<p>The area bounded by <span class="math inline">\(y = x^2 + 2\)</span> and <span class="math inline">\(y = 2|x| - \cos x\)</span> is of the form <span class="math inline">\(\frac{p}{q}\)</span> where <span class="math inline">\(p\)</span> and <span class="math inline">\(q\)</span> are relatively prime. Find <span class="math inline">\(p - q\)</span>.</p>
Step-by-Step Solution
Key Concept: Find intersection points by analyzing the equation and using properties of absolute value and cosine; then integrate the difference between the upper and lower curves.
<p><strong>Step 1:</strong> Find points of intersection by solving <span class="math inline">$x^2 + 2 = 2|x| - \cos x$</span>.</p><p>This simplifies to <span class="math inline">$(|x| - 1)^2 + 1 = -\cos x$</span>.</p><p><strong>Step 2:</strong> Since the left side is always <span class="math inline">$\geq 1$</span> and <span class="math inline">$-\cos x$</span> ranges in <span class="math inline">$[-1, 1]$</span>, equality holds only when both sides equal 1.</p><p>This gives <span class="math inline">$|x| = 1$</span> and <span class="math inline">$\cos x = -1$</span>, so <span class="math inline">$x = \pm 1$</span>.</p><p><strong>Step 3:</strong> For <span class="math inline">$-1 < x < 1$</span>, we have <span class="math inline">$x^2 + 2 > 2|x| - \cos x$</span>.</p><p><strong>Step 4:</strong> Calculate the area:</p><p><span class="math display">$$\text{Area} = 2\int_0^1 (x^2 + 2 - 2|x| + \cos x)\,dx = 2\int_0^1 (x^2 + 2 - 2x + \cos x)\,dx$$</span></p><p><span class="math display">$$= 2\left[\frac{x^3}{3} + 2x - x^2 + \sin x\right]_0^1 = 2\left(\frac{1}{3} + 2 - 1 + 0\right) = 2 \cdot \frac{4}{3} = \frac{8}{3}$$</span></p><p><strong>Step 5:</strong> Thus <span class="math inline">$p = 8$</span> and <span class="math inline">$q = 3$</span>, so <span class="math inline">$p - q = 8 - 3 = 5$</span>.</p>
Correct Answer: 5