jennychan12
  • jennychan12
piecewise function. integrals. see below.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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jennychan12
  • jennychan12
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?
anonymous
  • anonymous
Riemann integrals are not affected by singular points. So you can do the integral as though the point x=0 doesn't exist.
anonymous
  • anonymous
Plus, the point x=0 doesn't even come up in the integral, since you're taking the integral from 1 to 4.

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anonymous
  • anonymous
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)
jennychan12
  • jennychan12
so basically \[\int\limits_{1}^{4} \frac{ x^2 }{ \left| x \right| } dx\] ?
anonymous
  • anonymous
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|
jennychan12
  • jennychan12
but what about the absolute value?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 \[ x = | x | \]
anonymous
  • anonymous
And so where you see \[ |x | \] you can just write \[ x \]
jennychan12
  • jennychan12
wait. the abs. value doesnt matter cuz it's always positive from 1-4 right?
anonymous
  • anonymous
that's right.
jennychan12
  • jennychan12
wait. typo....
jennychan12
  • jennychan12
\[\int\limits_{-4}^{2} f(x)dx\]
jennychan12
  • jennychan12
wait. but i think it's the same concept
anonymous
  • anonymous
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: \[\int\limits_{-4}^{2}f(x) dx = \int\limits_{-4}^{0} f(x) dx + \int\limits_{0}^{2} f(x) dx\]
jennychan12
  • jennychan12
but how would you integrate that? just assume it's x^2/x ??
anonymous
  • anonymous
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 \[ |x| = x \]
jennychan12
  • jennychan12
why from 0 to 2 ?
jennychan12
  • jennychan12
oh wait nvm i got it
anonymous
  • anonymous
|dw:1355978372554:dw|
jennychan12
  • jennychan12
i got -6
anonymous
  • anonymous
I got 10. I might have been a little rough in calculation though.
anonymous
  • anonymous
wolfram alpha confirms an answer of 10.
jennychan12
  • jennychan12
stupid negative sign... either way, the answer's 2
jennychan12
  • jennychan12
ohh can i see the wolfram alpha one? do u have a link?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=integral+from+x+%3D+-4+to+x%3D2+of+x^2+%2F+abs%28x%29+dx
jennychan12
  • jennychan12
thanks.
anonymous
  • anonymous
\[\int\limits_{-4}^{2} \frac{x^2}{|x|} dx = \int\limits_{-4}^{0} \frac{x^2}{|x|} dx + \int\limits_{0}^{2} \frac{x^2}{|x|} dx\]
jennychan12
  • jennychan12
oh just that ok then yeah i understand
anonymous
  • anonymous
\[= \int\limits_{-4}^{0} - x dx + \int\limits_{0}^{2} x dx\]
jennychan12
  • jennychan12
but what about when x = 0?
anonymous
  • anonymous
Oh yeah just use limits like the other person did. And it'll come out to what I wrote up there.
jennychan12
  • jennychan12
ok then thanks.
anonymous
  • anonymous
\[\lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2} {|x|}dx = \lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2}{-x} dx\]
anonymous
  • anonymous
You'd probably write it like that if you wanted to be heaps formal.
jennychan12
  • jennychan12
just kidding my teacher printed all the answers wrong -_-
anonymous
  • anonymous
huh? Kidding about what?

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