int_{frac{1}{3}}^3 frac{1}{x} sin left(frac{1 | vedclass

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(A) Let $I = \int_{\frac{1}{3}}^3 \frac{1}{x} \sin \left(\frac{1}{x}-x\right) d x$. Substitute $x = \frac{1}{t}$, which implies $dx = -\frac{1}{t^2} d

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(A) Let $I = \int_{\frac{1}{3}}^3 \frac{1}{x} \sin \left(\frac{1}{x}-xight) d x$. Substitute $x = \frac{1}{t}$, which implies $dx = -\frac{1}{t^2} dt$. When $x = \frac{1}{3}$, $t = 3$, and when $x = 3$, $t = \frac{1}{3}$. Substituting these into the integral: $I = \int_3^{\frac{1}{3}} t \sin \left(t - \frac{1}{t}ight) \left(-\frac{1}{t^2}ight) dt$ $I = \int_3^{\frac{1}{3}} -\frac{1}{t} \sin \left(t - \frac{1}{t}ight) dt$ Using the property $\sin(- 	heta) = -\sin(	heta)$, we have $\sin(t - \frac{1}{t}) = -\sin(\frac{1}{t} - t)$. $I = \int_3^{\frac{1}{3}} \frac{1}{t} \sin \left(\frac{1}{t} - tight) dt$ Reversing the limits of integration changes the sign: $I = -\int_{\frac{1}{3}}^3 \frac{1}{t} \sin \left(\frac{1}{t} - tight) dt$ Since the variable of integration is a dummy variable, we can replace $t$ with $x$: $I = -\int_{\frac{1}{3}}^3 \frac{1}{x} \sin \left(\frac{1}{x} - xight) dx$ $I = -I$ $2I = 0 \implies I = 0$.

$\int_{\frac{1}{3}}^3 \frac{1}{x} \sin \left(\frac{1}{x}-x\right) d x=$

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