1. **Problem 1: Solve for $b$ given the integral equation** Given: $$\frac{1}{b-a} \int_a^b f(x) \, dx = 2$$ with $a=0$ and $f(x) = 2 + 7x - x^3$, and the equation: $$(2 + 7b - b^3) = 16$$ 2. **Step 1: Write the integral and evaluate it** $$\int_0^b (2 + 7x - x^3) \, dx = \left[2x + \frac{7x^2}{2} - \frac{x^4}{4}\right]_0^b = 2b + \frac{7b^2}{2} - \frac{b^4}{4}$$ 3. **Step 2: Use the average value formula** $$\frac{1}{b-0} \int_0^b f(x) \, dx = 2 \implies \frac{1}{b} \left(2b + \frac{7b^2}{2} - \frac{b^4}{4}\right) = 2$$ 4. **Step 3: Simplify the equation** $$2 + \frac{7b}{2} - \frac{b^3}{4} = 2$$ 5. **Step 4: Subtract 2 from both sides** $$\frac{7b}{2} - \frac{b^3}{4} = 0$$ 6. **Step 5: Multiply both sides by 4 to clear denominators** $$4 \times \left(\frac{7b}{2} - \frac{b^3}{4}\right) = 4 \times 0$$ $$2 \times 7b - b^3 = 0$$ $$14b - b^3 = 0$$ 7. **Step 6: Factor out $b$** $$b(14 - b^2) = 0$$ 8. **Step 7: Solve for $b$** Either $b=0$ (not valid since denominator would be zero) or $$14 - b^2 = 0 \implies b^2 = 14 \implies b = \sqrt{14} \approx 3.74$$ --- **Final answer for problem 1:** $$b = \sqrt{14} \approx 3.74$$ --- 1. **Problem 2: Find the position of the particle at $t = \frac{3\pi}{2}$ given velocity and initial position** Given velocity: $$v(t) = 1 - \sin t$$ Initial position: $$x(\pi) = \pi$$ 2. **Step 1: Write the position function as an integral of velocity** $$x(t) = x(\pi) + \int_{\pi}^t v(s) \, ds = \pi + \int_{\pi}^t (1 - \sin s) \, ds$$ 3. **Step 2: Compute the integral** $$\int_{\pi}^t (1 - \sin s) \, ds = \int_{\pi}^t 1 \, ds - \int_{\pi}^t \sin s \, ds = (t - \pi) + (\cos t - \cos \pi)$$ 4. **Step 3: Substitute $t = \frac{3\pi}{2}$** $$x\left(\frac{3\pi}{2}\right) = \pi + \left(\frac{3\pi}{2} - \pi\right) + \left(\cos \frac{3\pi}{2} - \cos \pi\right)$$ 5. **Step 4: Evaluate cosines** $$\cos \frac{3\pi}{2} = 0, \quad \cos \pi = -1$$ 6. **Step 5: Simplify** $$x\left(\frac{3\pi}{2}\right) = \pi + \frac{\pi}{2} + (0 - (-1)) = \pi + \frac{\pi}{2} + 1 = \frac{3\pi}{2} + 1$$ --- **Final answer for problem 2:** $$x\left(\frac{3\pi}{2}\right) = \frac{3\pi}{2} + 1$$