Answer to Question #201514 in Quantitative Methods for Dhruv rawat

Question

Use simpson's method to approximate

8

∫ (x^2-x+3)dx with 8 sub-interval

Answer 1

Approximate the integral

$\displaystyle\int_{0}^{8}(x^2-x+3)dx$

with $n=8$ using the Simpson's rule.

$\displaystyle\int_{a}^{b}f(x)dx\approx\dfrac{\Delta x}{3}(f(x_0)+4f(x_1)+2f(x_2)+...$

$+2f(x_{n-2})+4f(x_{n-1}+f(x_n))$

where $\Delta x=\dfrac{b-a}{n}$

We have that $a=0, b=8, n=8.$ Therefore $\Delta x=\dfrac{8-0}{8}=1.$

Divide the interval$[0, 8]$into $n=8$ subintervals of the length $\Delta x=1$ with the following endpoints: $a=0, 1, 2, 3, 4, 5, 6, 7, 8=b.$

Now, just evaluate the function at these endpoints.

$f(x_0)=f(0)=3$

$4f(x_1)=4f(1)=12$

$2f(x_2)=2f(2)=10$

$4f(x_3)=4f(3)=36$

$2f(x_4)=2f(4)=30$

$4f(x_5)=4f(5)=92$

$2f(x_6)=2f(6)=66$

$4f(x_7)=4f(7)=180$

$f(x_8)=f(8)=59$

$\dfrac{1}{3}(3+12+10+36+30+92+66+180+59)$

$=\dfrac{488}{3}\approx162.6667$

$\displaystyle\int_{0}^{8}(x^2-x+3)dx\approx\dfrac{488}{3}\approx162.6667$