(1/T)∫_t_1 ^t_2 Vsin ωt−V_γ dt=? t_1 and t_2 are solution to Vsin ωt=V_γ V≥V_γ V_γ ≥0 and t_1 <t


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Question Number 737 by 123456 last updated on 08/Mar/15

$\frac{1}{T} \underset{t_{1}}{\int^{t_{2}}} V \sin ω t - V_{γ} d t = ?$ $t_{1} a n d t_{2} a r e s o l u t i o n t o$ $V \sin ω t = V_{γ}$ $V ⩾ V_{γ}$ $V_{γ} ⩾ 0$ $a n d t_{1} < t_{2}$
Commented by prakash jain last updated on 08/Mar/15

$V \sin ω t = V_{γ}$ $t_{2} = t_{1} + 2 π k T ∵ ω = \frac{2 π}{T}$
	Answered by prakash jain last updated on 08/Mar/15

	$\frac{1}{T} \int V \sin ω t - V_{γ} d t$ $= \frac{1}{T} {[- \cos ω t - V_{γ} t]}_{t_{1}}^{t_{2}} = \frac{V_{γ}}{T} [t_{1} - t_{2}] = \frac{V_{γ}}{T} (- \frac{2 π k}{T})$ $Since \sin {w t}_{1} = \sin {w t}_{2} \Rightarrow \cos ω t_{1} = \cos {w t}_{2}$ $The given integral = - \frac{2 π k}{T^{2}} V_{γ}, k > 0$ $V is assumed to be constant amplitude .$
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