If intlimits_e^x {t,f(t),dt = sin x - xcos x | vedclass

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(D) Given the equation $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$.Applying the Leibniz integral rule,we differentiate both

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$-1/2$

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(D) Given the equation $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$.Applying the Leibniz integral rule,we differentiate both sides with respect to $x$:$\frac{d}{{dx}}\left[ {\int\limits_e^x {t\,f(t)\,dt} } ight] = \frac{d}{{dx}}\left[ {\sin x - x\cos x - \frac{{{x^2}}}{2}} ight]$Using the product rule on the right side:$x\,f(x) = \cos x - (\cos x - x\sin x) - x$$x\,f(x) = \cos x - \cos x + x\sin x - x$$x\,f(x) = x\sin x - x$Dividing by $x$ (since $x 
eq 0$):$f(x) = \sin x - 1$Now,substitute $x = \frac{\pi}{6}$:$f(\frac{\pi}{6}) = \sin(\frac{\pi}{6}) - 1$$f(\frac{\pi}{6}) = \frac{1}{2} - 1 = -\frac{1}{2}$.

If $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$ for all $x \in R - \{0\}$,then the value of $f(\frac{\pi}{6})$ is

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