Question

I could definitely use some help with the Trapezoid, Midpoint, and Simpson's rule (where n=6) for:
\[\large9\int\limits_{1}^{4}\sqrt{\ln \left| x \right|}\cdot dx\]
I think I must be confused by the instructions for these or something because it's not taking my evaluated decimal answers, six significant figures required. (+1 medal as always for steps shown, ty in advance to taking the time to help)

Answer 1

\[\triangle x = \frac{b-a}{n} = \frac{4-1}{6}=\frac{1}{2}\]

Answer 2

And so...

Answer 3

\[ 9 \times\sum_{n = 1}^{6} \frac 12\left ( \sqrt{ \ln \left (1 + (n-1) \frac 1 2 \right ) } + \sqrt{ \ln \left (1 + (n) \frac 1 2 \right ) } \right ) \frac 12 \]

Answer 4

Err what? I know the definition of an integral is the sum of areas as the width of each sample goes to zero, but I've never seen what you've written before like that @experimentX

Answer 5

isn't it asking for numerical integration??

Answer 6

\[T_n= \frac{\triangle x}{2} (f(x_1)+2f(x_2)+...2f(x_{n-1})+f(x_n)\]
\[M_n= \triangle x (f(x_1)+f(x_2)+...+f(x_n)\]
\[S_n= \frac{\triangle x}{3} (f(x_1)+4f(x_2)+2f(x_3)+2f(x_4)+...2f(x_{n-2})+4f(x_{n-1})+f(x_n)\]

Answer 7

True it is, but it's asking not for exact, it's asking for approximated by these things.

Answer 8

To six decimal places. I keep trying to see if I have an error in my formulas or a typo on the calculator or something

Answer 9

For Simpson's Rule, the coefficients go 1-4-2-4-2-...-2-4-1.

Answer 10

Also, I'm assuming it's a typo, but for Midpoint Rule, the sample points \(x_i^\star\) should be \(x_i^\star=\frac{1}{2}(x_{i-1}+x_i)\)

Answer 11

@blockcolder , Yay finally somebody who knows this stuff! :D So for the midpoint rule, each f(x) should be f(\(x_i^*\)) where that sample point what you wrote above? I'll have to try that again. And yeah, clearly Simpson's rule in my notes was wrong then if that's how it's supposed to go. In the example problem in the text they only worked out one with n=5, not any larger.

Answer 12

\[S_n= \frac{\triangle x}{3} (f(1)+4f(1.5)+2f(2)+4f(2.5)+2f(3)+4f(3.5)+f(4))\]
Like this?

Answer 13

You can't use Simpson's Rule for n=5, because it's odd.
For n=6, Simpson's Rule should be:
\[S_n=\frac{\Delta x}{3}(f(x_0)+2f(x_1)+4f(x_2)+2f(x_3)+4f(x_4)+2f(x_5)+f(x_6))\]
You're forgetting to include \(x_0\), btw. \(x_0=a\)

Answer 14

Ok I've gotten everything correct except for the midpoint gentlement

Answer 15

T\(_n \approx\) 23.32200512

S\(_n \approx\) 23.68778685

Ty for your help so far @blockcolder & @experimentX ;-)

Answer 16

I guess I need a bit of a better understanding as to what the terms of the midpoint rule should be?

Answer 17

haven't done midpoint rule yet ... let's see what's on google.

Answer 18

Meh Google didn't help much, it still giving me errors for some reason. There's a second problem tied to this one which also requires using the midpoint, right-endpoint, and left-endpoint based on a drawing posted, not an actual function.

Answer 19

Here take a look @experimentX

This is the other one the follows this:

Answer 20

the midpoint rule just says
\[ \sum \Delta x f(x + \Delta x/2)\]

Answer 21

Kind of a pain if you ask me, I tried using half-unit samples and as you can see it says NO!

Answer 22

Granted that's a separate problem there, but it's the same midpoint rule.

Answer 23

And you can't just average the right and left endpoints apparently

Answer 24

Which is what I THOUGHT the "midpoint" was all about, no?

Answer 25

seems all right ... L2<M2<R2

Answer 26

the area of that curve must be around 46's

Answer 27

But as you can see I'm getting a big fat NO from anything in the system regarding Midpoint rule, which I feel logical must mean I'm doing something wrong in my process.

Answer 28

logically*

Answer 29

can you give me the function?

Answer 30

You mean for that image one? No that's literally it. The function is what's visually specified, unfortunately

Answer 31

seems like a parabola ... what's your answer?? try to reduce the step size to get more accurate result.

Answer 32

For the one proposed initially for this question/topic where I needed to use 3 different methods, that's as written sqrt(ln|x|)).

Reduce to 0.25 samples instead of 0.5?

Answer 33

For \(\triangle x\) I mean

Answer 34

I'm currently trying to compare this topic with this one: http://www.physicsforums.com/showthread.php?t=157496

Answer 35

well ... that gives more accurate result for every numerical integration. For the original problem,
Sum[ 9 ( Sqrt[Log[1 + n * 0.5]]), { n, 1 , 5 }]
on mathematica gives 41.3457

Answer 36

Sorry,
Sum[ 9 ( Sqrt[Log[1 + n * 0.5]])*0.5, { n, 1 , 5 }] = 20.6728

Answer 37

Alas it says no, to this: http://www.wolframalpha.com/input/?i=Sum [+9+%28+Sqrt[Log[1+%2B+n+*+0.5]]%29*0.5%2C+{+n%2C+1+%2C+5+}]

Answer 38

Sorry ... i am not getting accurate results. wait for a while.

Answer 39

err that link broke

Answer 40

I can see why the coefficient out front is 9/2 easily enough, let me the terms...

Answer 41

http://www.wolframalpha.com/input/?i=Sum%5B+9+%28+Sqrt%5BLog%5B1+%2B+n+*+0.5+%2B+0.25%5D%5D%29*0.5%2C+%7B+n%2C+0+%2C+5+%7D%5D

Answer 42

9*\(\triangle x\)(\(\sqrt{\ln|1.25|}\)+\(\sqrt{\ln|1.75|}\)+\(\sqrt{\ln|2.25|}\)+\(\sqrt{\ln|2.75|}\)+\(\sqrt{\ln|3.25|}\)+\(\sqrt{\ln|3.75|}\))

Answer 43

this is for right sum
Sum[ 9 ( Sqrt[Log[1 + n * 0.5 ]])*0.5, { n, 1 , 6 }] = 25.9712
For mid sum
Sum[ 9 ( Sqrt[Log[1 + n * 0.5 + 0.25]])*0.5, { n, 0 , 5}] = 24.1294
For left sum
Sum[ 9 ( Sqrt[Log[1 + n * 0.5 ]])*0.5, { n, 0 , 5 }] = 20.6728

Answer 44

well ... what value did you get for that??

Answer 45

i think it's better to call it upper sum and lower sum rather than left sum or right sum.

Answer 46

AH HA! 24.12940953421 It wasn't taking 24.1294

Answer 47

Freaking computer, like 9 millionths matters...

Answer 48

online test??

Answer 49

1. How do I manipulate your syntax-based formula for other functions?
2. Can I give you a medal and you give the other guy a medal? I would love to give credit to both of you :D
3. Not quite, online homework stuff.

Answer 50

One of the few parts I'm stuck on

Answer 51

It's only a few points, but I'd rather get as much as I can and understand why, takes pressure off the test

Answer 52

sure ..where are you stuck??

Answer 53

Well it's this section, and part of it it I think is because of the computer being a punk and not taking it if it's not EXACTLY what it wanted

Answer 54

lol

Answer 55

I'll create two new topics @experimentX , so you can get extra medals for your extra effort. It's the same stuff so we're on the same page already, and nobody else seems to want to help out with this mess of calculus so you're GREATLY appreciated. One of the few you and @blockcolder are, and my deepest respects to you both :-)

Answer 56

well ... just check in for value of x
Sum[ function goes first, {value of increment, initial value , final value}]

in function place modify it by
f(x + n*increment step)* increment step.

Answer 57

no probs ... it's all right. BTW i didn't know midpoint formula until now. well, i have unusual learning methods.