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anonymous

  • one year ago

FInd the area of region enclosed by the given curves y=x+5 y^2=x y=2 y=-1 what I did was sliced it to parts. from o to -1 and 0 to 2 did I do it right? top half delta y and bottom delta x

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  1. freckles
    • one year ago
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    hello are you here? I would start off with a drawing first

  2. freckles
    • one year ago
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    \[\int\limits_{-1}^{2} (\text{ \right }-\text{ \left } )dy\]

  3. anonymous
    • one year ago
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    yes I understand that and I wanted to split it in half at y=0 and take areas separately would that work? @freckles

  4. freckles
    • one year ago
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    yes but why do that?

  5. freckles
    • one year ago
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    seems much easier just to do one integral \[\int\limits_{-1}^2 (y^2-(y-5)) dy\]

  6. anonymous
    • one year ago
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    I guess I am just over thinking it. because today my teacher was talking about splitting integrals too for some equations

  7. freckles
    • one year ago
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    well you have a x=y^2 is right of x=y-5 so x=y^2 is greater than x=y-5

  8. freckles
    • one year ago
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    |dw:1442380700153:dw|

  9. freckles
    • one year ago
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    it is clear from the picture between y=-1 and y=2 we have x=y^2 is greater than x=y-5 you only need one integral

  10. freckles
    • one year ago
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    if the functions had crossed passed and therefore you had some kind of switching of which was greater than you may need more than one integral

  11. freckles
    • one year ago
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    for example if you had: |dw:1442380823519:dw| you would need two integrals here

  12. freckles
    • one year ago
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    |dw:1442380886729:dw| \[\int\limits_{-1}^c (g(y)-f(y))dy+\int\limits_c^2 (f(y)-g(y))\]

  13. anonymous
    • one year ago
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    ahhhhhhhh I see and yeeah that looks like two different areas for the second example lol thank you!!

  14. freckles
    • one year ago
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    oops forgot to write a dy at the end of that one integral

  15. freckles
    • one year ago
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    \[\int\limits\limits_{-1}^c (g(y)-f(y))dy+\int\limits\limits_c^2 (f(y)-g(y)) \color{red}{ dy}\]

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