The topic of this problem is energy storage elements. And the problem is to find V out in the op amp circuits shown below. We have a source V in, we have two elements, we have a resistor and a capacitor and in our op amp circuit, and we're measuring the output of the op amp. So we're going to use the properties of an op amp. First of all, we have the ideal op amp symbol, and we know that the current at the inverting input and the non-inverting input are both equal to 0. There's no current flowing into the ideal op amp, and we know that the voltage at the inverting and non-inverting inputs are equal to one another. So we're going to use those properties to solve our circuit. So if we look at the voltage at our non-inverting input, we know it's going to be 0 volts because it's tied to ground. We also know that the voltage at the inverting input is 0 volts because of our properties of the ideal op amp, that is that V- is equal to V+. Now we're going to use those properties, and we're going to use Kirchhoff's current law, and we're going to sum the currents into this node. Let's call it Node 1, and again, we're summing the currents into that node. So if we do that, it'll give us a relationship between the input voltage and the output voltage. So let's do that. First of all, the current flowing through the resistor is V in- 0 / R. The current through the capacitor, which is our second current that we have to add up, using our relationship between Vni for capacitor, our current through the capacitor is Cdv, and our voltage is V sub 0- 0, dt, and we have the current, which is flowing out of the inverting input of the op amp. And we know that current is equal to 0. So, the sum of those currents is equal to 0. So, we have an equation now that relates V out and V in. Namely, we have V in / R = -CdVo (t) / dt. And so we can solve for Vo in terms of V in. So if we do that, we have a V out (t) = -1 / R times C and the integral from- infinity to t of V in(x)dx. So if we continue to solve for that, we have Vo(t), = -1 / RC. And we can break this integral into two different integrals. We have an integral from- infinity to t0 of V sub n (x)dx. And we also have another integral from t0 to t of Vin, (x)dx. The second term is our integral expression, After we start up our circuit. This is an initial condition for our circuit. So ultimately, we have a V out, which is a function of V in integrated and the elements R and C. So this circuit is known as an integrator. Because it's taking the input signal, it's integrating it over time, resulting in an output voltage, which is a function of that integration along with a constant, which is -1 / R times C. So this is our basic example of an integrator circuit.