1. **Problem (a):** Determine $\int (1 - t)^2 \, dt$. 2. **Formula and rules:** Use the power rule for integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$. 3. **Step-by-step solution (a):** - Expand the integrand: $(1 - t)^2 = 1 - 2t + t^2$. - Integrate term-by-term: $$\int (1 - 2t + t^2) \, dt = \int 1 \, dt - 2 \int t \, dt + \int t^2 \, dt$$ - Calculate each integral: $$= t - 2 \cdot \frac{t^2}{2} + \frac{t^3}{3} + C = t - t^2 + \frac{t^3}{3} + C$$ 4. **Answer (a):** $$\int (1 - t)^2 \, dt = t - t^2 + \frac{t^3}{3} + C$$ --- 5. **Problem (b)(i):** Evaluate $\int_1^3 (x^2 - 4x + 3) \, dx$. 6. **Step-by-step solution (b)(i):** - Integrate the function: $$\int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x + C$$ - Evaluate definite integral from 1 to 3: $$\left[ \frac{x^3}{3} - 2x^2 + 3x \right]_1^3 = \left( \frac{27}{3} - 2 \cdot 9 + 9 \right) - \left( \frac{1}{3} - 2 + 3 \right)$$ $$= (9 - 18 + 9) - \left( \frac{1}{3} + 1 \right) = 0 - \frac{4}{3} = -\frac{4}{3}$$ 7. **Answer (b)(i):** $$\int_1^3 (x^2 - 4x + 3) \, dx = -\frac{4}{3}$$ --- 8. **Problem (b)(ii):** Evaluate $\int_0^2 3 \sin t \, dt$. 9. **Step-by-step solution (b)(ii):** - Integrate: $$\int 3 \sin t \, dt = -3 \cos t + C$$ - Evaluate definite integral from 0 to 2: $$\left[-3 \cos t \right]_0^2 = -3 \cos 2 + 3 \cos 0 = -3 \cos 2 + 3$$ 10. **Answer (b)(ii):** $$\int_0^2 3 \sin t \, dt = 3 - 3 \cos 2$$ --- 11. **Problem (c):** Sketch curves $y = x^2 + 3$ and $y = 7 - 3x$ and find the area enclosed by integration. 12. **Step 1: Find intersection points by solving:** $$x^2 + 3 = 7 - 3x$$ $$x^2 + 3x - 4 = 0$$ Factor: $$(x + 4)(x - 1) = 0$$ So, $x = -4$ or $x = 1$. 13. **Step 2: Set up the integral for the area between curves:** The line is above the parabola between $x = -4$ and $x = 1$, so area: $$A = \int_{-4}^1 \left[(7 - 3x) - (x^2 + 3)\right] \, dx = \int_{-4}^1 (4 - 3x - x^2) \, dx$$ 14. **Step 3: Integrate:** $$\int (4 - 3x - x^2) \, dx = 4x - \frac{3x^2}{2} - \frac{x^3}{3} + C$$ 15. **Step 4: Evaluate definite integral:** $$\left[4x - \frac{3x^2}{2} - \frac{x^3}{3}\right]_{-4}^1 = \left(4(1) - \frac{3(1)^2}{2} - \frac{1^3}{3}\right) - \left(4(-4) - \frac{3(-4)^2}{2} - \frac{(-4)^3}{3}\right)$$ $$= \left(4 - \frac{3}{2} - \frac{1}{3}\right) - \left(-16 - \frac{3 \cdot 16}{2} + \frac{64}{3}\right)$$ $$= \left(4 - 1.5 - 0.3333\right) - \left(-16 - 24 + 21.3333\right)$$ $$= 2.1667 - (-18.6667) = 2.1667 + 18.6667 = 20.8334$$ 16. **Answer (c):** The area enclosed by the curves is approximately $20.833$ square units. --- **Summary:** - (a) $\int (1 - t)^2 dt = t - t^2 + \frac{t^3}{3} + C$ - (b)(i) $\int_1^3 (x^2 - 4x + 3) dx = -\frac{4}{3}$ - (b)(ii) $\int_0^2 3 \sin t dt = 3 - 3 \cos 2$ - (c) Area enclosed by $y = x^2 + 3$ and $y = 7 - 3x$ is approximately $20.833$.