Let's solve the problem step by step: Given functions: - \( f(x) = \frac{x^3 - 4x^2 - 12x}{12} \) - \( g(x) = \frac{-x^2 + 4x + 12}{4} \) - Interval: \( -2 \leq x \leq 6 \) --- ### **a. Sketch/graph the two curves on the same axes** To sketch: - Both functions are cubic/quadratic, so expect some curvature. - \( f(x) \) is a cubic divided by 12. - \( g(x) \) is a downward-facing parabola shifted and scaled. For actual plotting, you can use a graphing calculator or software, but here's a rough sketch idea: - **Plot \( f(x) \) and \( g(x) \) on the same axes from \( x = -2 \) to \( x = 6 \).** - The area between the curves is what we want to find. --- ### **b. Use the Fundamental Theorem of Calculus to calculate the area between the curves** The area between the curves from \( x = -2 \) to \( x = 6 \) is: \[ A = \int_{-2}^{6} [g(x) - f(x)] \, dx \] Let's compute \( g(x) - f(x) \): First, express both functions with common denominators: - \( f(x) = \frac{x^3 - 4x^2 - 12x}{12} \) - \( g(x) = \frac{-x^2 + 4x + 12}{4} = \frac{-3x^2 + 12x + 36}{12} \) So, \[ g(x) - f(x) = \frac{-3x^2 + 12x + 36 - (x^3 - 4x^2 - 12x)}{12} \] \[ = \frac{-3x^2 + 12x + 36 - x^3 + 4x^2 + 12x}{12} \] \[ = \frac{-x^3 + (-3x^2 + 4x^2) + (12x + 12x) + 36}{12} \] \[ = \frac{-x^3 + x^2 + 24x + 36}{12} \] So, \[ A = \int_{-2}^{6} \frac{-x^3 + x^2 + 24x + 36}{12} dx \] Let's integrate: \[ \int \frac{-x^3 + x^2 + 24x + 36}{12} dx = \frac{1}{12} \int (-x^3 + x^2 + 24x + 36) dx \] \[ = \frac{1}{12} \left( -\frac{x^4}{4} + \frac{x^3}{3} + 12x^2 + 36x \right) \] Evaluate from \( x = -2 \) to \( x = 6 \): 1. **At \( x = 6 \):** \[ -\frac{6^4}{4} + \frac{6^3}{3} + 12 \cdot 6^2 + 36 \cdot 6 = -\frac{1296}{4} + 72 + 432 + 216 \] \[ = -324 + 72 + 432 + 216 = 396 \] 2. **At \( x = -2 \):** \[ -\frac{(-2)^4}{4} + \frac{(-2)^3}{3} + 12 \cdot (-2)^2 + 36 \cdot (-2) \] \[ = -\frac{16}{4} - \frac{8}{3} + 48 - 72 \] \[ = -4 - \frac{8}{3} + 48 - 72 \] \[ = (-28) - \frac{8}{3} \] \[ = -\frac{84}{3} - \frac{8}{3} = -\frac{92}{3} \] 3. **Subtract:** \[ A = \frac{1}{12} [396 - (-\frac{92}{3})] \] \[ = \frac{1}{12} [396 + \frac{92}{3}] \] \[ = \frac{1}{12} [\frac{1188}{3} + \frac{92}{3}] \] \[ = \frac{1}{12} \cdot \frac{1280}{3} \] \[ = \frac{1280}{36} \] \[ = \frac{320}{9} \approx \boxed{35.56} \] --- ### **c. Approximate the area using Riemann sums with \( n = 4 \)** - Interval: \( -2 \) to \( 6 \) ⇒ width \( = 8 \) - \( \Delta x = \frac{6 - (-2)}{4} = 2 \) - Left endpoints: \( -2, 0, 2, 4 \) For each \( x_i \), compute \( g(x_i) - f(x_i) \): Let’s calculate for each \( x_i \): | \( x \) | \( f(x) \) | \( g(x) \) | \( g(x) - f(x) \) | |--------|------------|------------|-------------------| | -2 | \(\frac{-8 - 16 + 24}{12} = 0\) | \(\frac{-4 -8 + 12}{4} = 0\) | 0 | | 0 | 0 | 3 | 3 | | 2 | \(\frac{8 - 16 - 24}{12} = \frac{-32}{12} = -\frac{8}{3}\) | \(\frac{-4 + 8 + 12}{4} = 4\) | \(4 - (-\frac{8}{3}) = \frac{20}{3}\) | | 4 | \(\frac{64 - 64 - 48}{12} = \frac{-48}{12} = -4\) | \(\frac{-16 + 16 + 12}{4} = 3\) | \(3 - (-4) = 7\) | Now, sum the heights and multiply by \( \Delta x = 2 \): \[ \text{Area} \approx 2 \left[ 0 + 3 + \frac{20}{3} + 7 \right] \] \[ = 2 \left[ 10 + \frac{20}{3} \right] \] \[ = 2 \left[ \frac{30}{3} + \frac{20}{3} \right] = 2 \left[ \frac{50}{3} \right] = \frac{100}{3} \approx \boxed{33.33} \] --- ### **d. Approximate the area using the Trapezoidal Rule with \( n = 4 \)** Trapezoidal Rule: \[ A \approx \frac{\Delta x}{2} \left[ y_0 + 2y_1 + 2y_2 + 2y_3 + y_4 \right] \] Endpoints for \( n=4 \): \( x = -2, 0, 2, 4, 6 \) Calculate \( y = g(x) - f(x) \) at each: - \( x = -2 \): \( 0 \) - \( x = 0 \): \( 3 \) - \( x = 2 \): \( \frac{20}{3} \) - \( x = 4 \): \( 7 \) - \( x = 6 \): For \( f(6): \frac{216 - 144 - 72}{12} = 0 \) For \( g(6): \frac{-36 + 24 + 12}{4} = 0 \) So \( 0 - 0 = 0 \) Plug in: \[ A \approx \frac{2}{2}[0 + 2 \cdot 3 + 2 \cdot \frac{20}{3} + 2 \cdot 7 + 0] \] \[ = 1[0 + 6 + \frac{40}{3} + 14 + 0] \] \[ = 6 + 14 + \frac{40}{3} = 20 + \frac{40}{3} = \frac{60}{3} + \frac{40}{3} = \frac{100}{3} \approx \boxed{33.33} \] --- ### **e. Trapezoidal Rule with \( n = 8 \)** - \( \Delta x = 1 \) - Points: \( x = -2, -1, 0, 1, 2, 3, 4, 5, 6 \) Calculate \( g(x) - f(x) \) for all these points (you can use the formula above). For brevity, let's denote these as \( y_0, y_1, \dots, y_8 \). \[ A \approx \frac{1}{2} [y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + 2y_5 + 2y_6 + 2y_7 + y_8] \] Compute and sum each value. --- ### **Summary Table** | Part | Answer | |------|--------| | (a) | Graph as above | | (b) | \( \frac{320}{9} \approx 35.56 \) | | (c) | \( \frac{100}{3} \approx 33.33 \) | | (d) | \( \frac{100}{3} \approx 33.33 \) | | (e) | Compute as above using more points; value will be closer to 35.56 | Let me know if you want all the detailed calculations for part (e)!

Let's carefully solve the problem step-by-step, including detailed calculations for each part. --- ## Given Functions: \[ f(x) = \frac{x^3 - 4x^2 - 12x}{12} \] \[ g(x) = \frac{-x^2 + 4x + 12}{4} \] Interval: \( x \in [-2, 6] \) --- ## Part (a): Sketch the curves - To sketch, find key points (intercepts and critical points): - For \( f(x) \), plug in some values. - For \( g(x) \), plug in some values. - Since actual plotting isn't feasible here, note that both are polynomial functions: - \( f(x) \) is cubic/degree 3. - \( g(x) \) is quadratic (parabola). We'll use specific points in subsequent parts. --- ## Part (b): Area between the curves using the Fundamental Theorem of Calculus ### Step 1: Find the difference \( g(x) - f(x) \) Express both functions with common denominator 12: \[ f(x) = \frac{x^3 - 4x^2 - 12x}{12} \] \[ g(x) = \frac{-x^2 + 4x + 12}{4} = \frac{-3x^2 + 12x + 36}{12} \] Subtract: \[ g(x) - f(x) = \frac{-3x^2 + 12x + 36 - (x^3 - 4x^2 - 12x)}{12} \] \[ = \frac{-3x^2 + 12x + 36 - x^3 + 4x^2 + 12x}{12} \] \[ = \frac{-x^3 + (-3x^2 + 4x^2) + (12x + 12x) + 36}{12} \] \[ = \frac{-x^3 + x^2 + 24x + 36}{12} \] ### Step 2: Set up the integral \[ A = \int_{-2}^{6} [g(x) - f(x)] dx = \int_{-2}^{6} \frac{-x^3 + x^2 + 24x + 36}{12} dx \] Pull out the constant: \[ A = \frac{1}{12} \int_{-2}^{6} (-x^3 + x^2 + 24x + 36) dx \] ### Step 3: Integrate term-by-term \[ \int -x^3 dx = -\frac{x^4}{4} \] \[ \int x^2 dx = \frac{x^3}{3} \] \[ \int 24x dx = 12x^2 \] \[ \int 36 dx = 36x \] So, the indefinite integral: \[ I(x) = -\frac{x^4}{4} + \frac{x^3}{3} + 12x^2 + 36x \] ### Step 4: Evaluate at bounds \[ A = \frac{1}{12} [I(6) - I(-2)] \] Calculate \( I(6) \): \[ I(6) = -\frac{6^4}{4} + \frac{6^3}{3} + 12 \cdot 6^2 + 36 \cdot 6 \] \[ 6^4 = 1296 \] \[ I(6) = -\frac{1296}{4} + \frac{216}{3} + 12 \cdot 36 + 216 \] \[ = -324 + 72 + 432 + 216 \] \[ = (-324 + 72) + 432 + 216 = -252 + 432 + 216 = 180 + 216 = 396 \] Calculate \( I(-2) \): \[ I(-2) = -\frac{(-2)^4}{4} + \frac{(-2)^3}{3} + 12 \cdot (-2)^2 + 36 \cdot (-2) \] \[ (-2)^4 = 16 \] \[ I(-2) = -\frac{16}{4} + \frac{-8}{3} + 12 \cdot 4 - 72 \] \[ = -4 - \frac{8}{3} + 48 - 72 \] \[ = (-4 + 48 - 72) - \frac{8}{3} = (-28) - \frac{8}{3} \] Expressing \( -28 \) as fraction: \[ -28 = -\frac{84}{3} \] So, \[ I(-2) = -\frac{84}{3} - \frac{8}{3} = -\frac{92}{3} \] ### Step 5: Final calculation \[ A = \frac{1}{12} \left( 396 - (-\frac{92}{3}) \right) = \frac{1}{12} \left( 396 + \frac{92}{3} \right) \] Express 396 as fraction with denominator 3: \[ 396 = \frac{1188}{3} \] Sum: \[ \frac{1188}{3} + \frac{92}{3} = \frac{1280}{3} \] Divide by 12: \[ A = \frac{1}{12} \times \frac{1280}{3} = \frac{1280}{36} = \frac{320}{9} \approx 35.56 \] --- ## **Final answer for (b):** \[ \boxed{ \text{Area} \approx \frac{320}{9} \approx 35.56 } \] --- ## Part (c): Approximate the area using **Riemann sums with \( n=4 \)** ### Step 1: Determine \(\Delta x\) Interval: \( -2 \) to \( 6 \) \[ \Delta x = \frac{6 - (-2)}{4} = \frac{8}{4} = 2 \] ### Step 2: Left endpoints: \[ x_0 = -2,\quad x_1=0,\quad x_2=2,\quad x_3=4 \] ### Step 3: Compute \( g(x_i) - f(x_i) \): Recall: \[ g(x) - f(x) = \frac{-x^3 + x^2 + 24x + 36}{12} \] At each \( x_i \): - **At \( x = -2 \):** \[ g(-2) - f(-2) = \frac{-(-2)^3 + (-2)^2 + 24(-2) + 36}{12} \] \[ = \frac{-(-8) + 4 - 48 + 36}{12} = \frac{8 + 4 - 48 +36}{12} = \frac{0}{12} = 0 \] - **At \( x=0 \):** \[ \frac{0 + 0 + 0 + 36}{12} = \frac{36}{12} = 3 \] - **At \( x=2 \):** \[ \frac{-8 + 4 + 48 + 36}{12} = \frac{80}{12} = \frac{20}{3} \] - **At \( x=4 \):** \[ \frac{-64 + 16 + 96 + 36}{12} = \frac{84}{12} = 7 \] ### Step 4: Riemann sum: \[ A \approx \Delta x \times \left[ y_0 + y_1 + y_2 + y_3 \right] \] But for left Riemann sum: \[ A \approx \Delta x \times ( y_0 + y_1 + y_2 + y_3 ) \] which simplifies to: \[ A \approx 2 \times ( 0 + 3 + \frac{20}{3} + 7 ) \] \[ = 2 \times \left( 10 + \frac{20}{3} \right) = 2 \times \left( \frac{30}{3} + \frac{20}{3} \right) = 2 \times \frac{50}{3} = \frac{100}{3} \approx 33.33 \] --- ## **Final answer for (c):** \[ \boxed{ \text{Approximate area} \approx \frac{100}{3} \approx 33.33 } \] --- ## Part (d): Use the **Trapezoidal Rule with \( n=4 \)** ### Step 1: Endpoints \( x = -2, 0, 2, 4, 6 \) Calculate \( g(x) - f(x) \): - **At \( x=-2 \):** 0 (from previous) - **At \( x=0 \):** 3 - **At \( x=2 \):** \( \frac{20}{3} \) - **At \( x=4 \):** 7 - **At \( x=6 \):** \[ g(6)-f(6) = 0 \quad (\text{from previous calculation}) \] ### Step 2: Trapezoidal sum: \[ A \approx \frac{\Delta x}{2} \left[ y_0 + 2 y_1 + 2 y_2 + 2 y_3 + y_4 \right] \] where: \[ x_0 = -2,\quad y_0=0 \] \[ x_1=0,\quad y_1=3 \] \[ x_2=2,\quad y_2=\frac{20}{3} \] \[ x_3=4,\quad y_3=7 \] \[ x_4=6,\quad y_4=0 \] Calculate: \[ A \approx \frac{2}{2} \left[ 0 + 2 \times 3 + 2 \times \frac{20}{3} + 2 \times 7 + 0 \right] \] \[ = 1 \times \left[ 0 + 6 + \frac{40}{3} + 14 \right] \] \[ = 6 + 14 + \frac{40}{3} = 20 + \frac{40}{3} = \frac{60}{3} + \frac{40}{3} = \frac{100}{3} \approx 33.33 \] ### **Final for (d):** \[ \boxed{ \text{Approximate area} \approx \frac{100}{3} \approx 33.33 } \] --- ## **Part (e): Using **\( n=8 \)** subintervals for better approximation** - \(\Delta x = 1\) - Points: \( x=-2, -1, 0, 1, 2, 3, 4, 5, 6 \) Calculate \( g(x_i) - f(x_i) \) at each point: | \( x \) | \( g(x)-f(x) \) calculation | |---------|------------------------------| | -2 | 0 (from previous) | | -1 | \[ g(-1)-f(-1) = \frac{-(-1)^3 + (-1)^2 + 24(-1) +36}{12} = \frac{1 + 1 -24 +36}{12} = \frac{14}{12}=\frac{7}{6} \] | 0 | 3 (from previous) | | 1 | \[ g(1)-f(1) = \frac{-1 + 1 + 24(1) +36}{12} = \frac{0 + 24 +36}{12} = \frac{60}{12} = 5 \] | 2 | \( \frac{20}{3} \) (from previous) | | 3 | \[ g(3)-f(3) = \frac{-27 + 9 + 72 +36}{12} = \frac{90}{12} = 7.5 \] | 4 | 7 (from previous) | | 5 | \[ g(5)-f(5) = \frac{-125 + 25 + 120 + 36}{12} = \frac{56}{12} = \frac{14}{3} \] | 6 | 0 (from previous) | --- ### Step 1: Sum for Trapezoidal Rule: \[ A \approx \frac{\Delta x}{2} \left[ y_0 + 2 y_1 + 2 y_2 + 2 y_3 + 2 y_4 + 2 y_5 + 2 y_6 + 2 y_7 + y_8 \right] \] where: \[ x_i: -2, -1, 0, 1, 2, 3, 4, 5, 6 \] \[ y_i: 0,\ \frac{7}{6}, 3,\ 5,\ \frac{20}{3},\ 7.5,\ 7,\ \frac{14}{

Apologies for the interruption. Continuing from where I left off: We are calculating the area using the **Trapezoidal Rule with \( n=8 \)** subintervals. ### Step 2: List the function values \( y_i = g(x_i) - f(x_i) \): | \( x_i \) | Calculation of \( y_i = g(x_i) - f(x_i) \) | Result | |------------|----------------------------------------------|---------| | -2 | 0 (from previous calculation) | 0 | | -1 | \(\frac{1 + 1 - 24 + 36}{12} = \frac{14}{12} = \frac{7}{6}\) | \( \frac{7}{6} \) | | 0 | 3 | 3 | | 1 | \(\frac{0 + 24 + 36}{12} = \frac{60}{12} = 5\) | 5 | | 2 | \(\frac{ -8 + 4 + 48 + 36 }{12} = \frac{80}{12} = \frac{20}{3}\) | \( \frac{20}{3} \) | | 3 | \(\frac{ -27 + 9 + 72 + 36 }{12} = \frac{90}{12} = 7.5\) | 7.5 | | 4 | 7 (from previous) | 7 | | 5 | \(\frac{ -125 + 25 + 120 + 36 }{12} = \frac{56}{12} = \frac{14}{3}\) | \( \frac{14}{3} \) | | 6 | 0 (from previous) | 0 | ### Step 3: Apply the Trapezoidal Rule formula: \[ A \approx \frac{\Delta x}{2} \left[ y_0 + 2 y_1 + 2 y_2 + 2 y_3 + 2 y_4 + 2 y_5 + 2 y_6 + 2 y_7 + y_8 \right] \] where \(\Delta x = 1\). Now, sum the weighted function values: \[ \text{Sum} = y_0 + 2 y_1 + 2 y_2 + 2 y_3 + 2 y_4 + 2 y_5 + 2 y_6 + 2 y_7 + y_8 \] Plugging in the values: \[ = 0 + 2 \times \frac{7}{6} + 2 \times 3 + 2 \times 5 + 2 \times \frac{20}{3} + 2 \times 7.5 + 2 \times 7 + 2 \times \frac{14}{3} + 0 \] Calculate each term: - \( 2 \times \frac{7}{6} = \frac{14}{6} = \frac{7}{3} \) - \( 2 \times 3 = 6 \) - \( 2 \times 5 = 10 \) - \( 2 \times \frac{20}{3} = \frac{40}{3} \) - \( 2 \times 7.5 = 15 \) - \( 2 \times 7 = 14 \) - \( 2 \times \frac{14}{3} = \frac{28}{3} \) - Last term: 0 Now sum all: \[ \text{Sum} = 0 + \frac{7}{3} + 6 + 10 + \frac{40}{3} + 15 + 14 + \frac{28}{3} + 0 \] Combine fractions: \[ \frac{7}{3} + \frac{40}{3} + \frac{28}{3} = \frac{7 + 40 + 28}{3} = \frac{75}{3} = 25 \] Sum the integer parts: \[ 6 + 10 + 15 + 14 = 45 \] Total sum: \[ 45 + 25 = 70 \] Multiply by \(\frac{\Delta x}{2} = \frac{1}{2}\): \[ A \approx \frac{1}{2} \times 70 = 35 \] --- ## **Final approximation with \( n=8 \):** \[ \boxed{ \text{Area} \approx 35 } \] --- ## **Summary of Results:** | Part | Description | Final Result | |---------|--------------|--------------| | (b) | Exact area using calculus | \(\frac{320}{9} \approx 35.56\) | | (c) | Riemann sum with \( n=4 \) | \(\frac{100}{3} \approx 33.33\) | | (d) | Trapezoidal rule with \( n=4 \) | \(\frac{100}{3} \approx 33.33\) | | (e) | Trapezoidal rule with \( n=8 \) | **Approximate area = 35** | --- **Note:** The small differences are due to approximation methods. As \( n \) increases, the approximation gets closer to the exact value. If you'd like any further details or clarifications, please let me know!