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anonymous

  • 5 years ago

region bounded by the curves y=x^2+1 and y=2. I know the formula, just don't know how to proceed through it

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  1. anonymous
    • 5 years ago
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    Are you looking for the area?

  2. anonymous
    • 5 years ago
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    yup

  3. anonymous
    • 5 years ago
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    the formula confuses me, do you do the derivative or the integral first?

  4. anonymous
    • 5 years ago
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    So find where they intersect, then integrate the top function - the bottom function between the two x values for their intersection.

  5. anonymous
    • 5 years ago
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    they go from x=-1 to 1

  6. anonymous
    • 5 years ago
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    Ok, so You're finding the area above y=x^2+1 and below y=2.

  7. anonymous
    • 5 years ago
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    yup

  8. anonymous
    • 5 years ago
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    If you just integrated the function y=2 from -1 to 1 what area would you be finding?

  9. anonymous
    • 5 years ago
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    so far I just graphed it out, and they intersect at -1 and 1...looking for the area between -1 and 1, with y=2 on top and y=x^2+1 on the bottom

  10. anonymous
    • 5 years ago
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    So to answer my question. integrating y=2 from -1 to 1 would give the area of a square lying on the x axis centered around the line x=0.

  11. anonymous
    • 5 years ago
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    And the region we care about is inside that square, but is smaller than that whole area.

  12. anonymous
    • 5 years ago
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    \[A=\int\limits_{-1}^{1}[x ^{2}-1]dx \] is where I'm at....I follow you so far...just didnt want to delete this equation

  13. anonymous
    • 5 years ago
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    That is the area of the region _under_ the parabola but above y=0

  14. anonymous
    • 5 years ago
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    the area would be 4 then, without the parabola

  15. anonymous
    • 5 years ago
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    If we take the whole square, and we subtract the area of the region between the parabola and y=0 won't that give us the area we care about?

  16. anonymous
    • 5 years ago
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    yes it would

  17. anonymous
    • 5 years ago
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    So \[ A_{between} = A_{undertop} - A_{underbottom} \] \[ = \int\limits_{x_0}^{x_1} f_{top}(x)-f_{bottom}(x)\ dx\]

  18. anonymous
    • 5 years ago
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    right, I knew that equation, it was how to go through it, but you also answered that question...it's the integral that I had forgotten how to do

  19. anonymous
    • 5 years ago
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    Oh, you were confused about how to do the integral?

  20. anonymous
    • 5 years ago
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    yeah, I didn't think so, could you show me how to do the integral of the x^2+1?

  21. anonymous
    • 5 years ago
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    ahh I can look it up, thanks for the help though!

  22. anonymous
    • 5 years ago
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    \[\int\limits_{-1}^{1}2 - (x^2+1)dx = \int\limits_{-1}^{1}1dx - \int\limits_{-1}^{1}x^2dx = x|_{-1}^{1} - \frac{x^3}{3}|_{-1}^{1}\]

  23. anonymous
    • 5 years ago
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    I got 4/3...but I have no idea how to do what you just did

  24. anonymous
    • 5 years ago
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    Well you know that the integral of a sum is the sum of the integrals right? \[\int\limits_{a}^{b} (f + g)dx = \int\limits_{a}^{b}f\ dx + \int\limits_{a}^{b}g\ dx\]

  25. anonymous
    • 5 years ago
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    sorry, this site has bugs...yes I do know that, what is the dx doing at the end though?

  26. anonymous
    • 5 years ago
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    that's the variable you're integrating with. assume that f and g are functions of x

  27. anonymous
    • 5 years ago
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    So you have f = 2, and g = -(x^2 + 1) in this case.

  28. anonymous
    • 5 years ago
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    All I did was take the -1 from g, and add it to the 2 in f, then break it up into two separate integrals.

  29. anonymous
    • 5 years ago
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    ok with you so far, do you have to graph it out for the integrals?

  30. anonymous
    • 5 years ago
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    you can simply find where they intersect and use those for your limits, then subtract the integral of the bottom function from the integral of the top function.

  31. anonymous
    • 5 years ago
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    so for the integral of x^2 from -1 to 1 (which is area under the curve) how do you do that?

  32. anonymous
    • 5 years ago
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    Oh. \[\int x^adx = \frac{x^{a+1}}{a+1} + C\]

  33. anonymous
    • 5 years ago
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    ah good thanks

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