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We were given the two functions $f_1(x) = |x|$ and $f_2(x) = x^2-2$ of which we needed to find the surface of. My solution this far is: My solution

However, I don't know which bounds I have to use for my integral. When I plotted the graph on my graphic calculator the two bounds were $-2$ and $2$, but I don't know how I could possibly find them without.

Any help is appreciated.

LLScheme
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1 Answers1

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You have already found the bounds. You have written:

  • First case, $x\ge 0$: $$x^2-x-2=0\implies\begin{cases}x=2\\x=-1\end{cases}$$ But this first case requires $x\ge 0$ (see two lines above), so the only valid solution is $x=2$.

  • Second case, $x\le 0$: $$x^2+x-2=0\implies\begin{cases}x=1\\x=-2\end{cases}$$ But this second case requires $x\le 0$ (see two lines above), so the only valid solution is $x=-2$.

There go your two bounds.

wimi
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