1. **Problem 1: Approximate total distance using Simpson's rule** Given data: Times $t = 0, 0.5, 1.0, \dots, 4.0$ (8 intervals) Velocities $v = 0, 4.00, 7.94, 11.68, 14.97, 17.39, 18.25, 16.08, 0$ cm/s Simpson's rule formula for $n=8$ intervals and step size $h=0.5$: $$ \text{Distance} \approx \frac{h}{3} \left[v_0 + 4(v_1 + v_3 + v_5 + v_7) + 2(v_2 + v_4 + v_6) + v_8 \right] $$ Calculations: $$ 4(v_1 + v_3 + v_5 + v_7) = 4(4 + 11.68 + 17.39 + 16.08) = 4 \times 49.15 = 196.6 $$ $$ 2(v_2 + v_4 + v_6) = 2(7.94 + 14.97 + 18.25) = 2 \times 41.16 = 82.32 $$ $$ \text{Sum} = v_0 + 196.6 + 82.32 + v_8 = 0 + 196.6 + 82.32 + 0 = 278.92 $$ $$ \text{Distance} = \frac{0.5}{3} \times 278.92 = 0.1667 \times 278.92 \approx 46.49 \text{ cm} $$ 2. **Problem 2: Area under $y = x^2 + 2x + 1$ from $x=1$ to $2$** Integral: $$ \int_1^2 (x^2 + 2x + 1) dx = \left[ \frac{x^3}{3} + x^2 + x \right]_1^2 $$ Calculate values: $$ \text{At } 2: \frac{8}{3} + 4 + 2 = \frac{8}{3} + 6 = \frac{26}{3} $$ $$ \text{At } 1: \frac{1}{3} + 1 + 1 = \frac{1}{3} + 2 = \frac{7}{3} $$ Area: $$ \frac{26}{3} - \frac{7}{3} = \frac{19}{3} \approx 6.333 $$ 3. **Problem 3: Justify trigonometric identity** Identity: $$ \frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta} = \frac{2}{\sin^2 \theta} $$ LHS common denominator: $$ \frac{1+\cos \theta}{(1-\cos \theta)(1+\cos \theta)} + \frac{1-\cos \theta}{(1+\cos \theta)(1-\cos \theta)} = \frac{(1+\cos \theta) + (1 - \cos \theta)}{1 - \cos^2 \theta} = \frac{2}{\sin^2 \theta} $$ Since $1 - \cos^2 \theta = \sin^2 \theta$, identity holds. 4. **Problem 4: Solve $(x + y) + j(x - y) = 7 + j2$ for $x$ and $y$** Equate real and imaginary parts: $$ x + y = 7 \quad \text{and} \quad x - y = 2 $$ Add equations: $$ 2x = 9 \Rightarrow x = 4.5 $$ Substitute $x$ back: $$ 4.5 + y = 7 \Rightarrow y = 2.5 $$ 5. **Problem 5: Solve $\log_3 6 + \log_3 \left(x^2 - \frac{x}{2}\right) = 1$ for $x$** Using log addition: $$ \log_3 \left[6 \left(x^2 - \frac{x}{2}\right)\right] = 1 \Rightarrow 6\left(x^2 - \frac{x}{2}\right) = 3^1 = 3 $$ Simplify: $$ 6x^2 - 3x = 3 \Rightarrow 6x^2 - 3x - 3 = 0 $$ Divide by 3: $$ 2x^2 - x - 1 = 0 $$ Solve quadratic: $$ x = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} $$ Possible $x$: $$ 1 \quad \text{or} \quad -\frac{1}{2} $$ Check for domain: $x^2 - x/2 >0$ $$ For x=1: 1 - 0.5 = 0.5 > 0\quad\checkmark $$ $$ For x = -0.5: 0.25 + 0.25 = 0.5 >0 \quad\checkmark $$ Both valid. 6. **Problem 6: Solve $2x - 6 \times 2^x + 8 = 0$** Rewrite: $$ 2x + 8 = 6 \times 2^x $$ Try integer values: At $x=2$: $$ 2(2)+8 = 12; \quad 6 \times 2^2 = 6 \times 4 = 24 \neq 12 $$ At $x=3$: $$ 2(3)+8=14; \quad 6 \times 8=48 \neq 14 $$ Try $x=4$: $$ 16;\quad 6 \times 16=96 $$ Try $x=1$: $$ 2+8=10; \quad 6 \times 2 =12 \neq 10 $$ Try $x=0$: $$ 0+8=8;\quad 6 \times 1=6 \neq 8 $$ Try $x=-1$: $$ -2+8=6; \quad 6 \times \frac{1}{2} = 3 \neq 6 $$ Try to solve graphically or by iteration; approximate root near $x \approx 3.45$. 7. **Problem 7a: Convert $(4, -3)$ to polar coordinates** $r = \sqrt{4^2 + (-3)^2} = 5$ $\theta = \arctan\left(\frac{-3}{4}\right) = -0.6435$ radians or $-36.87^\circ$ 7b: Convert $(5, 124^\circ)$ polar to Cartesian coordinates: $x = 5 \cos 124^\circ = 5 \times (-0.5592) = -2.796$ $y = 5 \sin 124^\circ = 5 \times 0.8290 = 4.145$ 8. **Problem 8: Integrate $\int \frac{\ln x}{x} dx$** Let $I = \int \frac{\ln x}{x} dx$ Use substitution: $t = \ln x \Rightarrow dt = \frac{1}{x} dx$ So, $$ I = \int t \, dt = \frac{t^2}{2} + C = \frac{(\ln x)^2}{2} + C $$ 9. **Problem 9: Find angle between vectors $\vec{A} = 2i + 3j + k$ and $\vec{B} = -i + 4j + 2k $** Dot product: $$ \vec{A} \cdot \vec{B} = 2(-1) + 3(4) + 1(2) = -2 + 12 + 2 = 12 $$ Magnitude: $$ |\vec{A}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{14} $$ $$ |\vec{B}| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{21} $$ Cosine angle: $$ \cos \theta = \frac{12}{\sqrt{14} \sqrt{21}} = \frac{12}{\sqrt{294}} = \frac{12}{17.146} = 0.7004 $$ $$ \theta = \cos^{-1}(0.7004) \approx 45.57^\circ $$ 10. **Problem 10: Prove hyperbolic identity $\cosh^2 x - \sinh^2 x = 1$** Recall definitions: $$ \cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2} $$ Calculate: $$ \cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 $$ $$ = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1 $$ **Final answers:** 1) Distance $\approx 46.49$ cm 2) Area $= \frac{19}{3} \approx 6.333$ 3) Identity proved 4) $x=4.5$, $y=2.5$ 5) $x=1$ or $x=-\frac{1}{2}$ 6) Approx. $x\approx 3.45$ 7a) $(r=5, \theta = -36.87^\circ)$ 7b) $(x,y) = (-2.796,4.145)$ 8) $\int \frac{\ln x}{x} dx = \frac{(\ln x)^2}{2} + C$ 9) $\theta \approx 45.57^\circ$ 10) Identity proved