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.

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

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

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

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

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

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

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

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

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

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

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

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