1. **Problem 5.1:** Given $f(2)=5$ and $f(0)=1$, evaluate $$\int_0^2 f'(x) U(f(x)) \, dx$$ where $U$ is the unit step function. 2. **Step 1:** Recall the Fundamental Theorem of Calculus and substitution rule. If $U$ is the Heaviside step function, $U(f(x))=0$ if $f(x)<0$ and $1$ if $f(x)\geq0$. 3. **Step 2:** Since $f(0)=1>0$ and $f(2)=5>0$, $f(x)$ is positive on $[0,2]$, so $U(f(x))=1$ on $[0,2]$. 4. **Step 3:** The integral simplifies to $$\int_0^2 f'(x) \cdot 1 \, dx = \int_0^2 f'(x) \, dx = f(2)-f(0) = 5-1 = 4.$$ --- 5. **Problem 5.2:** Show that if $$\int_0^2 f'(c) \, dc = p$$ then $$\int_0^2 f'(c) \, dc = \frac{p}{2}$$. 6. **Step 1:** The statement as given is contradictory unless it means something else. Possibly a typo or missing context. 7. **Step 2:** By the Fundamental Theorem of Calculus, $$\int_0^2 f'(c) \, dc = f(2)-f(0) = p.$$ So the integral equals $p$, not $p/2$. 8. **Step 3:** Without additional context, this cannot be shown as stated. --- 9. **Problem 5.3:** Given $f(2)=5$ and $f(0)=1$, evaluate $$\int_0^2 f'(x) U(f(x))^2 \, dx$$. 10. **Step 1:** Since $U(f(x))$ is 1 on $[0,2]$ (as above), $U(f(x))^2 = 1^2 = 1$. 11. **Step 2:** The integral reduces to $$\int_0^2 f'(x) \, dx = f(2)-f(0) = 5-1 = 4.$$ --- 12. **Problem 5.3.1:** Calculate the magnitude of the bounded area between curves $$y=3x^2+5x+2$$ and $$y=-x$$ between their points of intersection. 13. **Step 1:** Find points of intersection by solving $$3x^2+5x+2 = -x$$. 14. **Step 2:** Rearranged: $$3x^2 + 6x + 2 = 0$$. 15. **Step 3:** Use quadratic formula: $$x = \frac{-6 \pm \sqrt{36 - 24}}{6} = \frac{-6 \pm \sqrt{12}}{6} = \frac{-6 \pm 2\sqrt{3}}{6} = -1 \pm \frac{\sqrt{3}}{3}.$$ 16. **Step 4:** The intersection points are approximately $x_1 = -1.577$ and $x_2 = -0.422$. 17. **Step 5:** The bounded area is $$A = \int_{x_1}^{x_2} \left| (3x^2 + 5x + 2) - (-x) \right| \, dx = \int_{x_1}^{x_2} (3x^2 + 6x + 2) \, dx$$ since $3x^2+6x+2 \geq 0$ in this interval. 18. **Step 6:** Compute the integral: $$A = \left[ x^3 + 3x^2 + 2x \right]_{x_1}^{x_2} = (x_2^3 + 3x_2^2 + 2x_2) - (x_1^3 + 3x_1^2 + 2x_1).$$ 19. **Step 7:** Substitute values: $x_1 \approx -1.577$, $x_2 \approx -0.422$. Calculate: $x_2^3 = (-0.422)^3 \approx -0.075$, $3x_2^2 = 3 \times 0.178 = 0.534$, $2x_2 = -0.844$. Sum: $-0.075 + 0.534 - 0.844 = -0.385$. $x_1^3 = (-1.577)^3 \approx -3.924$, $3x_1^2 = 3 \times 2.487 = 7.461$, $2x_1 = -3.154$. Sum: $-3.924 + 7.461 - 3.154 = 0.383$. 20. **Step 8:** Area magnitude: $$A = -0.385 - 0.383 = -0.768.$$ The area is positive magnitude $$0.768$$ (approx). --- 21. **Problem 5.3.2:** Calculate the volume of the solid of revolution formed when the area in 5.3.1 is revolved about the x-axis. 22. **Step 1:** Use the disk method: $$V = \pi \int_{x_1}^{x_2} \left[ (3x^2 + 5x + 2)^2 - (-x)^2 \right] \, dx = \pi \int_{x_1}^{x_2} \left[ (3x^2 + 5x + 2)^2 - x^2 \right] \, dx.$$ 23. **Step 2:** Expand: $$(3x^2 + 5x + 2)^2 = 9x^4 + 30x^3 + 37x^2 + 20x + 4.$$ 24. **Step 3:** Subtract $x^2$: $$9x^4 + 30x^3 + 37x^2 + 20x + 4 - x^2 = 9x^4 + 30x^3 + 36x^2 + 20x + 4.$$ 25. **Step 4:** Integrate term-by-term: $$\int (9x^4 + 30x^3 + 36x^2 + 20x + 4) dx = \frac{9}{5}x^5 + \frac{30}{4}x^4 + 12x^3 + 10x^2 + 4x + C.$$ 26. **Step 5:** Evaluate from $x_1$ to $x_2$ and multiply by $\pi$: $$V = \pi \left[ \frac{9}{5}x^5 + \frac{30}{4}x^4 + 12x^3 + 10x^2 + 4x \right]_{x_1}^{x_2}.$$ 27. **Step 6:** Substitute $x_1 \approx -1.577$, $x_2 \approx -0.422$ and compute numerically for volume. --- 28. **Problem 5.4:** Calculate the moment of inertia $I$ of a circular lamina of radius $\frac{7}{2}$ and mass $m$ about an axis through its center and perpendicular to the plane. 29. **Step 1:** The moment of inertia for a solid disk about its center axis is $$I = \frac{1}{2} m r^2.$$ 30. **Step 2:** Substitute $r = \frac{7}{2} = 3.5$: $$I = \frac{1}{2} m (3.5)^2 = \frac{1}{2} m \times 12.25 = 6.125 m.$$ 31. **Final answers:** - 5.1: $4$ - 5.2: Cannot prove as stated; integral equals $p$ - 5.3: $4$ - 5.3.1: Area $\approx 0.768$ - 5.3.2: Volume $= \pi \left[ \frac{9}{5}x^5 + \frac{30}{4}x^4 + 12x^3 + 10x^2 + 4x \right]_{-1.577}^{-0.422}$ - 5.4: $I = 6.125 m$