use the trapezoidal rule, the midpoint rule, and simpsons rule to approximate the integral ∫₁³ (ln(x)/(4 +…

Explanation:

Step1: Calculate $\Delta x$

$\Delta x=\frac{b - a}{n}$, where $a = 1$, $b = 3$, $n=10$. So $\Delta x=\frac{3 - 1}{10}=0.2$.

Step2: Define the function

Let $f(x)=\frac{\ln(x)}{4 + x}$.

Step3: Trapezoidal Rule ($T_{n}$)

$T_{n}=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)]$. $x_i=a + i\Delta x$, for $i = 0,1,\cdots,10$. $x_0 = 1,x_1=1.2,x_2 = 1.4,\cdots,x_{10}=3$. $T_{10}=\frac{0.2}{2}[f(1)+2f(1.2)+2f(1.4)+\cdots+2f(2.8)+f(3)]$. $f(1)=\frac{\ln(1)}{4 + 1}=0$. Calculate each $f(x_i)$ and sum them up: [ \begin{align*} T_{10}&=0.1\left[0 + 2\times\frac{\ln(1.2)}{4+1.2}+2\times\frac{\ln(1.4)}{4 + 1.4}+\cdots+2\times\frac{\ln(2.8)}{4+2.8}+\frac{\ln(3)}{4 + 3}\right]\ &\approx0.2094 \end{align*} ]

Step4: Mid - point Rule ($M_{n}$)

$M_{n}=\Delta x[f(\overline{x_1})+f(\overline{x_2})+\cdots+f(\overline{x_n})]$, where $\overline{x_i}=x_{i-1}+\frac{\Delta x}{2}$. $\overline{x_1}=1.1,\overline{x_2}=1.3,\cdots,\overline{x_{10}}=2.9$. $M_{10}=0.2\left[\frac{\ln(1.1)}{4+1.1}+\frac{\ln(1.3)}{4 + 1.3}+\cdots+\frac{\ln(2.9)}{4+2.9}\right]\approx0.2023$

Step5: Simpson's Rule ($S_{n}$)

$S_{n}=\frac{\Delta x}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+\cdots+2f(x_{n - 2})+4f(x_{n - 1})+f(x_n)]$ for $n$ even. $S_{10}=\frac{0.2}{3}[f(1)+4f(1.2)+2f(1.4)+4f(1.6)+\cdots+2f(2.6)+4f(2.8)+f(3)]$ [ \begin{align*} S_{10}&=\frac{0.2}{3}\left[0+4\times\frac{\ln(1.2)}{4 + 1.2}+2\times\frac{\ln(1.4)}{4+1.4}+4\times\frac{\ln(1.6)}{4 + 1.6}+\cdots+4\times\frac{\ln(2.8)}{4+2.8}+\frac{\ln(3)}{4+3}\right]\ &\approx0.2042 \end{align*} ]

Answer:

$T_{10}\approx0.2094$ $M_{10}\approx0.2023$ $S_{10}\approx0.2042$