# error bound for trapezoidal and simpson rule

• Nov 5th 2013, 11:30 AM
ubhutto
error bound for trapezoidal and simpson rule
hey everyone

The question i got was that estimate the minimum number of subintervals with an error of magnitude less than 10^-4 by the trapezoidal rule and the simpson rule for the integral( x^3 + x) from 0 to 2

i did the trapezoidal rule

10^-4 = 6t(2-0)^3 / (12n^2)
f''x is greatest at t=2

n = 282.84 which if approximated is 283
so n = 283

for Simpson rule i did:

10^-4 = 0(2-0)^5 / (180n^4)
but i get n = 0 which i think either isn't right or the simpson rule does not have an error.

can someone tell me if i did it right and also that whenever i get f''x or f''''x, do i always have to get a t between [a,b] that gives me the greatest value for f''x or f''''x

thanks alot
• Nov 5th 2013, 12:02 PM
SlipEternal
Re: error bound for trapezoidal and simpson rule
You did it correctly with the Trapezoidal Rule. For Simpson's Rule, $f^{(4)}(x) = 0$ for all $x$. So, the error is zero, no matter how many subintervals you use. In other words, Simpson's Rule does not just give an approximation, it gives the actual value of the integral, so

\begin{align*}\int_0^2(x^3+x)dx = \left.\left(\dfrac{1}{4}x^4 + \dfrac{1}{2}x^2\right) \right]_0^2 & = 6 \\ \dfrac{2-0}{6}\left[(0^3+0) + 4\left(\left(\dfrac{0+2}{2}\right)^3+\left(\dfrac{ 0+2}{2}\right) \right) + (2^3+2)\right] & = \dfrac{1}{3}\left[ 4(1^3+1)+2^3+2\right] \\ & = \dfrac{18}{3} = 6\end{align*}