$E(c) = (c-f(x))^2*dx$, from x=0 to x=3 The function f(x) is not specified but $f(x)*dx=7$ from x=0 to x=3, and $f(x))^2*dx=14$ from x=0 to x=3 a) Find E(5) b) What value of c makes E(c) a minimum? Be sure to verify that the answer is a minimum.
$E(c) = \int_0^3 (c-f(x))^2dx =\int_0^3 (c^2*dx) +\int_0^3 (f(x))^2*dx) +2 c\int_0^3 (f(x)*dx)=$
$E(5) =3*5*5+14-14*5 = 19$
$dE(c)/dc =0$ for minimum values
$c =14/6 =7/3$
this value is a minimum since the graph of $E(c)$ is concave (the coefficient of $c^2$ is positive!)
$E(7/3) =3*49/9 +14-14*7/3 =-2.33$