqrt{5})