<p>Let <em>y</em><sub>1</sub> = <em>f</em>(<em>x</em>) = 2<em>x</em> − 1 and <em>y</em><sub>2</sub> = <em>g</em>(<em>x</em>) = <em>x</em><sup>2</sup> − 4. Find the area of the region enclosed between the two curves.</p>
Step-by-Step Solution
Key Concept: Find intersection points by setting f(x) = g(x), then integrate the difference of the upper and lower curves over the bounded region. The curve that lies above changes at different intervals, so split the integral accordingly.
<p><strong>Step 1: Find intersection points</strong></p><p>Set f(x) = g(x): 2x - 1 = x² - 4</p><p>x² - 2x - 3 = 0</p><p>(x - 3)(x + 1) = 0</p><p>x = -1 and x = 3</p><p><strong>Step 2: Determine which curve is above</strong></p><p>Test x = 0: f(0) = -1, g(0) = -4</p><p>Since -1 > -4, f(x) is above g(x) on [-1, 3]</p><p><strong>Step 3: Set up the integral</strong></p><p>Area = ∫₋₁³ [f(x) - g(x)] dx = ∫₋₁³ [(2x - 1) - (x² - 4)] dx</p><p>= ∫₋₁³ [-x² + 2x + 3] dx</p><p><strong>Step 4: Integrate</strong></p><p>= [-x³/3 + x² + 3x]₋₁³</p><p>= [(-27/3 + 9 + 9) - (1/3 + 1 - 3)]</p><p>= [(-9 + 18) - (1/3 - 2)]</p><p>= [9 - (1/3 - 6/3)]</p><p>= [9 - (-5/3)]</p><p>= 9 + 5/3 = 27/3 + 5/3 = 32/3</p><p>∴ Answer: <strong>10.67</strong> (or 32/3 square units)</p>
Correct Answer: 10.67