If 24 int_0^{frac{pi}{4}} left( sin left| 4x | vedclass

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(B) Let $I = 24 \int_0^{\frac{\pi}{4}} \sin \left| 4x - \frac{\pi}{12} \right| dx + 24 \int_0^{\frac{\pi}{4}} [2 \sin x] dx$. First,evaluate $I_1 = 24

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Let $I = 24 \int_0^{\frac{\pi}{4}} \sin \left| 4x - \frac{\pi}{12} \right| dx + 24 \int_0^{\frac{\pi}{4}} [2 \sin x] dx$. First,evaluate $I_1 = 24 \int_0^{\frac{\pi}{4}} \sin \left| 4x - \frac{\pi}{12} \right| dx$. The expression inside the absolute value changes sign at $4x = \frac{\pi}{12}$,i.e.,$x = \frac{\pi}{48}$. $I_1 = 24 \left( \int_0^{\frac{\pi}{48}} -\sin \left( 4x - \frac{\pi}{12} \right) dx + \int_{\frac{\pi}{48}}^{\frac{\pi}{4}} \sin \left( 4x - \frac{\pi}{12} \right) dx \right)$. $I_1 = 24 \left( \left[ \frac{\cos(4x - \frac{\pi}{12})}{4} \right]_0^{\frac{\pi}{48}} + \left[ -\frac{\cos(4x - \frac{\pi}{12})}{4} \right]_{\frac{\pi}{48}}^{\frac{\pi}{4}} \right)$. $I_1 = 6 \left( (\cos(0) - \cos(-\frac{\pi}{12})) + (-\cos(\frac{11\pi}{12}) + \cos(0)) \right) = 6(1 - \cos(\frac{\pi}{12}) - \cos(\frac{11\pi}{12}) + 1) = 6(2 - \cos(\frac{\pi}{12}) + \cos(\frac{\pi}{12})) = 12$. Next,evaluate $I_2 = 24 \int_0^{\frac{\pi}{4}} [2 \sin x] dx$. Since $0 \le x \le \frac{\pi}{6}$,$0 \le 2 \sin x Since $\frac{\pi}{6} $I_2 = 24 \left( \int_0^{\frac{\pi}{6}} 0 dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 1 dx \right) = 24 \left( \frac{\pi}{4} - \frac{\pi}{6} \right) = 24 \left( \frac{\pi}{12} \right) = 2 \pi$. Thus,$I = I_1 + I_2 = 12 + 2 \pi$. Comparing with $2 \pi + \alpha$,we get $\alpha = 12$.

If $24 \int_0^{\frac{\pi}{4}} \left( \sin \left| 4x - \frac{\pi}{12} \right| + [2 \sin x] \right) dx = 2 \pi + \alpha$,where $[\cdot]$ denotes the greatest integer function,then $\alpha$ is equal to . . . . . .

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