Suppose R is the region bounded by the curves y = any function f(x) and y = c, on the interval a to b. Find the value of c that minimizes the volume of the solid that is generated by revolving R about the line y = c.
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$\displaystyle V(c)=\int_a^b \pi[f(x)-c]^2 dx$ $\displaystyle \frac{dV}{dc}=\int_a^b 2\pi[f(x)-c](-1) dx$ $\displaystyle =-2\pi\int_a^b [f(x)-c] dx$ resolve dV/dt = 0, we get $\displaystyle c = \frac{\int_a^b f(x)dx}{b-a}$
Where does the (-1) in step two come from?
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