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mr570
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How do I find the integral from 4 to 4 of (3*sqrt(16x^2)^2 dx ?
I think you have to use usubstitution but I don't entirely understand how that works in this case.
 one year ago
 one year ago
mr570 Group Title
How do I find the integral from 4 to 4 of (3*sqrt(16x^2)^2 dx ? I think you have to use usubstitution but I don't entirely understand how that works in this case.
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
use nothing
 one year ago

Umangiasd Group TitleBest ResponseYou've already chosen the best response.0
well the sqrt gets eliminated with the power 2, so...Its a simple integral
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[y=\sqrt{16x^2}\] is the upper half of a circle centered at the origin with radius 4 so \[6\int_{4}^4\sqrt{16x^2}dx\] is 6 times the area of the upper half of a circle with radius 4
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
oh sorry i didn't see the square at the end if it is really the square root squared, then it is easy enough, although a rather strange way to write it
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
it is really \[3\int_{4}^4\sqrt{16x^2}^2dx\]?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Assuming it's what sattelite said then this becomes very simple.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354850130324:dw Then this become easy to integrate.
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
i think the typo from asker
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
I think so too.
 one year ago
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