Average Value of a Function
f(x) =(1)/(X-3)^2 in the interval [0,2]
The average value?
average value= 1/interval *INT f(x)dx
Wouldn't the integral of that function be
(-1/3)(x-3)^-3 evaluated at the endpoints?
f(x) =(1)/(X-3)^2 in the interval [0,2]
average value= 1/interval *INT f(x)dx
Wouldn't the integral of that function be
(-1/3)(x-3)^-3 evaluated at the endpoints?