Find the center of mass of the triangle with vertices `(a,0), (0,b), (0,c) (a>0, b>c)`

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In order to evaluate the integral for `y_c` , we need

`(f(x))^2 = (b-b/ax)^2 = b^2 - 2b^2/ax + b^2/a^2 x^2`

Similarly

`(g(x))^2 = (c - c/ax)^2 = c^2 - 2c^2/ax + c^2/a^2x^2`

Then

`(f(x))^2 - (g(x))^2 = (b^2 - c^2) -2(b^2 - c^2)/a x + (b^2 - c^2)/a^2 x^2`

The antiderivative of this function is

`(b^2 -c^2)x - (b^2 - c^2)/a x^2 + (b^2 - c^2)/a^2 x^3/3`

Plugging this into integral for `y_c` and evaluating it at the limits a and 0, we get

`y_c =1/A int^a_0 1/2((f(x))^2 - (g(x))^2)dx = 1/2(b^2 - c^2)a(1 - 1 + 1/3)`

`` `y_c = 1/A1/6a(b^2 - c^2)` 

Since `A = 1/2a(b-c)` ,

`y_c = 1/3 (a(b^2 - c^2))/a(b-c)...

(The entire section contains 2 answers and 388 words.)

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