To estimate the area bound by the curves $y = 2x^2 + 1$ and $y = x^2 + 1$, we first find the difference function $f(x)$: $$f(x) = (2x^2 + 1) - (x^2 + 1) = x^2$$ a) Using the trapezium rule with 6 strips from $x=2$ to $x=5$. Step 1: Calculate the width of each strip, $h$. The interval is $[a, b] = [2, 5]$ and the number of strips $n=6$. $$h = \frac{b-a}{n} = \frac{5-2}{6} = \frac{3}{6} = 0.5$$ Step 2: Determine the x-coordinates for the ordinates. $$x_0 = 2$$ $$x_1 = 2 + 0.5 = 2.5$$ $$x_2 = 2.5 + 0.5 = 3$$ $$x_3 = 3 + 0.5 = 3.5$$ $$x_4 = 3.5 + 0.5 = 4$$ $$x_5 = 4 + 0.5 = 4.5$$ $$x_6 = 4.5 + 0.5 = 5$$ Step 3: Calculate the corresponding $y$ values using $f(x) = x^2$. $$y_0 = f(2) = 2^2 = 4$$ $$y_1 = f(2.5) = (2.5)^2 = 6.25$$ $$y_2 = f(3) = 3^2 = 9$$ $$y_3 = f(3.5) = (3.5)^2 = 12.25$$ $$y_4 = f(4) = 4^2 = 16$$ $$y_5 = f(4.5) = (4.5)^2 = 20.25$$ $$y_6 = f(5) = 5^2 = 25$$ Step 4: Apply the Trapezium Rule formula. The formula for the Trapezium Rule is: $$\text{Area} \approx \frac{h}{2} [ (y_0 + y_n) + 2(y_1 + y_2 + \dots + y_{n-1}) ]$$ $$\text{Area} \approx \frac{0.5}{2} [ (4 + 25) + 2(6.25 + 9 + 12.25 + 16 + 20.25) ]$$ $$\text{Area} \approx 0.25 [ 29 + 2(63.75) ]$$ $$\text{Area} \approx 0.25 [ 29 + 127.5 ]$$ $$\text{Area} \approx 0.25 [ 156.5 ]$$ $$\text{Area} \approx \boxed{39.125}$$ b) Using the mid-ordinate rule with 6 strips from $x=2$ to $x=6$. Step 1: Calculate the width of each strip, $h$. The interval is $[a, b] = [2, 6]$ and the number of strips $n=6$. $$h = \frac{b-a}{n} = \frac{6-2}{6} = \frac{4}{6} = \frac{2}{3}$$ Step 2: Determine the x-coordinates for the mid-points of each strip. The mid-points are $x_i + \frac{h}{2}$. $$m_1 = 2 + \frac{1}{2} \left(\frac{2}{3}\right) = 2 + \frac{1}{3} = \frac{7}{3}$$ $$m_2 = \frac{7}{3} + \frac{2}{3} = \frac{9}{3} = 3$$ $$m_3 = 3 + \frac{2}{3} = \frac{11}{3}$$ $$m_4 = \frac{11}{3} + \frac{2}{3} = \frac{13}{3}$$ $$m_5 = \frac{13}{3} + \frac{2}{3} = \frac{15}{3} = 5$$ $$m_6 = 5 + \frac{2}{3} = \frac{17}{3}$$ Step 3: Calculate the corresponding $y$ values using $f(x) = x^2$ at the mid-points. $$y_{m1} = f\left(\frac{7}{3}\right) = \left(\frac{7}{3}\right)^2 = \frac{49}{9}$$ $$y_{m2} = f(3) = 3^2 = 9$$ $$y_{m3} = f\left(\frac{11}{3}\right) = \left(\frac{11}{3}\right)^2 = \frac{121}{9}$$ $$y_{m4} = f\left(\frac{13}{3}\right) = \left(\frac{13}{3}\right)^2 = \frac{1