Question

Let
\(I = ∫^{\frac{π}{3}}_{\frac{π}{4}}(\frac{8sinx-sin2x}{x})dx\)
Then

• A. \(\frac{π}{2}<I<\frac{3π}{4}\)
• B. \(\frac{π}{5}<I<\frac{5π}{12}\)
• C. \(\frac{5π}{12}<I<\frac{\sqrt3}{3}π\)
• D. \(\frac{3π}{4}<I<π\)

Answer 1

The Correct Option is A

To evaluate the integral \(I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left(\frac{8\sin x - \sin 2x}{x}\right) dx\), we first need to break down the expression inside the integral. Let's proceed step-by-step to find the correct range for the integral's value. 

1. Rewrite the integrand:
   • The expression \(\sin 2x\) can be rewritten using the double-angle identity: \(\sin 2x = 2 \sin x \cos x\).
   • The integrand becomes \(\frac{8 \sin x - 2 \sin x \cos x}{x} = \frac{\sin x(8 - 2 \cos x)}{x}\).
2. Estimate the behavior of the function:
   • Observe that \(\sin x\) varies between \(\frac{\sqrt{2}}{2}\) at \(\frac{\pi}{4}\) and \(\frac{\sqrt{3}}{2}\) at \(\frac{\pi}{3}\).
   • The cosine value \(\cos x\) ranges from \(\frac{\sqrt{2}}{2}\) to \(\frac{1}{2}\) over the same interval.
   • Hence, the factor \(8 - 2 \cos x\) will be bounded between approximately \(6.83\) and \(7\).\)
3. Estimate the integral:
   • Estimate the integral \(I\) over the given limits: \(\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin x(8 - 2 \cos x)}{x} dx\).
   • Since \(\sin x/x\) is positive and decreases from \(\frac{\sqrt{2}/2}{\pi/4}\) to \(\frac{\sqrt{3}/2}{\pi/3}\), this implies the integral value \(I\) is positive but varies closer to the bounds determined by \(\sin x\).
4. Compare with given options:
   • Based on our estimations, \(\frac{\pi}{2} \approx 1.57\) and \(\frac{3\pi}{4} \approx 2.36\).
   • The computed integral lies well between these bounds since it involves the product of a less than unity decreasing function and a bounded result.

Finally, based on this analysis, the correct range for integral \(I\) is:

\(\frac{\pi}{2} < I < \frac{3\pi}{4}\)