Energy dissipation of a resistor in an AC circuit
Calculate voltage at t=0 and t=1, then use P=V^2/R?


Nope, power (or voltage) at t=0 and t=1 gives no information about what happens between t=0 and t=1.


  Use P=V^2/R and integrate between the limits to calculate energy dissipation


Yes, you have to calculate

\$ Energy = \int_{t=t_0}^{t=t_1} power dt = \frac{1}{R} \int_{t=t_0}^{t=t_1} v^2 dt\$

Also \$ P_{avg} = \frac{Energy}{t_1-t_0} = \frac{1}{t_1-t_0} \int_{t=t_0}^{t=t_1} \frac{v^2}{R} dt = \frac{1}{t_1-t_0} \frac{1}{R} \int_{t=t_0}^{t=t_1} v^2 dt\$

\$ P_{avg} = V_{rms}^2/R\$ with Vrms being the RMS (root mean square) value of voltage. 

\$ V_{rms} = \sqrt{ \frac{1}{t_1-t_0} \int_{t=t_0}^{t=t_1} v^2 dt }\$

If you combine these two equations you will end up with the one on top using average power.

When an AC voltage (or any periodic voltage) is considered, its "RMS voltage" is the DC voltage that would result in the same dissipation in a resistor.