Let’s solve each part step by step with explanations and final answers. --- ### **1. Evaluate the integral:** \[ \int_{1}^{2} \frac{8}{x^3} \, dx \] **Step 1:** Rewrite the integrand: \[ \frac{8}{x^3} = 8x^{-3} \] So, \[ \int_{1}^{2} 8x^{-3} \, dx \] **Step 2:** Integrate: \[ \int 8x^{-3} dx = 8 \int x^{-3} dx = 8 \left( \frac{x^{-2}}{-2} \right) = -4x^{-2} \] **Step 3:** Evaluate from 1 to 2: \[ \left[ -4x^{-2} \right]_{1}^{2} = -4 \left(\frac{1}{2^2}\right) + 4 \left(\frac{1}{1^2}\right) = -4 \left(\frac{1}{4}\right) + 4 \times 1 \] \[ = -1 + 4 = 3 \] **Final Answer:** \[ \boxed{3} \] --- ### **2. Marginal Cost Function** Given: \( C'(x) = 0.0009x^2 - 0.18x + 24 \) Fixed Cost = $800 #### **(a) Total cost for the first 300 units** The total cost for the first 300 units: \[ C(300) = C(0) + \int_0^{300} C'(x) dx \] Here, \(C(0) = 800\) (fixed cost) Calculate the integral: \[ \int_0^{300} (0.0009x^2 - 0.18x + 24) dx = 0.0009\int_0^{300} x^2 dx - 0.18\int_0^{300} x dx + 24\int_0^{300} dx \] Evaluate each: - \(\int x^2 dx = \frac{x^3}{3}\) - \(\int x dx = \frac{x^2}{2}\) - \(\int dx = x\) Plug in limits: \[ \int_0^{300} x^2 dx = \frac{300^3}{3} - 0 = \frac{27,000,000}{3} = 9,000,000 \] \[ \int_0^{300} x dx = \frac{300^2}{2} - 0 = \frac{90,000}{2} = 45,000 \] \[ \int_0^{300} dx = 300 - 0 = 300 \] Plug back: \[ 0.0009 \times 9,000,000 - 0.18 \times 45,000 + 24 \times 300 \] \[ = 8,100 - 8,100 + 7,200 = 7,200 \] Add fixed cost: \[ C(300) = 800 + 7,200 = \boxed{8,000} \] --- #### **(b) Total cost from 201st to 300th unit** This is \(C(300) - C(200)\). First, compute \(\int_0^{200} C'(x) dx\): \[ \int_0^{200} x^2 dx = \frac{200^3}{3} = \frac{8,000,000}{3} \approx 2,666,666.67 \] \[ \int_0^{200} x dx = \frac{200^2}{2} = 20,000 \] \[ \int_0^{200} dx = 200 \] Plug back: \[ 0.0009 \times 2,666,666.67 - 0.18 \times 20,000 + 24 \times 200 \] \[ = 2,400 - 3,600 + 4,800 = 3,600 \] So, \(C(200) = 800 + 3,600 = 4,400\) Thus, \[ C(300) - C(200) = 8,000 - 4,400 = \boxed{3,600} \] --- ### **3. Marginal Revenue Function** Given: \( R'(x) = -0.1x + 40 \) #### **(a) Total revenue from sale of 150 units** \[ R(150) = \int_0^{150} (-0.1x + 40) dx \] Integrate: \[ \int (-0.1x + 40) dx = -0.05x^2 + 40x \] Evaluate from 0 to 150: \[ (-0.05 \times 150^2 + 40 \times 150) - (0) \] \[ = (-0.05 \times 22,500) + 6,000 = (-1,125) + 6,000 = \boxed{4,875} \] --- #### **(b) Additional revenue from 150 to 250 units** \[ \int_{150}^{250} (-0.1x + 40) dx = \left[ -0.05x^2 + 40x \right]_{150}^{250} \] \[ = \left(-0.05 \times 250^2 + 40 \times 250 \right) - \left(-0.05 \times 150^2 + 40 \times 150 \right) \] Calculate values: - \(250^2 = 62,500\) - \(150^2 = 22,500\) Plug in: - For \(x = 250\): \(-0.05 \times 62,500 + 10,000 = -3,125 + 10,000 = 6,875\) - For \(x = 150\): Already calculated above = 4,875 Difference: \[ 6,875 - 4,875 = \boxed{2,000} \] --- ## **Summary of Answers** 1. **\(\int_{1}^{2} \frac{8}{x^3} dx = 3\)** 2. **(a) Total cost for first 300 units: $8,000** **(b) Cost for 201st to 300th units: $3,600** 3. **(a) Total revenue from 150 units: $4,875** **(b) Additional revenue from 150th to 250th unit: $2,000** Let me know if you need the step-by-step written for any individual part!