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amistre64Best ResponseYou've already chosen the best response.0
what does your graph look like?
 one year ago

jocelynevsqBest ResponseYou've already chosen the best response.0
dw:1359922854861:dw
 one year ago

jocelynevsqBest ResponseYou've already chosen the best response.0
wait nevermind it's not that
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
dw:1359922919356:dw thats what i was thinking it was like
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
then the only pertinent equations are: x^2x^4 and y=sqrt(x)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
they meet at x=0, and what is the other x value?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
sqrt(x) = x^2  x^4 0 = x^2 sqrt(x) x^4 that does seem a bit convoluted :/
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
hmmm, that is one representation i can see of it :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
so if thats the case, would changing it to parametric form help out?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.2
I'd: Write y as function of x: y=xsqrt(1x²) Integrate using u=1x² Multiply by 4, because of the symmetry...
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
sounds crazy enough to work :)
 one year ago

jocelynevsqBest ResponseYou've already chosen the best response.0
what do I multiply by 4?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.2
If you integrate as I did, you only get the area between the positive xaxis and the graph (upper right part of the whole thing) There a 4 such areas, so multiply by 4.
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.2
I've got to 8/3, hope this helps ;)
 one year ago
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