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  1. The voltage-current relationship for a capacitor is described by Eq. \eqref{1}: \begin{equation} \mathrm{v}\left(\, t\, \right) = \frac{1}{C}\int_{0}^{1}\mathrm{i}\left(\, \tau\, \right)\,\mathrm{d}\tau \label{1}\tag{1} \end{equation}
  2. The voltage-charge relationship for a capacitor is described by Eq. \eqref{2}: \begin{equation} \mathrm{v}\left(\, t\, \right) = \frac{1}{C}\,\mathrm{q}\left(\, t\, \right) \label{2}\tag{2} \end{equation}
  3. How can I prove that $\,\mathrm{q}\left(\, t\, \right) = \int_{0}^{1}\mathrm{i}\left(\, \tau\, \right)\,\mathrm{d}\tau$ ?.
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The equation you want is $$ q(t)= \int_0^t i(\tau) d\tau $$ This comes from integrating both sides of the definition $$ \frac{dq}{dt}= i(t) $$

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  • $\begingroup$ Before performing the integration of both sides do I need to multiply both sides by $dt$ to get $dq = i(t)dt$? $\endgroup$ – SpaceBeer Aug 28 '16 at 6:13
  • $\begingroup$ @SpaceBeer Both "yes" and "no" are possible here. You can do it without multiplying by $dt$, just integrating both sides as functions of $t$ and applying Newton-Leibnitz formula $\int \limits_{a}^{b} F'(x) dx = F(b) - F(a)$ to the $\frac{dq}{dt}$. $\endgroup$ – Evgeny Aug 28 '16 at 9:24
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You can use your equations $(1)$ and $(2)$ to answer your question. From equation $(1)$, you have $$ v(t) = \frac{1}C\int_0^1 i(\tau)\,d\tau\label{a}\tag{1}. $$

From your equation $(2)$, you have $$ v(t) = \frac{1}{C}q(t)\label{b}\tag{2}. $$

Therefore, multiply equation $(2)$ by the capacitance $C$ on each side to obtain $$ q(t) = Cv(t)\label{c}\tag{3}. $$ Substitute the value of $v(t)$ from equation $(1)$ into $(3)$, and you're done: $$ q(t) = C\left(\frac{1}C\int_0^1 i(\tau)\,d\tau\right) = \int_0^1 i(\tau)\,d\tau. $$

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