so let ln(x) be f^{-1}(x) so e^{x}=f(x)
F^{-1}(x)=xf^{-1}(x)-F(f^{-1}(x))
so = x(ln(x)- e^{ln(x)} = xln(x)-x evaluated at e and 1
which should be 1
if you haven't seen the equation before look at the geometry between a function and its inverse... they add up to make a rectangle.
I also don't know what the partition thingamabobs are but the answer is 1
Yes, that is easy to integrate but I suspect Simple Man is asking about the Riemann sum that will give that integral. The partition is . Those values for x make it easy to find the corresponding "f(x)": . However, it is also makes the interval lengths variable- the length of the first interval is , the second , etc.
The Riemann sum will be which is, of course, the same as