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anonymous
 5 years ago
i'm trying to find the centroid of the region bonded by the given curves
y= sinx; y=cosx, x=o and x=pi/4
anonymous
 5 years ago
i'm trying to find the centroid of the region bonded by the given curves y= sinx; y=cosx, x=o and x=pi/4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The first thing you need is the mass so if we set f(x) = sin(x) and g(x) = cos(x) to get the mass we want: \[\int\limits_{0}^{π/4}[\cos(x)\sin(x)]dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know thats part i think im getting the wrong numbers when i plug the integral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This integral should be equal to √(2) 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i got 2/ sqrt(2)1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but how do you do the integrarion by parts?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This first integral isn't by parts. ∫cos(x)dx = sin(x) ∫sin(x)dx = cos(x) sin(x)+cos(x) evaluated from 0 to π/4. At π/4 we get √2 and at 0 we get 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Next we have to find the x and y coordinate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For the x coordinate take: \[\int\limits_{0}^{π/4}x[\cos(x)\sin(x)]dx\] and divide by √2  1 and the y coordinate take: \[\int\limits\limits_{0}^{π/4}(1/2)[\cos ^{2}(x)  \sin ^{2}(x)]dx\] and divide that by √2  1
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