Hard Definite Integral: (x^2)/(x^4 + 6) from 2 to 7

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kidfromoz

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May 27, 2006
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Hi, I'm having some difficulties with the way to approach this definite integral:

I'm meant to use a few different methods of integration (when putting through an integral calculator on the computer, it seems far more complicated then expected)

Here is the integral

\(\displaystyle \ \int_2^7 \! \frac{x^2}{x^4+6}\,dx\\)

I tried something weird involving \(\displaystyle \frac{x^2}{(x^2 + i\sqrt6)(x^2 - i\sqrt6)}\\)

Any help would be much appreciated.. Thanks!
 
This isn't a pretty integral to deal with.

Here's what Maple gave me:

hardintegral211hz.gif
hardintegral224as.gif


It could be done, but it would be a booger.

Hey Soroban, work some of that magic.

*edit: Sorry, I threw a stray 2 in there. Well, you get the idea. It's still rather daunting.
 
I went to wolfram, an integral calculator.. and go that as well.
My professor also showed me the program mathematica and I got that.

Hmmm. The question asks me to solve this with three different methods, like simpons rule, trapeziodail rule and that. But i think it would be a very long and tedious process??

Oh and by the way sorry if My post was wrong.. But the integral was:
yuck4ec.th.png


hmmm this is really frustrating :!
 
Okay, I got this
Using advanced grapher (trial program)
By plotting (x^2/(x^4 + 6))
and integreating between 2 and 7
0.3259456 was the approximation:

defint5vx.th.png




I'm a little confused how to do the rectangle method.... hmm Does the graph look about right? I think that get's me somewhere.
 
OKay I've done some work using both trapezoidal method, and simpson's method
and have come up with very similar values as the graphing method
between 0.313537 ish and 0.330714 ish

I'm guessing because I only used 3 intervals with that, (2, 9/2, 7) I only got a rough answer with a fair margin of error?
So? By using more intervals say at like every 1/2 step... so
(2, 5/2, 3.... 7)
or even
(2, 9/4, 5/2, 11/4, 3.... 7)
I would get a way better answer?
Anyway to tell what answer is best?

Also, are there any other methods for this I know of trapezoidal and simpson's rule.. but is there any others?
Thanks again for your help so far!! its been quite helpful (as implied?)
 
By rectangle, I just meant what you're doing. The trapezoid

rule ,Simpson's rule, midpoint, etc. are a summation of the areas of

rectangles; Just different approaches.

Let's do it by hand for, say, our 10 partitions using Simpson:

\(\displaystyle \L\\\int_{2}^{7}\frac{x^2}{x^4+6}dx\)

\(\displaystyle \L\\\frac{7-2}{3(10)}=\frac{1}{6}\)

Code: 
        endpoint       f(x)               multiplier           result

0         2              2/11                    1                  2/11
1         5/2           100/721                  4                  400/721
2         3              3/29                    2                  6/29
3         7/2           196/2497                 4                  784/2497
4         4              8/131                   2                  16/131
5        9/2            108/2219                 4                  432/2219
6         5              25/631                  2                   50/631
7        11/2          484/14737                 4                   1936/14737
8         6               6/217                  2                   12/217
9        13/2          676/28657                 4                   2704/28657
10        7               49/2407                1                    49/2407
                                                                       ----------------------
                                                                            1.95

Add up the results and multiply bu 1/6:

1.95/6=.325
 
Thanks heaps!

I started doing that but only had a few different intervals.
Thanks again for all your help
 
Galactus

Using an applet to work this out with 100 partition I get the answer to be
0.5669...this is way different from your answer?

Why is this?
 
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