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anonymous

  • 5 years ago

OK SOME CALCULUS! I forgot how to do this, area of the region enclosed by the graphs of y=e^(x^2) - 2 and y= (4-x^2)^1/2 is?

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  1. anonymous
    • 5 years ago
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    i believe you integrate each individually and then subtract

  2. anonymous
    • 5 years ago
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    what perameters?

  3. anonymous
    • 5 years ago
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    meaning limits of integration? which parameters are you referring to?

  4. anonymous
    • 5 years ago
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    yeah limits, we call them both, or I think we do, ha..

  5. anonymous
    • 5 years ago
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    in that case, there should be two intersections between the two functions. use those

  6. anonymous
    • 5 years ago
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    so their x's?

  7. anonymous
    • 5 years ago
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    A = integration of upper function - integration of of lower function. a & b are the x values of the intersections between the 2 functions

  8. anonymous
    • 5 years ago
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    First draw a rough to visualize what it looks like

  9. anonymous
    • 5 years ago
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    yeah i have all that then, just had forgotten...x's are +/- 1.137 so those are the a and b?

  10. anonymous
    • 5 years ago
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    yep -1.137 to +1.137

  11. anonymous
    • 5 years ago
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    ok got it, thanks! final answer, 5.050

  12. anonymous
    • 5 years ago
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    Very complicated functions; difficult to find points of intersection by algebra. How did you find the points of intersection?

  13. anonymous
    • 5 years ago
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    Strikes me this is a higher level first-year Cal question. Double integrals First integral from x nought to x, second integral from (e^x)-2 to square root (4-x^2) dy dx

  14. anonymous
    • 5 years ago
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    Got em mixed up second integral from sq root (4-x^2) to (e^x)-2

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spraguer (Moderator)
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