Use simpson's rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. compare your answer with the value of the integral produced by a calculator. (round your answers to six decimal places.) y = 5xex, 0 ≤ x ≤ 1
a. simpson's rule
b. calculator approximation

= (1/5) x^5 => dy/dx = x^4

dS = dx * sqrt[1 + (dy/dx)^2]

= dx * sqrt(1 + x^4).

So the exact surface area is the integral from x=0 to x=5 of

sqrt(1+x^4) dx.

Since this cannot be integrated analytically in terms of "elementary" functions (it comes out including an elliptic integral, which is a tabulated function) -- the treatment by a numerical method makes sense.

Simpson's Rule dictates that each subinterval [a,b] be treated by approximating the area there as

[(b-a)/6] * [f(a) + 4f(0.5a + 0.5b) + f(b)]

In your problem, it's not obvious whether the subintervals should be chosen as equal intervals of length 1/2, or whether they should be crowded towards the left end of [0,5], since the curvature is concentrated at the left end. What militates against the latter choice is that the area is so much larger towards the right end. Since you're answering a "school" question, I'll play it safe and choose 1/2 as the uniform interval-length.

[0,1/2] area = (1/12) [1 + 4*sqrt(1 + 1/256) + sqrt(1+1/16)]

= (1/12)*6.03858 = 0.50643

[1/2,1] area = (1/12) [sqrt(1+1/16) + 4*sqrt(1 + 81/256) + sqrt(2)]

= (1/12)*7.03438 = 0.58640

[1,3/2] area = (1/12) [sqrt(2) + 4*sqrt(1+625/256) + sqrt(1+81/16)]

= (1/12)*11.29684 = 0.94160

OK, those are the first three (of 10) subintervals. You can do the other 7, and then add all ten results together.

For a check on reasonableness, I used fooplot to graph y=sqrt(1+x^4) from (0,1) to (5,25.02), and then also plotted y=(24/5)+1 from (0,1) to (5,25). The area under the latter is of course 5*13 = 65, and it can be seen from the plot that the area under sqrt(1+x^4) is substantially smaller, but not ENORMOUSLY smaller; it might be somewhere around 50.

16.9 times 2.85

step-by-step explanation: