Integrator and differentiator - op-amps vs. RC networks
The lower circuit, RC

 

only works when:


The output impedance of Vin source is 0Ω


and at the same time:


Vin is much larger than VC . Ideally the capacitor voltage is 0V, but that defeats the whole function as an integrator. The voltage across the capacitor influences the current through the resistor. 


and at the same time:


The load of the circuit (VC) side is ∞


These requirements are not easily to met. The latter requirement is caused by the fact that any load on the capacitor C or resistor R will change the frequency responce of the integrator.

The opamp circuit relaxes these requirements, at least for the infinite load and for the ratio between input and output voltage.



Because the output is driven by an opamp, the output impedance of the circuit is resonably low (in practice maybe couple tens of ohms) and therefore it is sufficient if the output load is about 100 times higher than that; say > 1000Ω A requirement that is a lot easier to meet.

Also because the opamp will try to keep its inputs at the same voltage (V- is virtual ground), the voltage across the capacitor is no longer influencing the input current of the integrator. The input impedance of the opamp circuit is entirely specified by the resistor, hence an ideal integrator.