<p>Let <br/>
\[ m = \int_{-2}^{0} \frac{|\sin x|}{\left[\dfrac{x}{\pi}\right] + \dfrac{1}{2}}\, dx \]
and <br/>
\[ n = \int_{0}^{2} \frac{|\sin t|}{-\left[\dfrac{t}{\pi}\right] - 1 + \dfrac{1}{2}}\, dt \]
Find the value of \(-\dfrac{m}{n}\).</p>
Step-by-Step Solution
Key Concept: Recognize that the floor function [x/π] is constant on [-2,0) and (0,2], and use substitution t = -x to establish a relationship between the integrals m and n by transforming one into the other.
<p><strong>Step 1:</strong> Evaluate the floor function on given intervals.</p><p>For x ∈ [-2, 0): x/π ∈ [-2/π, 0) ≈ [-0.637, 0), so [x/π] = -1</p><p>For t ∈ (0, 2]: t/π ∈ (0, 2/π] ≈ (0, 0.637], so [t/π] = 0</p><p><strong>Step 2:</strong> Simplify m:</p><p>m = ∫₍₋₂₎⁰ |sin x|/(-1 + 1/2) dx = ∫₍₋₂₎⁰ |sin x|/(-1/2) dx = -2∫₍₋₂₎⁰ |sin x| dx</p><p><strong>Step 3:</strong> Simplify n:</p><p>n = ∫₀² |sin t|/(-(0) - 1 + 1/2) dt = ∫₀² |sin t|/(-1/2) dt = -2∫₀² |sin t| dt</p><p><strong>Step 4:</strong> Use substitution t = -x in integral for m:</p><p>m = -2∫₍₋₂₎⁰ |sin x| dx = -2∫₀² |sin(-u)|(-du) = -2∫₀² |sin u| du = -2∫₀² |sin t| dt</p><p><strong>Step 5:</strong> Compare m and n:</p><p>Since m = -2∫₀² |sin t| dt and n = -2∫₀² |sin t| dt, we have m = n</p><p><strong>Step 6:</strong> Calculate the ratio:</p><p>-m/n = -1</p><p>∴ Answer: <strong>-1</strong></p>
Correct Answer: 1