Op-amp differentiator and integrator circuit: resistor and capacitor values
Grace Firsta - it depends on your requirements, that means: What do you expect from the circuits? At first, it is not possible to realize IDEAL differentiating ot integrating circuits. Both are instable - because of different reasons. As a consequence, the desired mathematical operation is possible only in a limited frequency band. Therefore, it is wise to consider the frequency domain rather than the time domain (as you did).

Example integrator: Assuming an ideal opamp (infinite gain) the transfer function for your last circuit is H(s)=(R5/R6)/(1+jwR5C3). This is the classical first-order lowpass function with gain. The 3-dB lowpass corner frequency is at fc=1/(2*Pi*R5C3). Because the iDEAL integrator transfer function is 1/jwT you must compare both functions in order to see if and under which conditions you can use the lowpass circuit for integrating purposes.

Following this goal, you will see that for frequencies which fulfill the condition jwR5C3>>1 the lowpass approaches the integrating function: H(s)=(R5/R6)/jwR5C3=1/jwR6C3. Hence, the "integrator time constant" is T=R6C3.

Hence, you have a lower frequency limit for integration: w>>1/R5C3. In addition, you have an upper limit set by the opamp itself. Remember the general condition mentioned at the beginning: Open-loop gain of the opamp (nearly) infinity (in practice: larger than 100). This defines the upper frequency limit. As a consequence, integration of an input signal is possible with sufficient accuracy within a certain frequency band only.

Final remark: As you will see, the parallel resistor R5 "disturbs" the integration process (ideal: R5 infinite), however, this resistor is necessary to provide a minimum of dc feedback for stabilizing the operational point.