1. **Evaluate** $\int_0^6 (x^2 + 3x) \, dx$ using Simpson's 1/3 rule with 6 equal parts. - Interval $[0,6]$, divide into 6 parts: $h = \frac{6-0}{6} = 1$ - Points: $x_0=0, x_1=1, x_2=2, x_3=3, x_4=4, x_5=5, x_6=6$ - Function values: $f(x) = x^2 + 3x$ - $f_0=0^2+3\cdot0=0$ - $f_1=1+3=4$ - $f_2=4+6=10$ - $f_3=9+9=18$ - $f_4=16+12=28$ - $f_5=25+15=40$ - $f_6=36+18=54$ - Simpson's 1/3 rule formula for even $n$: $$\int_a^b f(x) dx \approx \frac{h}{3} \left[f_0 + 4(f_1 + f_3 + f_5) + 2(f_2 + f_4) + f_6\right]$$ - Substitute values: $$= \frac{1}{3} [0 + 4(4 + 18 + 40) + 2(10 + 28) + 54]$$ $$= \frac{1}{3} [0 + 4 \times 62 + 2 \times 38 + 54]$$ $$= \frac{1}{3} [0 + 248 + 76 + 54] = \frac{1}{3} [378] = 126$$ 2. **Evaluate** $\int_0^1 (x^3 + 2x^2 + 3x + 1) \, dx$ using Newton–Cotes formula for $n=3$. - For $n=3$, the Newton–Cotes formula (Simpson's 3/8 rule) is: $$\int_a^b f(x) dx \approx \frac{3h}{8} [f_0 + 3f_1 + 3f_2 + f_3]$$ - Divide $[0,1]$ into 3 equal parts: $h = \frac{1-0}{3} = \frac{1}{3}$ - Points: $x_0=0, x_1=\frac{1}{3}, x_2=\frac{2}{3}, x_3=1$ - Calculate $f(x)$: - $f_0 = 0^3 + 2\cdot0^2 + 3\cdot0 + 1 = 1$ - $f_1 = (\frac{1}{3})^3 + 2(\frac{1}{3})^2 + 3(\frac{1}{3}) + 1 = \frac{1}{27} + 2 \times \frac{1}{9} + 1 + 1 = \frac{1}{27} + \frac{2}{9} + 1 + 1 = \frac{1}{27} + \frac{6}{27} + 1 + 1 = \frac{7}{27} + 2 = \frac{7}{27} + \frac{54}{27} = \frac{61}{27} \approx 2.259$ - $f_2 = (\frac{2}{3})^3 + 2(\frac{2}{3})^2 + 3(\frac{2}{3}) + 1 = \frac{8}{27} + 2 \times \frac{4}{9} + 2 + 1 = \frac{8}{27} + \frac{8}{9} + 3 = \frac{8}{27} + \frac{24}{27} + 3 = \frac{32}{27} + 3 = \frac{32}{27} + \frac{81}{27} = \frac{113}{27} \approx 4.185$ - $f_3 = 1 + 2 + 3 + 1 = 7$ - Apply formula: $$\int_0^1 f(x) dx \approx \frac{3 \times \frac{1}{3}}{8} [1 + 3(2.259) + 3(4.185) + 7] = \frac{1}{8} [1 + 6.777 + 12.555 + 7] = \frac{1}{8} [27.332] = 3.4165$$ 3. **Find** $\frac{dy}{dx}$ at $x=2.0$ using Newton's forward difference formula. - Given data: $$x: 1.0, 1.5, 2.0, 2.5, 3.0$$ $$y: 2.718, 4.481, 7.389, 12.182, 20.086$$ - Step size $h = 0.5$ - Forward differences: $$\Delta y_0 = y_1 - y_0 = 4.481 - 2.718 = 1.763$$ $$\Delta y_1 = y_2 - y_1 = 7.389 - 4.481 = 2.908$$ $$\Delta y_2 = y_3 - y_2 = 12.182 - 7.389 = 4.793$$ $$\Delta y_3 = y_4 - y_3 = 20.086 - 12.182 = 7.904$$ - Second differences: $$\Delta^2 y_0 = \Delta y_1 - \Delta y_0 = 2.908 - 1.763 = 1.145$$ $$\Delta^2 y_1 = \Delta y_2 - \Delta y_1 = 4.793 - 2.908 = 1.885$$ $$\Delta^2 y_2 = \Delta y_3 - \Delta y_2 = 7.904 - 4.793 = 3.111$$ - Newton's forward difference formula for derivative at $x_2=2.0$ (which is the 3rd point, index 2): $$\frac{dy}{dx} \approx \frac{1}{h} \left( \Delta y_2 - \frac{1}{2} \Delta^2 y_2 \right)$$ - Substitute values: $$= \frac{1}{0.5} (4.793 - \frac{1}{2} \times 3.111) = 2 (4.793 - 1.5555) = 2 \times 3.2375 = 6.475$$ **Final answers:** - Simpson's 1/3 rule integral: $126$ - Newton–Cotes $n=3$ integral: approximately $3.4165$ - Derivative at $x=2.0$: approximately $6.475$