The value of mathop {lim }limits_{x to frac{p | vedclass

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(C) Let $y = \mathop {\lim }\limits_{x 	o \pi /2} \frac{{\int_{\pi /2}^x {t \,dt} }}{{\sin \left( {2x - \pi } ight)}}$.Using the Fundamental Theore

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$\frac{\pi}{4}$

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(C) Let $y = \mathop {\lim }\limits_{x 	o \pi /2} \frac{{\int_{\pi /2}^x {t \,dt} }}{{\sin \left( {2x - \pi } ight)}}$.Using the Fundamental Theorem of Calculus,the integral is $\left[ \frac{t^2}{2} ight]_{\pi/2}^x = \frac{x^2}{2} - \frac{\pi^2}{8}$.So,$y = \mathop {\lim }\limits_{x 	o \pi /2} \frac{\frac{x^2}{2} - \frac{\pi^2}{8}}{\sin(2x - \pi)} = \mathop {\lim }\limits_{x 	o \pi /2} \frac{4x^2 - \pi^2}{8 \sin(2x - \pi)}$.Factoring the numerator: $4x^2 - \pi^2 = (2x - \pi)(2x + \pi)$.Thus,$y = \frac{1}{8} \mathop {\lim }\limits_{x 	o \pi /2} \frac{(2x - \pi)(2x + \pi)}{\sin(2x - \pi)}$.Using the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,where $	heta = 2x - \pi$ as $x 	o \pi/2$,we get:$y = \frac{1}{8} 	imes (2(\frac{\pi}{2}) + \pi) 	imes 1 = \frac{1}{8} 	imes 2\pi = \frac{\pi}{4}$.

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\int_{\frac{\pi }{2}}^x t \,dt}}{{\sin (2x - \pi )}}$ is

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