Thread: 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.

MHF.pdf

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}

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})

will that solve your problem?

Thank you very much TheEmptySet. I've found the answer which is 8. Hopefully I'm right. I'm gonna make a scan and post it:



I apologize for the smudges you see. Those are from the previous page (I write very deep) =/