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jennychan12

  • 2 years ago

piecewise function. integrals. see below.

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  1. jennychan12
    • 2 years ago
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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?

  2. scarydoor
    • 2 years ago
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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.

  3. scarydoor
    • 2 years ago
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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.

  4. scarydoor
    • 2 years ago
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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)

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

  6. scarydoor
    • 2 years ago
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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|

  7. jennychan12
    • 2 years ago
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    but what about the absolute value?

  8. scarydoor
    • 2 years ago
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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.

  9. scarydoor
    • 2 years ago
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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 \[ x = | x | \]

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

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

  12. scarydoor
    • 2 years ago
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    that's right.

  13. jennychan12
    • 2 years ago
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    wait. typo....

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

  15. jennychan12
    • 2 years ago
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    wait. but i think it's the same concept

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

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

  18. scarydoor
    • 2 years ago
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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 \[ |x| = x \]

  19. jennychan12
    • 2 years ago
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    why from 0 to 2 ?

  20. jennychan12
    • 2 years ago
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    oh wait nvm i got it

  21. cinar
    • 2 years ago
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    |dw:1355978372554:dw|

  22. jennychan12
    • 2 years ago
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    i got -6

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

  24. scarydoor
    • 2 years ago
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    wolfram alpha confirms an answer of 10.

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

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

  27. scarydoor
    • 2 years ago
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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

  28. jennychan12
    • 2 years ago
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    thanks.

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

  30. jennychan12
    • 2 years ago
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    oh just that ok then yeah i understand

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

  32. jennychan12
    • 2 years ago
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    but what about when x = 0?

  33. scarydoor
    • 2 years ago
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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.

  34. jennychan12
    • 2 years ago
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    ok then thanks.

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

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

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

  38. scarydoor
    • 2 years ago
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    huh? Kidding about what?

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