Question

practice another use the trapezoidal rule, the mid - point rule, and simpsons rule to approximate the given integral with the specified value of n. (round your answers to six decimal places.) ∫₀⁴ln(7 + eˣ) dx, n = 8 (a) the trapezoidal rule (b) the mid - point rule (c) simpsons rule solution or explanation f(x)=ln(7 + eˣ), δx=(4 - 0)/8 = 1/2 (a) t₈ = 1/2·2{ f(0)+2f(0.5)+f(1)+⋯+f(3.5)+f(4)}≈11.291719 (b) m₈ = 1/2f(0.25)+f(0.75)+⋯+f(3.25)+f(3.75)≈11.267900 (c) s₈ = 1/2·3f(0)+4f(0.5)+2f(1)+4f(1.5)+2f(2)+4f(2.5)+2f(3)+4f(3.5)+f(4)≈11.275788

Explanation:

Step1: Recall trapezoidal - rule formula

The trapezoidal - rule for approximating $\int_{a}^{b}f(x)dx$ with $n$ sub - intervals is $T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)]$, where $\Delta x=\frac{b - a}{n}$. Here, $a = 0$, $b = 4$, $n = 8$, so $\Delta x=\frac{4-0}{8}=\frac{1}{2}$. And $f(x)=\ln(7 + e^x)$. Then $T_8=\frac{1}{2\times2}[f(0)+2f(0.5)+2f(1)+2f(1.5)+2f(2)+2f(2.5)+2f(3)+2f(3.5)+f(4)]\approx11.291719$.

Step2: Recall mid - point rule formula

The mid - point rule for approximating $\int_{a}^{b}f(x)dx$ with $n$ sub - intervals is $M_n=\Delta x[f(\overline{x_1})+f(\overline{x_2})+\cdots+f(\overline{x_n})]$, where $\Delta x=\frac{b - a}{n}$ and $\overline{x_i}$ is the mid - point of the $i$th sub - interval. For $n = 8$, $\Delta x=\frac{1}{2}$, and the mid - points are $x = 0.25,0.75,1.25,1.75,2.25,2.75,3.25,3.75$. So $M_8=\frac{1}{2}[f(0.25)+f(0.75)+f(1.25)+f(1.75)+f(2.25)+f(2.75)+f(3.25)+f(3.75)]\approx11.267900$.

Step3: Recall Simpson's rule formula

Simpson's rule for approximating $\int_{a}^{b}f(x)dx$ with $n$ (where $n$ is even) sub - intervals is $S_n=\frac{\Delta x}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+\cdots+2f(x_{n - 2})+4f(x_{n - 1})+f(x_n)]$. For $n = 8$, $\Delta x=\frac{1}{2}$, so $S_8=\frac{1}{2\times3}[f(0)+4f(0.5)+2f(1)+4f(1.5)+2f(2)+4f(2.5)+2f(3)+4f(3.5)+f(4)]\approx11.275788$.

Answer:

(a) The trapezoidal rule approximation is $11.291719$.
(b) The mid - point rule approximation is $11.267900$.
(c) The Simpson's rule approximation is $11.275788$.