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

Answer: (A)

$\frac{π}{2}<I<\frac{3π}{4}$

Answer 2

I comes out around 1.536 which is not satisfied by any given options.
$I = ∫^{\frac{π}{3}}_{\frac{π}{4}}(\frac{8sinx-sin2x}{x})dx>I>I = ∫^{\frac{π}{3}}_{\frac{π}{4}}(\frac{8sinx-sin2x}{x})dx$
$\frac{π}{2}>I> ∫^{\frac{π}{3}}_{\frac{π}{4}}(\frac{8sinx-sin2x}{x})dx$
$\frac{sinx}{x}$ is decreasing in $(\frac{π}{3},\frac{π}{4})$
so it attains maximum at
x = x/4
$I> ∫^{\frac{π}{3}}_{\frac{π}{4}}(\frac{8 sin\pi/3}{\pi/3}-2)dx$
$I>√3-\frac{π}{6}$