How do you Use Simpson's rule with #n=8# to approximate the integral #int_0^pix^2*sin(x)dx#?
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=(ba)/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;
Note that this integral can be solved using integration by parts twice to get an exact answer which is
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To use Simpson's rule with ( n = 8 ) to approximate the integral ( \int_{0}^{\pi} x^2 \sin(x) , dx ), follow these steps:

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

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 ).

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.

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

The result is the approximation of the integral ( \int_{0}^{\pi} x^2 \sin(x) , dx ) using Simpson's rule with ( n = 8 ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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