Thread: Midpoint rule

1. Midpoint rule

Use the Midpoint Rule to approximate the integral
with n=3

***

The problem I'm having is, isn't there a rule that states that n has to be an even number? So how does this work?

2. Originally Posted by Archduke01
Use the Midpoint Rule to approximate the integral
with n=3

***

The problem I'm having is, isn't there a rule that states that n has to be an even number? So how does this work?
Midpoint rule, n should be even you can't apply it with n odd
you can use Trapezoidal rule for odd n
Simpson's rule same as Midpoint n even

3. First, is the coefficient of x^2 supposed to be 0?

Second, is n just the number of subintervals you're supposed to use? What's all this about the parity? Just break the integral into three parts and use the approximation $\displaystyle \int_a^b f \,\mathrm{d}x\approx (b-a)f(\frac{a+b}2)$.

4. Originally Posted by Archduke01
Use the Midpoint Rule to approximate the integral
with n=3
***The problem I'm having is, isn't there a rule that states that n has to be an even number? So how does this work?
First I am quite sure there is a typo in this post $\displaystyle 0x^2???$
Now, with the midpoint rule n does not have to be even.
The division points are $\displaystyle x_n = - 8 + \frac{{8n}}{3};\;n = 0,1,2,3$.
The midpoints are $\displaystyle m_k = \frac{{x_{k - 1} + x_k }}{2};\;k = 1,2,3$.
The midpoint sum is $\displaystyle \sum\limits_{k = 1}^3 {\frac{8}{3}f(m_k )}$.

5. My book said n even

Theorem 4.6: let $\displaystyle f\in C^2 [a,b]$, n be even, $\displaystyle h=\frac{(b-a)}{n+2}$, and $\displaystyle x_j = a+(j+1)h$ for each $\displaystyle j=-1,0,...,n+1$. there exist a $\displaystyle \mu \in (a,b)$ for which composite Midpoint rule for n+2 subintervals can be written with error term as

$\displaystyle \int_{a}^{b} f(x) dx = 2h \sum_{j=0}^{n/2} f(x_{2j}) + \frac{b-a}{6} h^2 f''(\mu)$

Eighth Edition
Richard L.Burden
J.Douglas Faires
Youngstown State University

6. That is an absolutely atrocious formula.