Question

Picture Problem using Simpson's Rule

Using Simpsons Rule (n = 8) , approximate an integral

Answer 1

Where can I post pictures ?

Answer 2

http://www.intmath.com/integration/6-simpsons-rule.php

Answer 3

\(\large\color{black}{ \displaystyle {\rm Simpson's~Rule} \\[0.9em] {\rm Area~~of~f~~over~~[1,3]~~with~n=8}~\\[1.9em]~\displaystyle \normalsize \frac{3-1}{8}\left(f(1)+4f(1.25)+2f(1.50)+4f(1.75)+2f(2) \\ +4f(2.25)+2f(2.50)+4f(2.75)+f(3)\right) }\)

Answer 4

then the maximum error formula is

\(\large\color{black}{ \displaystyle {\rm E}_{S}=\frac{ k(b-a)^5 }{180n^4} }\)

where k is the bound on the 4th derivative.... (if you want this one)

Answer 5

@blackstreet23 @solomanzelman
Do you have everything you need to finish all four parts of the problem?

Answer 6

No I am stuck at part b

Answer 7

Did you try @SolomonZelman 's formula? (you must have seen that before)

Answer 8

@blackstreet23

Answer 9

I mean I know that maximum points in a closed interval picture occur at critical points and end points. so do I need to take the fifth derivative ?

Answer 10

isn't the 4th derivative given? (you don't need 5th derivative )

Answer 11

i need the highest point of the fourth derivative and extrema occur at end points and critical points.

Answer 12

I see... The fifth derivative doesn't look too easy to guess highest
you could differentiate the 4th derivative to find the 5th derivative

Answer 13

but that is correct right?

Answer 14

I see... The forth derivative doesn't look too easy to guess highest
*

Answer 15

http://www.wolframalpha.com/input/?i=f%28x%29%3D-3%285x%5E4-60x%5E2%2B10%29%2F%2816%28x%5E2%2B1%29%5E%2815%2F4%29%29+x%3D-10+to+10+and+y%3D-1+to+1 beautiful finally figured out how to play with zooming options

Answer 16

notice the max is just under 1
for approximately what x value does that occur
the 4th derivative is even (doesn't matter if you choose the positive or negative version of this particular number )

Answer 17

but the fifth derivative is still necessary just to show work right?

Answer 18

@blackstreet23

Answer 19

I would graph and estimate the max. instead of going through the 5th derivative AND solving the resulting equation for f5(x)=0.
Generally the actual error is very much lower than the upper bound estimate, so numerically, it should be ok.

Answer 20

gtg

Answer 21

Graphically, it looks like to be at x=0.8.
Poking around with a few iterations give x=0.787, and f4(.787) around .7149.
That should solve part D.
For part C, you don't need the relative max.

Answer 22

but because those values are not within the interval i do not need to worry of them until part d right?

Answer 23

@mathmate

Answer 24

Exactly. For part B, the maximum is at x=1, and it's a strictly decreasing function, so no need to find the relative maximum.

Answer 25

For part D, even a few casual guesses will give you the maximum within 3 decimal digits.
Have you done part C (find the value of n)?

Answer 26

yes n=6

Answer 27

on d the new value of K4 will be 0.7149345503 right?

Answer 28

For part C, I have 10.1 or 10.2, which is kind of awkward, because it will bring the n=12 when we know almost sure that even n=10 may be good.
Yes, the value of \(k_4\) is 0.7149 at about x=0.79.

Answer 29

Correction:
Yes, I have n=5.713 => 6 for part C.
I must have used the error bound ten times less when I calculated yesterday.