<p>The function <em>f(x)</em> is defined as follows: <br> \(f(x) = x^2,\quad -1 \le x < 1\)<br> \(f(x) = \sqrt{x},\quad 1 \le x < 2\)<br> \(f(x) = \sqrt{2},\quad 2 \le x \le 4\)<br> Evaluate \(\int_{-1}^{4} f(x)\,dx\).</p>
Step-by-Step Solution
Key Concept: Break the integral at the points where f(x) changes definition, then evaluate each piece separately using appropriate antiderivatives for polynomials, radicals, and constants.
<p><strong>Step 1:</strong> Split the integral at the discontinuity points x=1 and x=2:</p><p>$\int_{-1}^{4} f(x)\,dx = \int_{-1}^{1} x^2\,dx + \int_{1}^{2} \sqrt{x}\,dx + \int_{2}^{4} \sqrt{2}\,dx$</p><p><strong>Step 2:</strong> Evaluate $\int_{-1}^{1} x^2\,dx$:</p><p>$= \left[\frac{x^3}{3}\right]_{-1}^{1} = \frac{1}{3} - \frac{-1}{3} = \frac{2}{3}$</p><p><strong>Step 3:</strong> Evaluate $\int_{1}^{2} \sqrt{x}\,dx = \int_{1}^{2} x^{1/2}\,dx$:</p><p>$= \left[\frac{2x^{3/2}}{3}\right]_{1}^{2} = \frac{2(2)^{3/2}}{3} - \frac{2(1)^{3/2}}{3} = \frac{2 \cdot 2\sqrt{2}}{3} - \frac{2}{3} = \frac{2}{3}(2\sqrt{2}-1)$</p><p><strong>Step 4:</strong> Evaluate $\int_{2}^{4} \sqrt{2}\,dx$:</p><p>$= \sqrt{2}[x]_{2}^{4} = \sqrt{2}(4-2) = 2\sqrt{2}$</p><p><strong>Step 5:</strong> Add all three results:</p><p>$\frac{2}{3} + \frac{2}{3}(2\sqrt{2}-1) + 2\sqrt{2} = 2 - \sqrt{2} + \frac{2}{3}(2\sqrt{2}-1) + 2\sqrt{2}$</p><p>∴ Answer: $2 - \sqrt{2} + \frac{2}{3}(2\sqrt{2}-1) + 2\sqrt{2}$ or simplified: $2 + \frac{2(2\sqrt{2}-1)}{3} + 2\sqrt{2} - \sqrt{2}$</p>
Correct Answer: \(2 - \sqrt{2} + \frac{2}{3}(2\sqrt{2}-1) + 2\sqrt{2}\)