# Using simpsons rule, how to find the error bound?

Randall

## Homework Statement

Find the error bound using Simpsons Rule for integral of SQRT (x) dx from 1 to 9.[/B]

## Homework Equations

E = (M * (b-a)^5) / (180 * n^4), where M = max value of the 4th deriv of x dx
[/B]

## The Attempt at a Solution

see attached please - I can't figure out what to use for the value of M, which is supposed to be the maximum value of the 4th derivative. I graphed the 4th deriv (see attached) and the max y = 0 and the max x is infinity? I need help figuring out M please thanks.https://www.dropbox.com/s/3cn5q3n96iutgu4/IMG_6352.JPG?dl=0 [/B]

#### Attachments

• IMG_6352.JPG
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Homework Helper
Dearly Missed

## Homework Statement

Find the error bound using Simpsons Rule for integral of SQRT (x) dx from 1 to 9.[/B]

## Homework Equations

E = (M * (b-a)^5) / (180 * n^4), where M = max value of the 4th deriv of x dx
[/B]

## The Attempt at a Solution

see attached please - I can't figure out what to use for the value of M, which is supposed to be the maximum value of the 4th derivative. I graphed the 4th deriv (see attached) and the max y = 0 and the max x is infinity? I need help figuring out M please thanks.https://www.dropbox.com/s/3cn5q3n96iutgu4/IMG_6352.JPG?dl=0 [/B]

M is not the max value of the fourth derivative of x dx (although it is related to the fourth derivative of ##f(x)## = the integrand in ##\int_a^b f(x) \, dx##). In fact,
$$M = \max_{a \leq x \leq b} |f^{(4)} (x) |$$

What happens near x = 0 or out at x = ∞ is irrelevant, since the only values the matter here are the ones between x = 1 and x = 9.

Homework Helper
Assuming the 1/3 rule, the approximate error is given by:

$$E_a = - \frac{(b-a)^5}{180n^4} \space \bar f^{(4)}(\epsilon)$$

Where ##\bar f^{(4)}(\epsilon) = \frac{\int_a^b f^{(4)}(x) \space dx}{b - a}## is the average fourth derivative on the interval.

Randall
M is not the max value of the fourth derivative of x dx (although it is related to the fourth derivative of ##f(x)## = the integrand in ##\int_a^b f(x) \, dx##). In fact,
$$M = \max_{a \leq x \leq b} |f^{(4)} (x) |$$

What happens near x = 0 or out at x = ∞ is irrelevant, since the only values the matter here are the ones between x = 1 and x = 9.

Thank you!
-R

Randall
Assuming the 1/3 rule, the approximate error is given by:

$$E_a = - \frac{(b-a)^5}{180n^4} \space \bar f^{(4)}(\epsilon)$$

Where ##\bar f^{(4)}(\epsilon) = \frac{\int_a^b f^{(4)}(x) \space dx}{b - a}## is the average fourth derivative on the interval.

Thank you!
-R

Randall
Ok. So after calculating the error bound using M=15/16, I get 0.6667. Does this mean the approximation needs to be within 0.66% of actual? Or does it need to be within 66.67% of actual? 66.67% seems awfully big, and 0.66% seems to be awfully small....

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