Answer
$$ 0.4\rm L$$
Work Step by Step
$$V_{ave} = \frac{1}{5} \int^5_0 V(t) dt = \frac{1}{4\pi} \int^5_{0} [1-cos(\frac{2}{5} \pi t)] dt $$
$$=\frac{1}{4\pi} [t - \frac{5}{2\pi} sin (\frac{2}{5} \pi t)]^5_0 = \frac{5}{4\pi} \approx 0.4\rm L$$
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