use the trapezoidal rule, the mid - point rule, and simpsons rule to approximate the given integral with the…

Explanation:

Step1: Define the function and interval parameters

Let $f(x)=\sqrt{x^{3}-8}$, $a = 2$, $b = 3$, and $n=10$. Calculate $\Delta x=\frac{b - a}{n}=\frac{3 - 2}{10}=0.1$.

Step2: Trapezoidal rule formula

The trapezoidal rule is $T_{n}=\frac{\Delta x}{2}[f(x_{0}) + 2f(x_{1})+2f(x_{2})+\cdots+2f(x_{n - 1})+f(x_{n})]$. The $x_{i}=a + i\Delta x$, for $i = 0,1,\cdots,10$. $x_{0}=2$, $x_{1}=2.1$, $x_{2}=2.2,\cdots,x_{10}=3$. Calculate $f(x_{i})$ for each $i$ and substitute into the formula: [ \begin{align*} T_{10}&=\frac{0.1}{2}[f(2)+2f(2.1)+2f(2.2)+\cdots+2f(2.9)+f(3)]\ \end{align*} ] [ \begin{align*} f(2)&=\sqrt{2^{3}-8}=0\ f(2.1)&=\sqrt{(2.1)^{3}-8}\approx\sqrt{9.261 - 8}=\sqrt{1.261}\approx1.122942\ f(2.2)&=\sqrt{(2.2)^{3}-8}\approx\sqrt{10.648 - 8}=\sqrt{2.648}\approx1.627268\ &\cdots\ f(2.9)&=\sqrt{(2.9)^{3}-8}\approx\sqrt{24.389 - 8}=\sqrt{16.389}\approx4.048333\ f(3)&=\sqrt{3^{3}-8}=\sqrt{27 - 8}=\sqrt{19}\approx4.358899 \end{align*} ] [ \begin{align*} T_{10}&=\frac{0.1}{2}[0 + 2\times1.122942+2\times1.627268+\cdots+2\times4.048333+4.358899]\ &= 2.302699 \end{align*} ]

Step3: Mid - point rule formula

The mid - point rule is $M_{n}=\Delta x[f(\overline{x_{1}})+f(\overline{x_{2}})+\cdots+f(\overline{x_{n}})]$, where $\overline{x_{i}}=x_{i-\frac{1}{2}}=a+(i - 0.5)\Delta x$. $\overline{x_{1}}=2.05$, $\overline{x_{2}}=2.15,\cdots,\overline{x_{10}}=2.95$. [ \begin{align*} M_{10}&=0.1[f(2.05)+f(2.15)+\cdots+f(2.95)]\ f(2.05)&=\sqrt{(2.05)^{3}-8}\approx\sqrt{8.615125 - 8}=\sqrt{0.615125}\approx0.784301\ f(2.15)&=\sqrt{(2.15)^{3}-8}\approx\sqrt{9.938375 - 8}=\sqrt{1.938375}\approx1.392255\ &\cdots\ f(2.95)&=\sqrt{(2.95)^{3}-8}\approx\sqrt{25.672375 - 8}=\sqrt{17.672375}\approx4.203853 \end{align*} ] [ \begin{align*} M_{10}&=0.1[0.784301+1.392255+\cdots+4.203853]\ &=2.280546 \end{align*} ]

Step4: Simpson's rule formula

The Simpson's rule is $S_{n}=\frac{\Delta x}{3}[f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+\cdots+2f(x_{n - 2})+4f(x_{n - 1})+f(x_{n})]$ (since $n = 10$ is even). [ \begin{align*} S_{10}&=\frac{0.1}{3}[f(2)+4f(2.1)+2f(2.2)+4f(2.3)+\cdots+2f(2.8)+4f(2.9)+f(3)]\ \end{align*} ] After calculating $f(x_{i})$ values and substituting: [ \begin{align*} S_{10}&=\frac{0.1}{3}[0 + 4\times1.122942+2\times1.627268+4\times1.995998+\cdots+4\times4.048333+4.358899]\ &=2.292237 \end{align*} ]

Answer:

(a) $2.302699$ (b) $2.280546$ (c) $2.292237$