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jennychan12

piecewise function. integrals. see below.

  • one year ago
  • one year ago

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  1. jennychan12
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    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?

    • one year ago
  2. scarydoor
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    Riemann integrals are not affected by singular points. So you can do the integral as though the point x=0 doesn't exist.

    • one year ago
  3. scarydoor
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    Plus, the point x=0 doesn't even come up in the integral, since you're taking the integral from 1 to 4.

    • one year ago
  4. scarydoor
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    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)

    • one year ago
  5. jennychan12
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    so basically \[\int\limits_{1}^{4} \frac{ x^2 }{ \left| x \right| } dx\] ?

    • one year ago
  6. scarydoor
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    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|

    • one year ago
  7. jennychan12
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    but what about the absolute value?

    • one year ago
  8. scarydoor
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    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.

    • one year ago
  9. scarydoor
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    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 \[ x = | x | \]

    • one year ago
  10. scarydoor
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    And so where you see \[ |x | \] you can just write \[ x \]

    • one year ago
  11. jennychan12
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    wait. the abs. value doesnt matter cuz it's always positive from 1-4 right?

    • one year ago
  12. scarydoor
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    that's right.

    • one year ago
  13. jennychan12
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    wait. typo....

    • one year ago
  14. jennychan12
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    \[\int\limits_{-4}^{2} f(x)dx\]

    • one year ago
  15. jennychan12
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    wait. but i think it's the same concept

    • one year ago
  16. scarydoor
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    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: \[\int\limits_{-4}^{2}f(x) dx = \int\limits_{-4}^{0} f(x) dx + \int\limits_{0}^{2} f(x) dx\]

    • one year ago
  17. jennychan12
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    but how would you integrate that? just assume it's x^2/x ??

    • one year ago
  18. scarydoor
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    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 \[ |x| = x \]

    • one year ago
  19. jennychan12
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    why from 0 to 2 ?

    • one year ago
  20. jennychan12
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    oh wait nvm i got it

    • one year ago
  21. cinar
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    |dw:1355978372554:dw|

    • one year ago
  22. jennychan12
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    i got -6

    • one year ago
  23. scarydoor
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    I got 10. I might have been a little rough in calculation though.

    • one year ago
  24. scarydoor
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    wolfram alpha confirms an answer of 10.

    • one year ago
  25. jennychan12
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    stupid negative sign... either way, the answer's 2

    • one year ago
  26. jennychan12
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    ohh can i see the wolfram alpha one? do u have a link?

    • one year ago
  27. scarydoor
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    http://www.wolframalpha.com/input/?i=integral+from+x+%3D+-4+to+x%3D2+of+x^2+%2F+abs%28x%29+dx

    • one year ago
  28. jennychan12
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    thanks.

    • one year ago
  29. scarydoor
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    \[\int\limits_{-4}^{2} \frac{x^2}{|x|} dx = \int\limits_{-4}^{0} \frac{x^2}{|x|} dx + \int\limits_{0}^{2} \frac{x^2}{|x|} dx\]

    • one year ago
  30. jennychan12
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    oh just that ok then yeah i understand

    • one year ago
  31. scarydoor
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    \[= \int\limits_{-4}^{0} - x dx + \int\limits_{0}^{2} x dx\]

    • one year ago
  32. jennychan12
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    but what about when x = 0?

    • one year ago
  33. scarydoor
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    Oh yeah just use limits like the other person did. And it'll come out to what I wrote up there.

    • one year ago
  34. jennychan12
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    ok then thanks.

    • one year ago
  35. scarydoor
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    \[\lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2} {|x|}dx = \lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2}{-x} dx\]

    • one year ago
  36. scarydoor
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    You'd probably write it like that if you wanted to be heaps formal.

    • one year ago
  37. jennychan12
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    just kidding my teacher printed all the answers wrong -_-

    • one year ago
  38. scarydoor
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    huh? Kidding about what?

    • one year ago
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