An inverting integrator uses an op-amp with a resistor R at the input and a capacitor C in the feedback path. Exploiting the virtual-ground at the inverting input, the input current V_in/R must flow into the capacitor, producing the classic integrator relation:

V_out(t) = −(1/RC) ∫ V_in dt + V_out(0)

The output rate of change (slope) is dV_out/dt = −V_in/(RC), so a constant input yields a constant slope — a straight-line ramp.

Here V_in is a constant negative DC voltage, V_in = −V (V > 0), and the output starts at zero:

V_out(t) = −(1/RC) × (−V) × t = +Vt/(RC)

• The double negation — inverting integrator acting on a negative input — gives a positive slope.
• Because RC is constant, the slope is constant, so the trace is a straight positive-going linear ramp rising steadily from zero (until the op-amp eventually approaches saturation, which the problem excludes).

The other options do not fit. A negative-going ramp would result only from a positive DC input; here the input is negative, so the slope flips to positive. A constant DC output would require the input to be zero (no charging current), which is not the case. A square wave cannot arise from integrating a steady DC level, since integration of a constant is inherently a linear ramp, not an abruptly switching waveform.