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anonymous
 4 years ago
find the exact area of the regions bounded by y=x^3/(sqrt(4x^2), y=0,x=0,x=sqrt(2)
anonymous
 4 years ago
find the exact area of the regions bounded by y=x^3/(sqrt(4x^2), y=0,x=0,x=sqrt(2)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\sqrt{2}} x ^{3}/\sqrt{4x ^{2}}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1328082060815:dwall we're doing here is integrating up to the line x=sqrt2\[\int_{0}^{\sqrt2}\frac{x^3}{\sqrt{4x^2}}dx\]which can be done with the substitution\[x=2\sin\theta\to dx=2\cos\theta d\theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Or you can put 4x^2=t and then do it..

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1...how would that work? you have x^3 on top

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah so one x goes in dt and what is left is x^2 which is 4t denominator is sqrt(t) so the net integral reduces to (4t^(1/2)t^(1/2))/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's a trigonometric integral u substituion wont work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And a minus sign outside.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The substitution works pretty well only....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you use substitution your integral will be left in terms of both t and x, which will not work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if it were only x on top not x^3 you could but not in this form

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Dude read what i have written if you put 4x^2as t you are left with only x^2 in the numerator which is 4t.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First do it and see you'll get the answer...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok well what do you get by doing it your method?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Whatever you get by trignometric substitution you'll get the same thing only...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0matt: Did you get what i'm trying to tell or should i elaborate further and solve the whole thing?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's okay I think what you are saying does make sense, I don't actually have pen or paper right now but i'll cease to doubt you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}x^3/\sqrt{4x ^{2}}\] put 4x^2=t. 2x dx=dt. dx=dt/2x\[\int\limits_{}^{}x ^{3}/\sqrt{t} 2x\] cancelling x you get x^2 on top which is 4t\[\int\limits_{}^{}(4t)/\sqrt{t} = (4/\sqrt{t})\sqrt{t}\] This is easily integrable.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah k, yeah i see. Well thank you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i forgot a minus sign. :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahha it's alright I see what you are doing now

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Never would have thought of that, cool.
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