Another (mostly) calculus question

[Smaug]

In one of the solutions to the problems in my physics textbooks, it's written (without further explanation, as if it were somehow obvious) that the gravitational potential energy of a vertical rod is given by the formula U=1/2*m*g*l. How does that make any sense? The sum of the gravitational energy of the infinitesimally small parts of the rod is obviously U=integral(m*g*l,l,0,l)=1/2*m*g*l^2.

You've made a mistake in assuming the potential energy of an infinitesimally small element of the rod in the integral. It actually is the following:
dU=dm*g*x
where dm is an infinitely small element of mass and x is its height above an arbitrary zero energy level. Since dm=rho(x)*dx where rho(x) is lineal density of the rod, it can be rewritten as
dU=rho(x)*g*x*dx .
If the rod is uniform (rho(x)=m/l=const) we arrive at
dU=(m/l)*g*x*dx
and the integral is the following:
U = int((m/l)*g*x dx,  x=0 to x=l) = (m*g*l^2)/(2*l) = 1/2*m*g*l.

Also it's a good practice to give distinct names to your parameters and variables. In

U=integral(m*g*l , l ,0,l)

you use l for both the length of the rod (a constant parameter in this context) and the height above the ground (an integration variable). Even though in this very case they are measured along the same line they are different things. While it may not look as a big deal it will inevitably lead to confusion sooner or later.

P. S. I have to note that for an inclined rod the calculation will not be quite the same bacause dm=rho(x)*dx=m/l*dx is only valid if the rod is parallel to the axis x. Otherwise you have to introduce an inclination angle.