Numerical approximations to integrals typically get better -- i.e., their error goes down -- proportional to a power of N, the number of subintervals in the interval of integration. For the upper and lower sums, the error typically goes down like 1/N as N increases. For the midpoint and trapezoidal rules, the error typically goes down like 1/N^2. For Simpson's rule, the error typically goes down like 1/N^4.
(a) Demonstrate this behavior numerically, using the integral of x^3 on [0,1] as a typical integral. (b) Demonstrate that this normal behavior is NOT seen in the integral of sqrt(x) on [0,1]! Apparently the slightly bad behavior of sqrt(x) at 0 (it is not differentiable there) is to blame.© BrainMass Inc. brainmass.com March 4, 2021, 6:45 pm ad1c9bdddf
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(a) Demonstrate this behavior numerically, using the integral of x^3 on [0,1] as a typical integral.
1 Power rule x^n = x^n+1
x^3 . dx = [ x^4/4] --------
[ ¼ - 0] => ¼ => 0.25
Now, we have to use the Trapezoidal rule and simpson's rule, then we have to compare the result with typical integral and show that the error goes down when n increase (ie) If the subinterval increases the error will be reduce.
f(x) dx = h/2 [ f 1+ 2 f2 + 2 f3+......+2 fn+ fn+1]
Where h= (b-a)/n n- is the subinterval (we can take any number for n, when n-increases the error will be reduce)
Here the interval [ 0,1], we take n=4
h= (1-0)/4 = 0.25
So, we take subinterval 0, 0.25, 0.5, 0.75,1
x^3 dx = ...
This solution is comprised of a detailed explanation to demonstrate this behavior numerically, using the integral of x^3 on [0,1] as a typical integral.