How do you Use Simpson's rule with #n=8# to approximate the integral #int_0^pix^2*sin(x)dx#?

Question

Ezra Adams · Accepted Answer

The answer is 5.86924686.

For any numerical approximation of a function, you always start with a table of values. For your problem, we have:

#a=0#
#b=pi#
#n=8#

So,

#Delta x=(b-a)/n=pi/8#
#x_i=a+i Delta x, i in {0, 1, ..., 8}#

Now it is a matter of applying Simpson's Rule:

#int_0^pi x^2*sin x dx = int_0^pi f(x)dx ~~ (Delta x)/3(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+...+2f(x_6)+4f(x_7)+f(x_8))#

I'll skip the substitution of values because it's messy.
We get 5.86924686 as the approximation.

Using numerical integration on a calculator gets a value of 5.869604401 which means the approximation is good to 3 decimal places.

Notice the pattern of the coefficients for the sum is: 1, 4, 2, 4, ..., 2, 4, 1. This means that to use Simpson's Rule, we need an odd number of values or an even number of intervals; #n# is even.

Note that this integral can be solved using integration by parts twice to get an exact answer which is #pi^2-4#.

Ezra Adams · Answer

undefined To use Simpson's rule with $ n = 8 $ to approximate the integral $ \int_{0}^{\pi} x^2 \sin(x) \, dx $, follow these steps:

1. Divide the interval $[0, \pi]$ into $ n = 8 $ equal subintervals. The width of each subinterval will be $ \Delta x = \frac{\pi}{8} $.

2. Compute the function values at the endpoints and the midpoint of each subinterval. That is, evaluate $ f(x) = x^2 \sin(x) $ at $ x = 0, \frac{\pi}{8}, \frac{\pi}{4}, \frac{3\pi}{8}, \frac{\pi}{2}, \frac{5\pi}{8}, \frac{3\pi}{4}, \frac{7\pi}{8}, $ and $ \pi $.

3. Apply Simpson's rule formula for $ n = 8 $:
$$ \text{Approximation} = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8) \right] $$
Where $ x_0, x_1, \ldots, x_8 $ are the endpoints and midpoints of the subintervals.

4. Substitute the function values obtained in step 2 into the formula and compute the approximation.

5. The result is the approximation of the integral $ \int_{0}^{\pi} x^2 \sin(x) \, dx $ using Simpson's rule with $ n = 8 $.