int_{1/2}^{2} frac{1}{x} sin left( x - frac{1 | vedclass

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(A) Let $I = \int_{1/2}^{2} \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f\left( \frac

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(A) Let $I = \int_{1/2}^{2} \frac{1}{x} \sin \left( x - \frac{1}{x} ight) dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f\left( \frac{ab}{x} ight) \frac{ab}{x^2} dx$,where $a = 1/2$ and $b = 2$,we have $ab = 1$. So,$I = \int_{1/2}^{2} \frac{1}{(1/x)} \sin \left( \frac{1}{x} - x ight) \cdot \frac{1}{x^2} dx$. $I = \int_{1/2}^{2} x \sin \left( -\left( x - \frac{1}{x} ight) ight) \cdot \frac{1}{x^2} dx$. Since $\sin(-	heta) = -\sin(	heta)$,we get $I = -\int_{1/2}^{2} \frac{1}{x} \sin \left( x - \frac{1}{x} ight) dx$. Thus,$I = -I$,which implies $2I = 0$,so $I = 0$.

$\int_{1/2}^{2} \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx = $

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