1. Stating the problem: We are given acceleration $a = \frac{dv}{dt} = 6 - 2t$, velocity $v$ is a function of $t$, and displacement $S$ defined as the integral of velocity over time. 2. Correct the velocity expression given acceleration: Since $a = \frac{dv}{dt} = 6 - 2t$, integrating acceleration to get velocity $v$: $$v = \int a \, dt = \int (6 - 2t) \, dt = 6t - t^2 + C$$ Assuming initial velocity at $t=0$ is zero, $C=0$, so $$v = 6t - t^2$$ 3. Finding displacement $S$: Displacement is the integral of velocity: $$S = \int_0^t v \, dt = \int_0^t (6t - t^2) \, dt$$ Since the variable of integration should be different from the limit variable, replace integrand variable as $x$: $$S = \int_0^t (6x - x^2) \, dx = \left[3x^2 - \frac{x^3}{3} \right]_0^t = 3t^2 - \frac{t^3}{3}$$ 4. Final expressions: Velocity: $$v = 6t - t^2$$ Displacement: $$S = 3t^2 - \frac{t^3}{3}$$ The graph of $v = 2t + 6$ in your message seems to differ; the correct velocity based on provided acceleration is $v=6t - t^2$. The shaded area under the velocity curve from 0 to $t$ represents displacement $S$.