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I am wondering why my following steps fail to derive $\frac{1}{2}at^2+{v_0}t+x_0=x$

$Derivation\,of\,Equation\,1:$ $$a=\frac{dv}{dt}$$ $$a\,dt = dv$$ $$\int_{t_0}^{t}a\,dt=\int_{v_0}^{v}dv$$ $$at-at_0=v-v_0$$ $$at-a(0)=v-v_0$$ $$at=v-v_0$$ $$Equation\,1: at+v_0=v$$ $Derivation\,of\,Equation\,2:$ $$v=\frac{dx}{dt}$$ $$v\,dt = dx$$ $$\int_{t_0}^{t}v\,dt=\int_{x_0}^{x}dx$$ $$vt-vt_0=x-x_0$$ $$vt-v(0)=x-x_0$$ $$vt=x-x_0$$ $$Equation\,2:vt+x_0=x$$ $Substituting\,Equation\,1\,into\,Equation\,2:$ $$(at+v_0)t+x_0=x$$ $$at^2+{v_0}t+x_0=x$$

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Your integration in the "Derivation of Equation 2" is wrong. In this integral v is a function of t and not a constant. You integrate as v were a constant. You need to introduce the actual expression of v(t) under the integral and integrate over t.

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