If you found C to be 2/5, then the integral should bePhox said:Homework Statement
f(x) = C|x-2| for 0 <= x <= 3
f(x) = 0 otherwise
Homework Equations
The Attempt at a Solution
Solved for C, found it to be (2/5).
So.. I'm confused how to set up my integral here. I tried integral(2/3(x-2)dx) from m to 3 = 1/2. That didn't yield the correct result.
SammyS said:If you found C to be 2/5, then the integral should be[itex]\displaystyle \int_{M}^{3}\frac{2}{5}|x-2|\,dx \ .[/itex]
Set that equal to 1/2 and solve for M, although it might be easier to integrate from 0 to m.
Graph the integrand to see why.
Did you graph the function?Phox said:Yes, actually that's what I meant to say. I don't understand how I can integrate the function having the absolute value there. I understand how to do it if it were say.. integrating from 0 to 3. But since we don't know what m is I don't understand.
And neither does wolfram alpha lol
Phox said:Yes, and I understand that. What I don't understand is where m lies. Is it in the range of <2 or >2
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