The average value of sine wave is times the peak value?

Correct option is D

​$$V_i(t) = V_m \sin \omega t \\\therefore \text{Average value } (V_{\text{avg}}) = \frac{1}{\pi} \int_{0}^{\pi} V_i(t) dt \\V_{\text{avg}} = \frac{1}{\pi} \int_{0}^{\pi} V_m \sin \omega t \, dt \\V_{\text{avg}} = \frac{-V_m}{\pi} \Big[ \cos \omega t \Big]_{0}^{\pi} \\V_{\text{avg}} = \frac{-V_m}{\pi} \Big[ \cos \pi - \cos 0 \Big] \\V_{\text{avg}} = \frac{-V_m}{\pi} \Big[ -1 - 1 \Big] \\V_{\text{avg}} = \frac{+2 V_m}{\pi} \\V_{\text{avg}} = 0.637 V_m$$​​