How does an op amp integrator work?
The op-amp is going to try its best to keep the voltage between it's plus and minus input the same.  In an ideal op-amp, no current flows into the inputs, so the only way that it can do that is by changing its output voltage.

In the schematic below, \$v_+ = 0\mathrm{V}\$.  That means that the op-amp will try to hold \$v_-\$ at zero, also.

Whatever voltage is generated by V2 gets turned into a current by R1.  Because \$v_-\$ is being held at \$0\mathrm{V}\$, that same current has to flow in C1.  And because \$v_-\$ is being held at \$0\mathrm{V}\$, the op-amp has to drive the output voltage such that the current in C1 matches the current in R1.

So if \$v_2\$ is constant, then the current into the node around the negative input is constant, which means that the current out of that node from the cap must be constant -- and that can only happen if the output voltage is falling at a constant rate.  The end result is that the op-amp integrates the input voltage into the output voltage.

More complicated voltages at \$v_2\$ cause more complicated behavior, but the op-amp is always going to be trying to drive \$v_-\$ to \$0\mathrm{V}\$.  It can only do that by satisfying \$ \frac{d}{dt} C_1 v_{out} + \frac{v_2}{R_1} = 0 \$.  If you solve that differential equation, it says that $$ v_{out} = -\frac{1}{R_1 C_1} \int v_2 dt $$

HTH







simulate this circuit &ndash; Schematic created using CircuitLab