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jennychan12
piecewise function. integrals. see below.
the piecewise function f(x) = \[\frac{ x^2 }{ \left| x \right| }, x \ne 0\] \[0, x=0\] find \[\int\limits_{1}^{4} f(x)dx\] how would u do that?
Riemann integrals are not affected by singular points. So you can do the integral as though the point x=0 doesn't exist.
Plus, the point x=0 doesn't even come up in the integral, since you're taking the integral from 1 to 4.
and so then you know also that x > 0, so on the integral (1,4) f(x) = x^2 / x. (explicitly: you can take away the absolute value sign)
so basically \[\int\limits_{1}^{4} \frac{ x^2 }{ \left| x \right| } dx\] ?
Yeah that's right. This is intuitively why singular points don't really matter: In this picture, there's a singular point where the function is different at just one point. But, you remember when you do the Riemann approximation of the integral (the area under the curve) how you multiply the value of the function by the width? Well Singular points basically have no width... so the area underneath them, is effectively nothing. So you can ignore them. |dw:1355969105016:dw|
but what about the absolute value?
What you can do for the absolute value, at least in this case, is figure out whether x, over the interval of integration, is positive or negative, and then use that knowledge to remove the absolute value sign.
For example, you are integrating from x=1 to x=4. You know that x is positive for all of those values. So you know that across that whole integral \[ x = | x | \]
And so where you see \[ |x | \] you can just write \[ x \]
wait. the abs. value doesnt matter cuz it's always positive from 1-4 right?
\[\int\limits_{-4}^{2} f(x)dx\]
wait. but i think it's the same concept
Ah. okay. Well, you have something like this: |dw:1355970713671:dw| Where the integral is summing up that "shaded" area. So you can write it like this: \[\int\limits_{-4}^{2}f(x) dx = \int\limits_{-4}^{0} f(x) dx + \int\limits_{0}^{2} f(x) dx\]
but how would you integrate that? just assume it's x^2/x ??
That, in effect, is just breaking up the integral into two more. Now you know that from x=-4 to x=0 that x <= 0, so that \[|x| = -x \]. And on x=0 to x=2 \[ |x| = x \]
oh wait nvm i got it
I got 10. I might have been a little rough in calculation though.
wolfram alpha confirms an answer of 10.
stupid negative sign... either way, the answer's 2
ohh can i see the wolfram alpha one? do u have a link?
http://www.wolframalpha.com/input/?i=integral+from+x+%3D+-4+to+x%3D2+of+x^2+%2F+abs%28x%29+dx
\[\int\limits_{-4}^{2} \frac{x^2}{|x|} dx = \int\limits_{-4}^{0} \frac{x^2}{|x|} dx + \int\limits_{0}^{2} \frac{x^2}{|x|} dx\]
oh just that ok then yeah i understand
\[= \int\limits_{-4}^{0} - x dx + \int\limits_{0}^{2} x dx\]
but what about when x = 0?
Oh yeah just use limits like the other person did. And it'll come out to what I wrote up there.
\[\lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2} {|x|}dx = \lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2}{-x} dx\]
You'd probably write it like that if you wanted to be heaps formal.
just kidding my teacher printed all the answers wrong -_-
huh? Kidding about what?