1. **Problem:** Apply Simpson's rule to approximate the distance traveled for velocity data over 8 intervals between $t=0$ and $t=4$. Simpson's rule formula with $n=8$ intervals, $h=\frac{4-0}{8}=0.5$: $$\text{Distance} = \frac{h}{3}[v_0 + 4(v_1+v_3+v_5+v_7) + 2(v_2+v_4+v_6) + v_8]$$ Values from table: $v_0=0, v_1=4.00, v_2=7.94, v_3=11.68, v_4=14.97, v_5=17.39, v_6=18.25, v_7=16.08, v_8=0$ Calculate: $$4(v_1+v_3+v_5+v_7) = 4(4 + 11.68 + 17.39 + 16.08) = 4(49.15) = 196.6$$ $$2(v_2+v_4+v_6) = 2(7.94 + 14.97 + 18.25) = 2(41.16) = 82.32$$ Sum inside brackets: $$0 + 196.6 + 82.32 + 0 = 278.92$$ Distance: $$\frac{0.5}{3} \times 278.92 = \frac{0.5}{3} \times 278.92 = 46.49$$ 2. **Problem:** Find area under $y = x^2 + 2x + 1$ from $x=1$ to $x=2$. Integrate: $$\int_1^2 (x^2 + 2x + 1) dx = \left[\frac{x^3}{3} + x^2 + x \right]_1^2$$ Evaluate: $$\left(\frac{8}{3} + 4 + 2\right) - \left(\frac{1}{3} + 1 + 1\right) = \left(\frac{8}{3} + 6\right) - \left(\frac{1}{3} + 2\right) = \frac{8}{3} - \frac{1}{3} + 6 - 2 = \frac{7}{3} + 4 = \frac{19}{3} \approx 6.33$$ 3. **Problem:** Show that $\frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta} = \frac{2}{\sin^2 \theta}$. Start with left side: $$\frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta} = \frac{(1+\cos \theta) + (1-\cos \theta)}{(1-\cos \theta)(1+\cos \theta)} = \frac{2}{1 - \cos^2 \theta}$$ Use identity $\sin^2 \theta = 1 - \cos^2 \theta$, so $$\frac{2}{\sin^2 \theta}$$ Hence proven. 4. **Problem:** Solve for $x$ and $y$ given $(x + y) + j(x - y) = 7 + j2$. Equate real and imaginary parts: $$x + y = 7$$ $$x - y = 2$$ Add equations: $$2x = 9 \Rightarrow x = 4.5$$ Substitute back: $$4.5 + y = 7 \Rightarrow y = 2.5$$ 5. **Problem:** Find $x$ in $\log_3 6 + \log_3 (x^2 - \frac{x}{2}) = 1$. Use log sum: $$\log_3 \biggl(6 (x^2 - \frac{x}{2}) \biggr) = 1$$ Convert: $$6 \left(x^2 - \frac{x}{2}\right) = 3^1 = 3$$ Simplify: $$6 \left(x^2 - \frac{x}{2} \right) = 6x^2 - 3x = 3$$ Rearrange: $$6x^2 - 3x - 3 = 0 \Rightarrow 2x^2 - x - 1 =0$$ Solve quadratic: $$x = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}$$ Solutions: $$x = 1, \quad x = -\frac{1}{2}$$ 6. **Problem:** Solve $2x - 6x^2 + 8 = 0$. Rearrange: $$-6x^2 + 2x + 8 = 0 \Rightarrow 6x^2 - 2x - 8 = 0$$ Divide by 2: $$3x^2 - x - 4 = 0$$ Use quadratic formula: $$x = \frac{1 \pm \sqrt{1 + 48}}{6} = \frac{1 \pm 7}{6}$$ Solutions: $$x = \frac{8}{6} = \frac{4}{3}, \quad x = -1$$ 7. **Problem:** (a) Express $(4, -3)$ in polar coordinates. Calculate: $$r = \sqrt{4^2 + (-3)^2} = 5$$ $$\theta = \arctan\left(\frac{-3}{4}\right) \approx -0.6435 \text{ radians}$$ Polar coordinates: $$(5, -0.6435)$$ (b) Convert $(5, 124^\circ)$ to Cartesian: $$x = 5 \cos 124^\circ \approx 5 \times (-0.5592) = -2.796$$ $$y = 5 \sin 124^\circ \approx 5 \times 0.8290 = 4.145$$ Cartesian coordinates: $$(-2.796, 4.145)$$ 8. **Problem:** Compute integral $\int \frac{\ln x}{x} dx$. Use substitution: $$u = \ln x \Rightarrow du = \frac{1}{x} dx$$ So: $$\int u \, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C$$ 9. **Problem:** Find angle between vectors $\vec{A} = 2\vec{i} + 3\vec{j} + \vec{k}$ and $\vec{B} = -\vec{i} + 4\vec{j} + 2\vec{k}$. Dot product: $$\vec{A} \cdot \vec{B} = 2(-1) + 3(4) + 1(2) = -2 + 12 + 2 = 12$$ Magnitudes: $$|\vec{A}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{14}$$ $$|\vec{B}| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{21}$$ Angle $\theta$: $$\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{12}{\sqrt{14} \sqrt{21}} = \frac{12}{\sqrt{294}} = \frac{12}{17.1756} = 0.699$$ $$\theta = \arccos(0.699) \approx 45.57^\circ$$ 10. **Problem:** Prove the hyperbolic identity $\cosh^2 x - \sinh^2 x = 1$. Recall: $$\cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2}$$ Compute: $$\cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2$$ $$= \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1$$ **Final answers:** 1. Approximate distance $\approx 46.49$ cm 2. Area under curve $= \frac{19}{3} \approx 6.33$ 3. Identity proven 4. $x=4.5$, $y=2.5$ 5. $x=1$ or $x=-\frac{1}{2}$ 6. $x=\frac{4}{3}$ or $x=-1$ 7a. $(r,\theta)=(5, -0.6435)$ radians 7b. $(-2.796, 4.145)$ 8. $\int \frac{\ln x}{x} dx = \frac{(\ln x)^2}{2} + C$ 9. Angle between vectors $\approx 45.57^\circ$ 10. Identity proven