Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to "integral between 1 and 3 of" 1/sqrt(1+x^3) dx giving your answer to three significant figures.

Just as a note before I start, some of the symbols I use here will be horribly confused, this won't be an issue with a whiteboard but doing maths in a text editor is not great so I've had to make do.

Taking the expression as g(f(x)) where g is the integral function, Simpsons Rule gives us g(f(x)) roughly equals (h/3)[f(x0) + 4f(x1) + 2(f(x2)) + ... + 4f(x(n-1)) + f(x(n))] which is an approximation of g(f(x)), where n is the number of "strips" and h is the width of each strip.

To find h, we use the equation h = (b-a)/n, where a and b are the bounds of the integral. In this instance, the "integral from 1 to 3" has bounds 1 and 3, giving us a=3 and b=1 (on a whiteboard b would be the at the top of the integral and a at the bottom). So then, we have h = (3-1)/4 = 1/2. So substituting this in to Simpsons Rule, the value of x in f(x) goes up from 1 to 3 in increments of 1/2, giving us g(f(x)) = ((1/2)/3)[f(1) + 4f(1.5) + 2(f(2)) + 4f(2.5) + f(3)].

Now all we have to do is use our value of x to find each value of f(x) (f(x) = 1/sqrt(1+x^3)), write them out in a table, and substitute them into g(f(x)), which gives us g(f(x)) roughly equals 0.743 to 3dp