Question

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?

Answer 1

i believe you integrate each individually and then subtract

Answer 2

what perameters?

Answer 3

meaning limits of integration? which parameters are you referring to?

Answer 4

yeah limits, we call them both, or I think we do, ha..

Answer 5

in that case, there should be two intersections between the two functions. use those

Answer 6

so their x's?

Answer 7

A = integration of upper function - integration of of lower function. a & b are the x values of the intersections between the 2 functions

Answer 8

First draw a rough to visualize what it looks like

Answer 9

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

Answer 10

yep -1.137 to +1.137

Answer 11

ok got it, thanks!
final answer, 5.050

Answer 12

Very complicated functions; difficult to find points of intersection by algebra. How did you find the points of intersection?

Answer 13

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

Answer 14

Got em mixed up second integral from sq root (4-x^2) to (e^x)-2