1. **Problem 1:** Evaluate the integral for $t_{avg}^2 = \int_1^4 t (\sqrt{x} - \frac{1}{x}) \, dt$. 2. The integral expression and evaluation given are inconsistent and contain errors in variable usage and limits. Let's clarify and solve the integral correctly. 3. Since the integral is with respect to $t$ but the integrand contains $x$, we assume the integral should be with respect to $x$ from 1 to 4: $$t_{avg}^2 = \int_1^4 t (\sqrt{x} - \frac{1}{x}) \, dx$$ If $t$ is a constant or function of $x$, it must be specified. Assuming $t$ is constant or $t=x$, let's solve for $\int_1^4 (\sqrt{x} - \frac{1}{x}) \, dx$. 4. Compute: $$\int_1^4 \sqrt{x} \, dx = \int_1^4 x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_1^4 = \frac{2}{3} (4^{3/2} - 1^{3/2}) = \frac{2}{3} (8 - 1) = \frac{14}{3}$$ $$\int_1^4 \frac{1}{x} \, dx = [\ln x]_1^4 = \ln 4 - \ln 1 = \ln 4$$ 5. Therefore, $$\int_1^4 (\sqrt{x} - \frac{1}{x}) \, dx = \frac{14}{3} - \ln 4$$ 6. **Problem 2A:** Compute the area bounded by $y = 4x^3 - x^5$ and the x-axis. 7. Find roots of $y=0$: $$4x^3 - x^5 = x^3(4 - x^2) = 0 \implies x=0, x=\pm 2$$ 8. The function crosses the x-axis at $x=-2, 0, 2$. The area bounded is: $$A = \int_{-2}^0 |4x^3 - x^5| \, dx + \int_0^2 |4x^3 - x^5| \, dx$$ 9. Check sign of $y$ in intervals: - For $x$ in $(0,2)$, $4x^3 - x^5 > 0$ (since $x^2 < 4$) - For $x$ in $(-2,0)$, $4x^3 - x^5 < 0$ (since $x^3$ negative, $x^5$ negative, but $4x^3$ dominates) 10. So, $$A = -\int_{-2}^0 (4x^3 - x^5) \, dx + \int_0^2 (4x^3 - x^5) \, dx$$ 11. Compute each integral: $$\int (4x^3 - x^5) \, dx = x^4 - \frac{x^6}{6} + C$$ 12. Evaluate: $$-\int_{-2}^0 (4x^3 - x^5) \, dx = -\left[ x^4 - \frac{x^6}{6} \right]_{-2}^0 = -\left(0 - \left(16 - \frac{64}{6}\right)\right) = 16 - \frac{64}{6} = \frac{96}{6} - \frac{64}{6} = \frac{32}{6} = \frac{16}{3}$$ $$\int_0^2 (4x^3 - x^5) \, dx = \left[ x^4 - \frac{x^6}{6} \right]_0^2 = \left(16 - \frac{64}{6}\right) - 0 = \frac{96}{6} - \frac{64}{6} = \frac{32}{6} = \frac{16}{3}$$ 13. Total area: $$A = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$$ 14. **Problem 2B:** Compute the area between $g(x) = 15 - x^2$ and $h(x) = x$ from $x=-1$ to $x=3$. 15. The area is: $$A = \int_{-1}^3 [(15 - x^2) - x] \, dx = \int_{-1}^3 (15 - x^2 - x) \, dx$$ 16. Integrate: $$\int (15 - x^2 - x) \, dx = 15x - \frac{x^3}{3} - \frac{x^2}{2} + C$$ 17. Evaluate: $$A = \left[15x - \frac{x^3}{3} - \frac{x^2}{2}\right]_{-1}^3 = \left(45 - 9 - \frac{9}{2}\right) - \left(-15 + \frac{1}{3} - \frac{1}{2}\right)$$ 18. Simplify: $$45 - 9 - 4.5 = 31.5$$ $$-15 + 0.3333 - 0.5 = -15.1667$$ 19. Area: $$31.5 - (-15.1667) = 31.5 + 15.1667 = 46.6667 = \frac{140}{3}$$ 20. **Using cylindrical shell method:** For the area between curves, cylindrical shells are typically used for volumes of revolution. Since the problem asks for area, the integral method above suffices. **Final answers:** - Problem 1 integral: $\frac{14}{3} - \ln 4$ - Problem 2A area: $\frac{32}{3}$ square units - Problem 2B area: $\frac{140}{3}$ square units