The value of intlimits_{frac{1}{2}}^2 frac{1} | vedclass

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

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(A) Let $I = \int\limits_{\frac{1}{2}}^2 \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a = \frac{1}{2}$ and $b = 2$,we have $a+b = \frac{5}{2}$.Substituting $x = \frac{1}{t}$,we get $dx = -\frac{1}{t^2} dt$.When $x = \frac{1}{2}$,$t = 2$. When $x = 2$,$t = \frac{1}{2}$.$I = \int\limits_2^{\frac{1}{2}} t \sin \left( \frac{1}{t} - t \right) \left( -\frac{1}{t^2} \right) dt$$I = \int\limits_2^{\frac{1}{2}} \frac{1}{t} \sin \left( t - \frac{1}{t} \right) dt$$I = -\int\limits_{\frac{1}{2}}^2 \frac{1}{t} \sin \left( t - \frac{1}{t} \right) dt$$I = -I$Therefore,$2I = 0$,which implies $I = 0$.

The value of $\int\limits_{\frac{1}{2}}^2 \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx$ is equal to

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