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anonymous

  • one year ago

Calculate the area bounded:

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  1. anonymous
    • one year ago
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  2. freckles
    • one year ago
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    A little drawing always helps to setup the integral.

  3. freckles
    • one year ago
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    \[\int\limits_c^d (\text{ \right relation } - \text{ \left relation } ) dy\]

  4. anonymous
    • one year ago
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    I made the drawing. Cant identify which are is bounded thoough

  5. anonymous
    • one year ago
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    |dw:1441747573378:dw|

  6. freckles
    • one year ago
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    where is the other part of your parabola

  7. anonymous
    • one year ago
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    Dont you square the "x" to make it y=\[\sqrt{x}\]

  8. freckles
    • one year ago
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    |dw:1441747647427:dw|

  9. freckles
    • one year ago
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    |dw:1441747703759:dw|

  10. freckles
    • one year ago
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    no y^2=x doesn't mean just y=sqrt(x) \[x=y^2 \text{ is } \sqrt{x}=y \text{ if } y>0 \\ x=y^2 \text{ is } -\sqrt{x}=y \text{ if } y<0\]

  11. freckles
    • one year ago
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    in other words y^2=x is a parabola

  12. freckles
    • one year ago
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    anyways you have the right relation is given by x=y^2 and the left relation is given by x=y-5 and I think it should be clear what the limits are

  13. anonymous
    • one year ago
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    Aaaaah, I see. I missed the other half.

  14. anonymous
    • one year ago
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    Am in integrating in respect to Y?

  15. freckles
    • one year ago
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    yep because we are looking at right and left function

  16. Jhannybean
    • one year ago
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    Think about it, if we turned you sideways and half your body was missing and someone asked, "what would he look like if he was turned right side up?" and we responded "only what you can see", they wouldnt think of you as a complete person would they? xD

  17. anonymous
    • one year ago
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    So limits will be 2 and -1?

  18. Jhannybean
    • one year ago
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    Therefore in that respect, \(y^2 = x \iff y = \pm \sqrt{x}\) haha. sorry if that interrupted you.....

  19. anonymous
    • one year ago
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    Lol ,Jhannybean. You're good.

  20. freckles
    • one year ago
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    yes the limits are y=-1 to y=2

  21. anonymous
    • one year ago
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    \[\int\limits_{-1}^{2} \sqrt{y}-(y+5) dy\]

  22. freckles
    • one year ago
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    why sqrt(y)?

  23. freckles
    • one year ago
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    it it x=y^2 not x=sqrt(y)

  24. anonymous
    • one year ago
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    oopsyes i meant y^2

  25. freckles
    • one year ago
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    \[\int\limits_{-1}^{2} (y^2-(y-5)) dy\]

  26. anonymous
    • one year ago
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    \[∫y ^{2}−(y+5)dy\]

  27. freckles
    • one year ago
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    also you have x=y-5 not x=y+5

  28. anonymous
    • one year ago
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    When you flip it turns to a minus?

  29. freckles
    • one year ago
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    huhh

  30. freckles
    • one year ago
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    y=x+5 subtracting 5 on both sides gives y-5=x

  31. freckles
    • one year ago
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    x=y-5 is another way to say y=x+5 but we solved for x because we are looking at our functions in terms of y

  32. Jhannybean
    • one year ago
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    You're trying to keep everything consistent. If you're solving for x, then solve all your intersecting functions for x only.

  33. Jhannybean
    • one year ago
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    \[\int_a^b A(y)dy\]

  34. anonymous
    • one year ago
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    \[\int\limits_{-1}^{2} \frac{ y ^{3} }{ 3 } - \frac{ y ^{2} }{ 2 }-5y\]

  35. freckles
    • one year ago
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    too many signs and you are suppose to drop the integral sign after integrating

  36. freckles
    • one year ago
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    \[\int\limits\limits_{-1}^{2} (y^2-(y-5)) dy=[\frac{y^3}{3}-\frac{y^2}{2}+5y]_{-1}^{2}\]

  37. anonymous
    • one year ago
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    shouldn't that 5y be -5y?

  38. freckles
    • one year ago
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    no -(-5) =5 not -5

  39. anonymous
    • one year ago
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    oooooooh

  40. Jhannybean
    • one year ago
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    \[\int\limits_{-1}^{2} (y^2-(y-5)) dy = \int_{-1}^2 (y^2-y+5)dy\]

  41. Jhannybean
    • one year ago
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    Remember when you're integrating to simplify EVERYTHING inside your ( ) first before integrating. It makes integrating a lot easier and error free.

  42. anonymous
    • one year ago
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    Got it.

  43. anonymous
    • one year ago
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    I got a final answer of (32/3)-(35/6)=29/6

  44. freckles
    • one year ago
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    33/2 is what I have but I could have made an error

  45. freckles
    • one year ago
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    \[\int\limits\limits\limits_{-1}^{2} (y^2-(y-5)) dy=[\frac{y^3}{3}-\frac{y^2}{2}+5y]_{-1}^{2} \\ =[\frac{2^3}{3}-\frac{2^2}{2}+5(2)]-[\frac{(-1)^3}{3}-\frac{(-1)^2}{2}+5(-1)] \\ =[\frac{8}{3}-2+10]-[\frac{-1}{3}-\frac{1}{2}-5] \\ =\frac{8}{3}+\frac{1}{3}+\frac{1}{2}-2+10+5 \\ =\frac{9}{3}+\frac{1}{2}+13 \\ =3+13+\frac{1}{2} \\ =16+\frac{1}{2} \\ =\frac{32}{2}+\frac{1}{2} \\ =\frac{33}{2}\]

  46. freckles
    • one year ago
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    I think I see what you messed on

  47. anonymous
    • one year ago
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    I see what i did wrong. I got my signs mixed up on the right side. But yes its 33/2 thank you both!

  48. Jhannybean
    • one year ago
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    Good job.

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