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[tex]\Delta d = v_1\Delta t + \frac{1}{2}a\Delta t^2[/tex]

I am trying to find this equation using calculus.

So from what I understand, [itex]\Delta d[/itex] is the area under the curve for [itex]v(\Delta t)[/itex].

We can define [itex]v(\Delta t) = v_1 + a\Delta t[/itex] where [itex]\Delta t = t_2 - t_1[/itex]

I can take the integral of [itex]v(\Delta t)[/itex] to find [itex]\Delta d[/itex] call it [itex]D(\Delta t)[/itex]. Where, [itex]D'(\Delta t) = v(\Delta t)[/itex].

[tex]D(\Delta t) = \int v(\Delta t)\ d\Delta t \implies D(\Delta t) = \int (v_1 + a\Delta t)d\Delta t\implies D(\Delta t) = v_1\Delta t + \frac{1}{2}a\Delta t^2[/tex]

and suppose I know [itex]v(t)[/itex] could I say:

[tex]\Delta d = \int_{t_1}^{t_2}v(t)dt = P(t_2) - P(t_1)\ where\ P'(t) = v(t)[/tex]

[tex]\Delta d = D(\Delta t) \implies D(\Delta t) = P(t_2) - P(t_1)[/tex]

Is this all correct ? Any feedback would be appreciated.