Integral Calculus-2
Integral Calculus-2
Allen Star Batch
Grade 12
Question:
Let $I = \int_{\pi/4}^{\pi/3} \frac{\sin x}{x} dx$, then $I$ belongs to:
$\left(\frac{\sqrt{3}}{8}, \frac{\sqrt{2}}{6}\right)$
$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}\right)$
$\left(\frac{1}{2}, \frac{\sqrt{2}}{2}\right)$
None of these
Step-by-Step Solution
Key Concept: Monotonicity analysis via the derivative determines extrema on closed intervals.
For $f(x) = \frac{\sin x}{x}$, we compute $f'(x) = \frac{x\cos x - \sin x}{x^2} = \frac{(x-\tan x)\cos x}{x^2}$. Since $\tan x > x$ on $(\frac{\pi}{4}, \frac{\pi}{3})$, we have $f'(x) < 0$, so $f$ is decreasing. The minimum value is $m = f(\frac{\pi}{3}) = \frac{\sin(\pi/3)}{\pi/3} = \frac{3\sqrt{3}}{2\pi}$ and maximum is $M = f(\frac{\pi}{4}) = \frac{\sin(\pi/4)}{\pi/4} = \frac{2\sqrt{2}}{\pi}$.
Correct Answer: 1