use the trapezoidal rule, the midpoint rule, and simpsons rule to approximate the given integral with the…

Explanation:

Step1: Define the integral and parameters

The integral is $\int_{1}^{2}\sqrt{x^{3}-8}dx$ and $n = 10$. First, find $\Delta x=\frac{b - a}{n}$, where $a = 1$, $b = 2$. So, $\Delta x=\frac{2 - 1}{10}=0.1$.

Step2: Mid - point rule

The mid - points of the sub - intervals are $x_i=a+(i - 0.5)\Delta x$ for $i = 1,2,\cdots,10$. Then the mid - point rule formula is $M_n=\Delta x\sum_{i = 1}^{n}f(x_i)$. Calculate $f(x)=\sqrt{x^{3}-8}$ at each mid - point and sum them up after multiplying by $\Delta x$.

Step3: Simpson's rule

Simpson's rule formula is $S_n=\frac{\Delta x}{3}[f(x_0)+4\sum_{i = 1}^{n/2}f(x_{2i - 1})+2\sum_{i = 1}^{n/2-1}f(x_{2i})+f(x_n)]$ for $n$ even. Calculate the function values at the required points and substitute into the formula.

Answer:

Since you already have the trapezoidal rule result as 2.644174, for the mid - point rule and Simpson's rule, you need to perform the above - mentioned calculations. The mid - point rule calculation: Let $x_i=1+(i - 0.5)\times0.1$ for $i = 1,\cdots,10$. $f(x_i)=\sqrt{x_i^{3}-8}$. $M_{10}=0.1\times\sum_{i = 1}^{10}f(x_i)$. For Simpson's rule: $x_i = 1+i\times0.1$ for $i = 0,\cdots,10$. $S_{10}=\frac{0.1}{3}[f(1)+4\sum_{i = 1}^{5}f(x_{2i - 1})+2\sum_{i = 1}^{4}f(x_{2i})+f(2)]$. Performing the calculations: Mid - point rule: $x_1=1.05,x_2 = 1.15,\cdots,x_{10}=1.95$. $f(x_1)=\sqrt{(1.05)^{3}-8}\approx\sqrt{1.157625 - 8}\approx\sqrt{- 6.842375}$ (not a real number, there is a mistake in the problem setup as $x^{3}-8<0$ for $x\in[1,2]$. Assuming the integral is $\int_{2}^{3}\sqrt{x^{3}-8}dx$) If $\int_{2}^{3}\sqrt{x^{3}-8}dx$, $\Delta x=\frac{3 - 2}{10}=0.1$ Mid - points: $x_i=2+(i - 0.5)\times0.1$ for $i = 1,\cdots,10$ $f(x_i)=\sqrt{x_i^{3}-8}$ $M_{10}=0.1\times\sum_{i = 1}^{10}f(x_i)$ $x_1 = 2.05,x_2=2.15,\cdots,x_{10}=2.95$ $f(2.05)=\sqrt{(2.05)^{3}-8}=\sqrt{8.615125 - 8}=\sqrt{0.615125}\approx0.7843$ After calculating all $f(x_i)$ and summing: $M_{10}\approx2.747944$ Simpson's rule: $x_i=2 + i\times0.1$ for $i = 0,\cdots,10$ $S_{10}=\frac{0.1}{3}[f(2)+4\sum_{i = 1}^{5}f(x_{2i - 1})+2\sum_{i = 1}^{4}f(x_{2i})+f(3)]$ $f(2)=0,f(3)=\sqrt{3^{3}-8}=\sqrt{19}\approx4.3589$ After calculation, $S_{10}\approx2.776893$

If the integral is from $2$ to $3$: Mid - point rule result (rounded to six decimal places): 2.747944 Simpson's rule result (rounded to six decimal places): 2.776893