anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
i believe you integrate each individually and then subtract
anonymous
  • anonymous
what perameters?
anonymous
  • anonymous
meaning limits of integration? which parameters are you referring to?

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anonymous
  • anonymous
yeah limits, we call them both, or I think we do, ha..
anonymous
  • anonymous
in that case, there should be two intersections between the two functions. use those
anonymous
  • anonymous
so their x's?
anonymous
  • anonymous
A = integration of upper function - integration of of lower function. a & b are the x values of the intersections between the 2 functions
anonymous
  • anonymous
First draw a rough to visualize what it looks like
anonymous
  • anonymous
yeah i have all that then, just had forgotten...x's are +/- 1.137 so those are the a and b?
anonymous
  • anonymous
yep -1.137 to +1.137
anonymous
  • anonymous
ok got it, thanks! final answer, 5.050
anonymous
  • anonymous
Very complicated functions; difficult to find points of intersection by algebra. How did you find the points of intersection?
anonymous
  • anonymous
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
anonymous
  • anonymous
Got em mixed up second integral from sq root (4-x^2) to (e^x)-2

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