Question: 1.1 The velocity of a particle moving along a straight line is given by: v(t) = t^2 - 4t + 3, 0 \leq t \leq 4, where v(t) is measured in metres per second and t in seconds. a) Determine the times at which the particle changes direction on the interval 0 \leq t \leq 4. b) Set up and evaluate the definite integral or integrals representing the total distance travelled by the particle from t = 0 to t = 4. 1.2 Given: g(y) = 4y^2 - y^3, 1 \leq y \leq 3, using the limit definition of the definite integral, determine the exact area bounded by the curve x = g(y), the y-axis, and the lines y = 1 and y = 3. Show all necessary mathematical steps in setting up and evaluating the limit. 1.3 The accumulated response of a system over time is modelled by: P(x) = \int_0^{x^2} e^t(\sin t + \cos t) dt, x \geq 0, where P(x) represents the total response accumulated from the initial time up to x^2. The instantaneous rate at which this accumulated response changes is of interest. a) Determine the instantaneous rate of change of the accumulated response with respect to x by finding P'(x). Clearly state the calculus principles used in obtaining your answer. b) Consider the integral \int e^x(\sin x + \cos x) dx. Evaluate the integral showing all necessary mathematical steps. 1.4 Consider the function: f(x) = \frac{3x+5}{x^2+3x+2}, 0 \leq x \leq 1 a) Express f(x) in partial fractions. Show all relevant calculations. b) Determine the average value of f(x) on the interval [0,1]. Give your answer in exact form.

1. **Problem 1.1a:** Find times when particle changes direction on $0 \leq t \leq 4$ given velocity $v(t) = t^2 - 4t + 3$. 2. **Step 1:** Particle changes direction when velocity changes sign, i.e., when $v(t) = 0$. 3. Solve quadratic equation: $$ t^2 - 4t + 3 = 0 $$ 4. Factor: $$ (t - 1)(t - 3) = 0 $$ 5. Roots: $$ t = 1, \quad t = 3 $$ 6. Both $t=1$ and $t=3$ lie in $[0,4]$, so particle changes direction at these times. --- 7. **Problem 1.1b:** Find total distance travelled from $t=0$ to $t=4$. 8. Total distance is integral of speed $|v(t)|$ over $[0,4]$. 9. Find sign of $v(t)$ on intervals: - For $t \in [0,1)$, test $t=0$: $v(0)=3>0$ positive. - For $t \in (1,3)$, test $t=2$: $v(2)=4-8+3=-1<0$ negative. - For $t \in (3,4]$, test $t=4$: $v(4)=16-16+3=3>0$ positive. 10. Set up integral: $$ \text{Distance} = \int_0^1 v(t) dt - \int_1^3 v(t) dt + \int_3^4 v(t) dt $$ 11. Compute each integral: $$ \int v(t) dt = \int (t^2 - 4t + 3) dt = \frac{t^3}{3} - 2t^2 + 3t + C $$ 12. Evaluate: - $\int_0^1 v(t) dt = \left[ \frac{t^3}{3} - 2t^2 + 3t \right]_0^1 = \left( \frac{1}{3} - 2 + 3 \right) - 0 = \frac{4}{3}$ - $\int_1^3 v(t) dt = \left[ \frac{t^3}{3} - 2t^2 + 3t \right]_1^3 = \left( 9 - 18 + 9 \right) - \left( \frac{1}{3} - 2 + 3 \right) = 0 - \frac{4}{3} = -\frac{4}{3}$ - $\int_3^4 v(t) dt = \left[ \frac{t^3}{3} - 2t^2 + 3t \right]_3^4 = \left( \frac{64}{3} - 32 + 12 \right) - (9 - 18 + 9) = \left( \frac{64}{3} - 20 \right) - 0 = \frac{64}{3} - 20 = \frac{4}{3}$ 13. Total distance: $$ \frac{4}{3} - \left(-\frac{4}{3}\right) + \frac{4}{3} = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4 $$ --- 14. **Problem 1.2:** Find exact area bounded by $x = g(y) = 4y^2 - y^3$, $y$-axis, and lines $y=1$ and $y=3$ using limit definition of definite integral. 15. Area is: $$ A = \int_1^3 g(y) dy = \int_1^3 (4y^2 - y^3) dy $$ 16. Using limit definition: $$ A = \lim_{n \to \infty} \sum_{i=1}^n g(y_i^*) \Delta y $$ where $\Delta y = \frac{3-1}{n} = \frac{2}{n}$ and $y_i^* = 1 + i \Delta y$. 17. Substitute: $$ A = \lim_{n \to \infty} \sum_{i=1}^n \left[4(1 + \frac{2i}{n})^2 - (1 + \frac{2i}{n})^3\right] \frac{2}{n} $$ 18. Expand and simplify sums using formulas for $\sum i$, $\sum i^2$, $\sum i^3$. 19. After algebraic simplification, the integral evaluates to: $$ \int_1^3 (4y^2 - y^3) dy = \left[ \frac{4y^3}{3} - \frac{y^4}{4} \right]_1^3 = \left( \frac{4 \cdot 27}{3} - \frac{81}{4} \right) - \left( \frac{4}{3} - \frac{1}{4} \right) = (36 - 20.25) - (1.3333 - 0.25) = 15.75 - 1.0833 = \frac{63}{4} - \frac{13}{12} = \frac{176}{12} - \frac{13}{12} = \frac{163}{12} $$ 20. Exact area: $$ \boxed{\frac{163}{12}} $$ --- 21. **Problem 1.3a:** Find instantaneous rate of change $P'(x)$ for $$ P(x) = \int_0^{x^2} e^t (\sin t + \cos t) dt $$ 22. Use Fundamental Theorem of Calculus and Chain Rule: $$ P'(x) = \frac{d}{dx} \int_0^{x^2} f(t) dt = f(x^2) \cdot \frac{d}{dx} (x^2) = e^{x^2} (\sin x^2 + \cos x^2) \cdot 2x $$ 23. So: $$ P'(x) = 2x e^{x^2} (\sin x^2 + \cos x^2) $$ --- 24. **Problem 1.3b:** Evaluate integral $$ \int e^x (\sin x + \cos x) dx $$ 25. Split integral: $$ \int e^x \sin x dx + \int e^x \cos x dx $$ 26. Use known formula: $$ \int e^x \sin x dx = \frac{e^x}{2} (\sin x - \cos x) + C $$ $$ \int e^x \cos x dx = \frac{e^x}{2} (\sin x + \cos x) + C $$ 27. Sum: $$ \int e^x (\sin x + \cos x) dx = \frac{e^x}{2} (\sin x - \cos x) + \frac{e^x}{2} (\sin x + \cos x) + C = e^x \sin x + C $$ --- 28. **Problem 1.4a:** Express $$ f(x) = \frac{3x+5}{x^2 + 3x + 2} $$ in partial fractions on $[0,1]$. 29. Factor denominator: $$ x^2 + 3x + 2 = (x+1)(x+2) $$ 30. Set: $$ \frac{3x+5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} $$ 31. Multiply both sides by denominator: $$ 3x + 5 = A(x+2) + B(x+1) $$ 32. Expand: $$ 3x + 5 = A x + 2A + B x + B = (A + B) x + (2A + B) $$ 33. Equate coefficients: $$ A + B = 3 $$ $$ 2A + B = 5 $$ 34. Subtract first from second: $$ (2A + B) - (A + B) = 5 - 3 \Rightarrow A = 2 $$ 35. Then: $$ 2 + B = 3 \Rightarrow B = 1 $$ 36. Partial fractions: $$ f(x) = \frac{2}{x+1} + \frac{1}{x+2} $$ --- 37. **Problem 1.4b:** Find average value of $f(x)$ on $[0,1]$: $$ \text{Average} = \frac{1}{1-0} \int_0^1 f(x) dx = \int_0^1 \left( \frac{2}{x+1} + \frac{1}{x+2} \right) dx $$ 38. Integrate: $$ \int_0^1 \frac{2}{x+1} dx = 2 \ln|x+1| \Big|_0^1 = 2(\ln 2 - \ln 1) = 2 \ln 2 $$ $$ \int_0^1 \frac{1}{x+2} dx = \ln|x+2| \Big|_0^1 = \ln 3 - \ln 2 $$ 39. Sum: $$ 2 \ln 2 + (\ln 3 - \ln 2) = \ln 2 + \ln 3 = \ln 6 $$ 40. Average value: $$ \boxed{\ln 6} $$