Let's solve the problem step by step. ### **Step 1: Understand the Problem** You are given the function: \[ f(x) = 4x^3 - 3x^2 - 8x \] You are to find the area of the shaded region, which is the area between the curve and the x-axis (where \( f(x) \geq 0 \)). --- ### **Step 2: Find the Roots** Set \( f(x) = 0 \): \[ 4x^3 - 3x^2 - 8x = 0 \] Factor out \( x \): \[ x(4x^2 - 3x - 8) = 0 \] So, one root is \( x = 0 \). For the quadratic part: \[ 4x^2 - 3x - 8 = 0 \] Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 4, b = -3, c = -8 \): \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(4)(-8)}}{8} = \frac{3 \pm \sqrt{9 + 128}}{8} = \frac{3 \pm \sqrt{137}}{8} \] So, the roots are: \[ x_1 = 0, \quad x_2 = \frac{3 + \sqrt{137}}{8}, \quad x_3 = \frac{3 - \sqrt{137}}{8} \] --- ### **Step 3: Determine the Interval for the Shaded Region** From the graph, the shaded region is between \( x = -1 \) and \( x = 2 \). --- ### **Step 4: Set Up the Integral** The area is: \[ \text{Area} = \int_{-1}^{2} |f(x)|\, dx \] From the graph, between \( x = -1 \) and \( x = 0 \), \( f(x) < 0 \), and between \( x = 0 \) and \( x = 2 \), \( f(x) > 0 \). So, \[ \text{Area} = -\int_{-1}^{0} f(x) dx + \int_{0}^{2} f(x) dx \] --- ### **Step 5: Compute the Integrals** #### **First, find the antiderivative:** \[ \int f(x) dx = \int (4x^3 - 3x^2 - 8x) dx = x^4 - x^3 - 4x^2 + C \] #### **Calculate each part:** **A. \(\int_{-1}^0 f(x) dx\):** \[ F(x) = x^4 - x^3 - 4x^2 \] \[ F(0) = 0^4 - 0^3 - 4 \cdot 0^2 = 0 \] \[ F(-1) = (-1)^4 - (-1)^3 - 4 \cdot (-1)^2 = 1 + 1 - 4 = -2 \] \[ \int_{-1}^0 f(x) dx = F(0) - F(-1) = 0 - (-2) = 2 \] **B. \(\int_{0}^2 f(x) dx\):** \[ F(2) = (2)^4 - (2)^3 - 4 \cdot (2)^2 = 16 - 8 - 16 = -8 \] \[ F(0) = 0 \] \[ \int_{0}^2 f(x) dx = F(2) - F(0) = -8 - 0 = -8 \] --- ### **Step 6: Find the Total Area** \[ \text{Area} = -(2) + (-8) = 2 + 8 = 10 \] --- ### **Final Answer** \[ \boxed{10} \] --- **Summary of Steps:** 1. Set up the function and roots. 2. Set up the integral with the correct signs. 3. Integrate and evaluate. 4. Add the absolute values to find the area. **Final area = 10**.

Let's analyze the given function and find the area between the curve and the x-axis step by step. --- ### **Given Function:** \[ f(x) = 4x + 3x^2 - x^3 \] --- ### **Step 1: Find the roots of \(f(x) = 0\)** Set \(f(x) = 0\): \[ 4x + 3x^2 - x^3 = 0 \] Rewrite: \[ -x^3 + 3x^2 + 4x = 0 \] Factor out \(x\): \[ x(-x^2 + 3x + 4) = 0 \] So, one root is: \[ x = 0 \] Now, solve the quadratic: \[ -x^2 + 3x + 4 = 0 \] Multiply both sides by \(-1\) to make it standard: \[ x^2 - 3x - 4 = 0 \] Use quadratic formula: \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} \] \[ x = \frac{3 \pm 5}{2} \] Calculate both roots: - When \(+\): \[ x = \frac{3 + 5}{2} = \frac{8}{2} = 4 \] - When \(-\): \[ x = \frac{3 - 5}{2} = \frac{-2}{2} = -1 \] **Roots are:** \[ x = -1, \quad x = 0, \quad x = 4 \] --- ### **Step 2: Determine where \(f(x)\) is positive or negative** To find the sign of \(f(x)\) in each interval, choose test points: - For \(x < -1\), choose \(x = -2\): \[ f(-2) = 4(-2) + 3(-2)^2 - (-2)^3 = -8 + 3(4) - (-8) = -8 + 12 + 8 = 12 > 0 \] - For \(-1 < x < 0\), choose \(x = -0.5\): \[ f(-0.5) = 4(-0.5) + 3(0.25) - (-0.5)^3 = -2 + 0.75 + 0.125 = -2 + 0.75 + 0.125 = -1.125 < 0 \] - For \(0 < x < 4\), choose \(x=1\): \[ f(1) = 4(1) + 3(1) - 1 = 4 + 3 - 1 = 6 > 0 \] - For \(x > 4\), choose \(x=5\): \[ f(5) = 4(5) + 3(25) - 125 = 20 + 75 - 125 = -30 < 0 \] **Summary:** - \(f(x) > 0\) on \((-\infty, -1)\) and \((0, 4)\) - \(f(x) < 0\) on \((-1, 0)\) and \((4, \infty)\) --- ### **Step 3: Set up the integral for the area** Suppose the shaded region is between \(x = -1\) and \(x=4\), where the curve crosses the x-axis at these roots. The total area is: \[ \text{Area} = \int_{-1}^{0} |f(x)| dx + \int_{0}^{4} |f(x)| dx \] Since \(f(x) < 0\) on \((-1, 0)\), the absolute value becomes \(-f(x)\). And on \((0,4)\), \(f(x) > 0\), so the absolute value is just \(f(x)\). Thus: \[ \text{Area} = -\int_{-1}^{0} f(x) dx + \int_{0}^{4} f(x) dx \] --- ### **Step 4: Find the antiderivative \(F(x)\)** \[ f(x) = 4x + 3x^2 - x^3 \] \[ F(x) = \int f(x) dx = \int (4x + 3x^2 - x^3) dx \] Compute term-by-term: \[ F(x) = 2x^2 + x^3 - \frac{1}{4} x^4 + C \] --- ### **Step 5: Evaluate the integrals** **A. \(\int_{-1}^{0} f(x) dx\):** \[ F(0) = 2(0)^2 + (0)^3 - \frac{1}{4} (0)^4 = 0 \] \[ F(-1) = 2(-1)^2 + (-1)^3 - \frac{1}{4} (-1)^4 = 2(1) + (-1) - \frac{1}{4}(1) = 2 - 1 - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4} \] \[ \int_{-1}^{0} f(x) dx = F(0) - F(-1) = 0 - \frac{3}{4} = -\frac{3}{4} \] Since the integral is negative and we're taking the absolute value: \[ \text{Contribution} = - \left( -\frac{3}{4} \right) = \frac{3}{4} \] **B. \(\int_{0}^{4} f(x) dx\):** \[ F(4) = 2(4)^2 + (4)^3 - \frac{1}{4}(4)^4 = 2(16) + 64 - \frac{1}{4}(256) = 32 + 64 - 64 = 32 \] \[ F(0) = 0 \] \[ \int_{0}^{4} f(x) dx = F(4) - F(0) = 32 - 0 = 32 \] --- ### **Step 6: Calculate the total area** \[ \text{Total Area} = \frac{3}{4} + 32 = \frac{3}{4} + \frac{128}{4} = \frac{131}{4} \] ### **Final Answer:** \[ \boxed{\frac{131}{4}} \quad \text{or} \quad 32.75 \] --- ### **Summary:** - Roots at \(x = -1, 0, 4\) - Sign analysis shows where the function is positive or negative - Set up the integral accordingly - Calculate using the antiderivative \(F(x) = 2x^2 + x^3 - \frac{1}{4} x^4\) - Final area = \(\boxed{\frac{131}{4}}\) or 32.75 Let me know if you'd like any further clarification!