Capacitor's voltage, current, and energy

Problem-Solving Strategy for solving the Capacitor's voltage, current, and energy

STEP 1. The current of capacitor is: $i(t)=C\frac{dv(t)}{dt}$

STEP 2. The voltage of capacitor is: $v(t)=v(t_0)+\frac{1}{C} \intop\nolimits_{t_0}^{t}i(x) dx$

STEP 3. The energy of capacitor is: $w_C(t)=\frac{1}{2}Cv^2(t)$

PROBLEM:

Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed. 

If there are any errors in solution, please let me know in the comments

SOLUTION:

Capacitor's voltage $v(t)$ is

$v(t)= \begin{Bmatrix}0&(t<0)\\200t &(0 \leq t<10ms)\\-200(t-2\times10^{-2}) &(10ms \leq t<20ms)\\200(t-2\times10^{-2}) &(20ms \leq t<30ms)\\-200(t-4\times10^{-2}) &(30ms \leq t<40ms)\\200(t-4\times10^{-2}) &(40ms \leq t<50ms)\\2&(t \geq50ms) \end{Bmatrix}$

Hence the sketch can be shown as follows:

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