Numerical Integration: Midpoint, Trapezoidal & Simpson's Rules

$\Delta x = \dfrac{3 - 1}{2} = 1$.

Midpoints: $1.5$ and $2.5$.

$M_2 = 1\left(\dfrac{2}{3} + \dfrac{2}{5}\right) = \dfrac{16}{15}$.

$\Delta x = \dfrac{2}{4} = \dfrac{1}{2}$.

Midpoints: $\tfrac{1}{4},\ \tfrac{3}{4},\ \tfrac{5}{4},\ \tfrac{7}{4}$.

$M_4 = \dfrac{1}{2}\left[\dfrac{1}{1 + (1/4)^3} + \dfrac{1}{1 + (3/4)^3} + \dfrac{1}{1 + (5/4)^3} + \dfrac{1}{1 + (7/4)^3}\right]$.

$\Delta x = \dfrac{6}{3} = 2$: pieces $[0, 2]$, $[2, 4]$, $[4, 6]$.

Midpoints: $1,\ 3,\ 5$.

$M_3 = 2\,(4 + 7 + 3) = 28$.

$\Delta x = \dfrac{1}{2}$.

At $x = 1, 1.5, 2, 2.5, 3$: $\ 1,\ \tfrac{2}{3},\ \tfrac{1}{2},\ \tfrac{2}{5},\ \tfrac{1}{3}$.

$T_4 = \dfrac{1}{2}\left[\tfrac{1}{2}(1) + \tfrac{2}{3} + \tfrac{1}{2} + \tfrac{2}{5} + \tfrac{1}{2}\left(\tfrac{1}{3}\right)\right] = \dfrac{1}{2}\cdot\dfrac{67}{30} = \dfrac{67}{60}$.

$\Delta h = 10$, $\ n = 3$.

Areas $A = \pi\left(\tfrac{d}{2}\right)^2$: $\ 100\pi,\ 64\pi,\ 36\pi,\ 25\pi$.

$T_3 = 10\left[\tfrac{1}{2}(100\pi) + 64\pi + 36\pi + \tfrac{1}{2}(25\pi)\right] = 10(162.5\pi) = 1625\pi$.

$\Delta x = 1$.

At $x = 0, 1, 2$: $\ 1,\ e,\ e^2$.

$T_2 = 1\left[\tfrac{1}{2} + e + \tfrac{1}{2}e^2\right] \approx 6.913$. Since $(e^x)'' = e^x > 0$, the curve is concave up and the trapezoids lie above it.

$n = 2$ is even; $\Delta x = 1$.

At $x = 0, 1, 2$: $\ 0,\ 1,\ 8$.

$0 + 4(1) + 8 = 12$.

$S_2 = \dfrac{1}{3}(12) = 4$, exactly $\displaystyle\int_0^2 x^3\,dx$.

$n = 4$ is even; $\Delta x = \dfrac{1}{4}$.

At $x = 0, \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{3}{4}, 1$: $\ 1,\ e^{-1/16},\ e^{-1/4},\ e^{-9/16},\ e^{-1}$.

$1 + 4e^{-1/16} + 2e^{-1/4} + 4e^{-9/16} + e^{-1}$.

$S_4 = \dfrac{1}{12}\left[1 + 4e^{-1/16} + 2e^{-1/4} + 4e^{-9/16} + e^{-1}\right]$.

$n = 4$ is even; $\Delta h = 1$.

Areas $A = \dfrac{C^2}{4\pi}$: $\ \dfrac{4}{4\pi},\ \dfrac{3.24}{4\pi},\ \dfrac{2.56}{4\pi},\ \dfrac{1.44}{4\pi},\ \dfrac{0.64}{4\pi}$.

$\dfrac{1}{4\pi}\big[4 + 4(3.24) + 2(2.56) + 4(1.44) + 0.64\big] = \dfrac{28.48}{4\pi}$.

$S_4 = \dfrac{1}{3}\cdot\dfrac{28.48}{4\pi} = \dfrac{28.48}{12\pi}$.

$|E_S| \le \dfrac{L(b-a)^5}{180n^4}$.

$f^{(4)}(x) = e^{-x}$.

$e^{-x}$ is decreasing, so $|f^{(4)}(x)| \le e^{0} = 1$ on $[0, 2]$: $\ L = 1$.

$|E_S| \le \dfrac{1 \cdot 2^5}{180 \cdot 4^4} = \dfrac{32}{46080} = \dfrac{1}{1440}$.

$|E_T| \le \dfrac{M(b-a)^3}{12n^2}$.

$f'(x) = \sin x + x\cos x$, $\ f''(x) = 2\cos x - x\sin x$.

$|f''(x)| \le 2|\cos x| + |x||\sin x| \le 2 + 2 = 4$ on $[0, 2]$: $\ M = 4$.

$|E_T| \le \dfrac{4 \cdot 2^3}{12 \cdot 8^2} = \dfrac{32}{768} = \dfrac{1}{24}$.

$|E_M| \le \dfrac{M(b-a)^3}{24n^2}$ and $|E_S| \le \dfrac{L(b-a)^5}{180n^4}$.

$f''(x) = 12x^2$ and $f^{(4)}(x) = 24$.

On $[0, 2]$: $\ M = 12(2)^2 = 48$ and $L = 24$.

$|E_M| \le \dfrac{48 \cdot 8}{24 \cdot 400} = \dfrac{1}{25}$; $\ |E_S| \le \dfrac{24 \cdot 32}{180 \cdot 256} = \dfrac{1}{60}$.

$f''(x) = -\tfrac{1}{4}x^{-3/2}$, largest in size at $x = 1$: $\ M = \tfrac{1}{4}$.

$\dfrac{\tfrac{1}{4}(3)^3}{12n^2} \le 0.0005$.

$n^2 \ge \dfrac{27}{48 \times 0.0005} = 1125$, so $n \ge 33.54$.

$n = 34$.

$f^{(4)}(x) = e^{-x}$, so $L = 1$ on $[0, 2]$.

$\dfrac{1 \cdot 2^5}{180n^4} \le 10^{-6}$.

$n^4 \ge \dfrac{32}{180} \times 10^{6} \approx 177{,}778$, so $n \ge 20.53$.

Round up to an even number: $n = 22$.

Given: $M = 3$.

$\dfrac{3 \cdot 4^3}{24n^2} = \dfrac{8}{n^2} \le 0.01$.

$n^2 \ge 800$, so $n \ge 28.28$.

$n = 29$.