<p>The area bounded by the two curves \(y = x^2 + 3x + 5\) and \(y = -x^2 + 5x + 9\) for \(x\) between \(-1\) and \(4\), is ______ (up to two decimal places).</p>
Step-by-Step Solution
Key Concept: The area between two curves is found by integrating the difference of the upper and lower functions over the given interval. First, determine which curve is above the other, then integrate their difference.
<p><strong>Step 1: Identify the curves and interval</strong></p><p>Curve 1: $y_1 = x^2 + 3x + 5$</p><p>Curve 2: $y_2 = -x^2 + 5x + 9$</p><p>Interval: $x \in [-1, 4]$</p><p><strong>Step 2: Determine which curve is above</strong></p><p>Find the difference: $y_2 - y_1 = (-x^2 + 5x + 9) - (x^2 + 3x + 5)$</p><p>$= -x^2 + 5x + 9 - x^2 - 3x - 5$</p><p>$= -2x^2 + 2x + 4$</p><p>Check at $x = 0$: $-2(0)^2 + 2(0) + 4 = 4 > 0$, so $y_2 > y_1$ in this interval.</p><p><strong>Step 3: Set up the integral</strong></p><p>Area $= \int_{-1}^{4} (y_2 - y_1) \, dx = \int_{-1}^{4} (-2x^2 + 2x + 4) \, dx$</p><p><strong>Step 4: Find the antiderivative</strong></p><p>$\int (-2x^2 + 2x + 4) \, dx = -\frac{2x^3}{3} + x^2 + 4x + C$</p><p><strong>Step 5: Evaluate at the bounds</strong></p><p>At $x = 4$: $-\frac{2(4)^3}{3} + (4)^2 + 4(4) = -\frac{128}{3} + 16 + 16 = -\frac{128}{3} + 32 = \frac{-128 + 96}{3} = -\frac{32}{3}$</p><p>At $x = -1$: $-\frac{2(-1)^3}{3} + (-1)^2 + 4(-1) = -\frac{2(-1)}{3} + 1 - 4 = \frac{2}{3} - 3 = \frac{2 - 9}{3} = -\frac{7}{3}$</p><p><strong>Step 6: Calculate the definite integral</strong></p><p>Area $= \left[-\frac{32}{3}\right] - \left[-\frac{7}{3}\right] = -\frac{32}{3} + \frac{7}{3} = -\frac{25}{3}$</p><p>Since area must be positive: Area $= \left|{-\frac{25}{3}}\right| = \frac{25}{3} = 8.33$</p><p><strong>∴ Answer: 8.33</strong></p>
Correct Answer: 8