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use the trapezoidal rule, the midpoint rule, and simpsons rule to appro…

Question

use the trapezoidal rule, the midpoint rule, and simpsons rule to approximate the given integral with the specified value of n. (round your answers to six decimal places.) (int_{0}^{12}x^{3}sin(x)dx, n = 8) (a) the trapezoidal rule -1392.087125 good job. (b) the midpoint rule -1744.434609 nice job! (c) simpsons rule -1636.408033 good work.

Explanation:

Step1: Define the integral and parameters

The integral is $\int_{0}^{12}x^{3}\sin(x)dx$ and $n = 8$. First, find the width of each sub - interval $\Delta x=\frac{b - a}{n}$, where $a = 0$, $b = 12$ and $n=8$. So, $\Delta x=\frac{12-0}{8}=1.5$.

Step2: Trapezoidal rule

The trapezoidal rule formula is $T_n=\frac{\Delta x}{2}[f(x_0)+2\sum_{i = 1}^{n - 1}f(x_i)+f(x_n)]$.
Let $x_i=a + i\Delta x$. Calculate $f(x)=x^{3}\sin(x)$ at each $x_i$.
$x_0 = 0,f(x_0)=0$; $x_1=1.5,f(x_1)=(1.5)^{3}\sin(1.5)$; $x_2 = 3,f(x_2)=3^{3}\sin(3)$; $x_3=4.5,f(x_3)=(4.5)^{3}\sin(4.5)$; $x_4 = 6,f(x_4)=6^{3}\sin(6)$; $x_5=7.5,f(x_5)=(7.5)^{3}\sin(7.5)$; $x_6 = 9,f(x_6)=9^{3}\sin(9)$; $x_7=10.5,f(x_7)=(10.5)^{3}\sin(10.5)$; $x_8 = 12,f(x_8)=12^{3}\sin(12)$.
Then $T_8=\frac{1.5}{2}[0 + 2(f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5)+f(x_6)+f(x_7))+f(x_8)]\approx - 1392.087125$.

Step3: Mid - point rule

The mid - point rule formula is $M_n=\Delta x\sum_{i = 0}^{n - 1}f(\frac{x_i+x_{i + 1}}{2})$.
The mid - points are $m_0=\frac{0 + 1.5}{2}=0.75,m_1=\frac{1.5+3}{2}=2.25,m_2=\frac{3 + 4.5}{2}=3.75,m_3=\frac{4.5+6}{2}=5.25,m_4=\frac{6+7.5}{2}=6.75,m_5=\frac{7.5+9}{2}=8.25,m_6=\frac{9+10.5}{2}=9.75,m_7=\frac{10.5+12}{2}=11.25$.
Calculate $f(m_i)=m_i^{3}\sin(m_i)$ for each $m_i$ and sum them up: $M_8=1.5\sum_{i = 0}^{7}f(m_i)\approx - 1744.434609$.

Step4: Simpson's rule

The Simpson's rule formula is $S_n=\frac{\Delta x}{3}[f(x_0)+4\sum_{i = 1, i\text{ odd}}^{n - 1}f(x_i)+2\sum_{i = 2, i\text{ even}}^{n - 2}f(x_i)+f(x_n)]$.
Using the values of $x_i$ and $f(x)=x^{3}\sin(x)$ calculated before:
$S_8=\frac{1.5}{3}[0+4(f(x_1)+f(x_3)+f(x_5)+f(x_7))+2(f(x_2)+f(x_4)+f(x_6))+f(x_8)]\approx - 1636.408033$.

Answer:

(a) - 1392.087125
(b) - 1744.434609
(c) - 1636.408033