Evaluate $\displaystyle \int_{2}^{4} \frac{\sqrt{\ln(9-x)} \ dx}{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}} $.

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- Dec 3rd 2007, 08:25 PMshilz222Integral
Evaluate $\displaystyle \int_{2}^{4} \frac{\sqrt{\ln(9-x)} \ dx}{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}} $.

- Dec 4th 2007, 06:37 AMKrizalid
- Dec 4th 2007, 06:50 AMTKHunny
Is this for a statistics class?

How are your numerical methods?

The first derivative is nearly zero (0) and it is also symmetric on the interval. You should be able to exploit this to your distinct advantage to get very fast convergence. - Dec 4th 2007, 10:46 AMtopsquark
- Dec 4th 2007, 11:01 AMJhevon
- Dec 4th 2007, 11:54 AMKrizalid
First thoughts: I saw $\displaystyle 6-4=2$ & $\displaystyle 6-2=4,$ so the substitution will be helpful. Besides the logarithm quantities will be the same.

(Observe that there's no difference between two integrals which have the same integration limits & integrand but that they differ by the $\displaystyle dx$ & $\displaystyle du,$ the value of the integral is the same.)

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Let $\displaystyle u=6-x.$ (Note that integration limits will be reversed, we'll get the same integral.)

$\displaystyle \int_2^4 {\frac{{\sqrt {\ln (9 - x)} }}

{{\sqrt {\ln (9 - x)} + \sqrt {\ln (x + 3)} }}\,dx} = \int_2^4 {\frac{{\sqrt {\ln (u + 3)} }}

{{\sqrt {\ln (u + 3)} + \sqrt {\ln (9 - u)} }}\,du} .$

Now define

$\displaystyle \varphi=\int_2^4 {\frac{{\sqrt {\ln (9 - x)} }}

{{\sqrt {\ln (9 - x)} + \sqrt {\ln (x + 3)} }}\,dx} = \int_2^4 {\frac{{\sqrt {\ln (x + 3)} }}

{{\sqrt {\ln (x + 3)} + \sqrt {\ln (9 - x)} }}\,dx} .$

Add the last two integrals, divide the answer by 2 and we're done.

P.S.: Jhevon coming to Chile? Wow, that'd be nice! Let's learn some of spanish :D - Dec 4th 2007, 03:12 PMThePerfectHacker
The greatest integrator I have ever seen on a math forum is "Dr.Doe".

- Dec 4th 2007, 03:14 PMJhevon
- Dec 4th 2007, 03:15 PMThePerfectHacker
- Dec 4th 2007, 03:31 PMKrizalid
John Doe, 26 years old.

On his way to be a Dr.

He's huge, extremely huge.

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I like integration and all those stuff, but I'm not a Master or something. There're a lot people which knows more than I, but always glad to reply threads which involve integration.

Someone told me that I have to be patient :D (Thing that I really hate.) - Dec 4th 2007, 04:08 PMPlato
I realize that I am somewhat of a spoilsport.

I think that the skill to do arcane integrals will go the way of finding square roots by hand.

Given the availability of computer algebra systems on laptops and calculators (TI89 & TI92), it just seems to me that we can better spend what little class time we have exploring the real properties, theories, and applications of the integral. For example, I see no need to teach partial fractions. Let the software do it.

How many mathematicians know that many derivatives do not have so-called integrals until we employ a Denjoy/Henstock type definition? What are the implications of that? I once posed such a problem to a numerical analyst. His attempt to solve it crashed the university’s computer system. My aim was to point out “what is the point of trying to evaluate something that does not exist?”.

So who cares if you can find the answer to obscure problems?

We all care if someone crashes an entire system! - Dec 4th 2007, 04:22 PMThePerfectHacker
No, being able to do integral manipulations does become useful. When solving a specific integral problem, integration ideas can become very useful.

Note, all (or most) the great mathematicians from the past (take Jacobi for example) were amazing integral manipulators/algebra manipulators. - Dec 4th 2007, 05:08 PMPlato
Note that they lived before lived before computers.

Before Newton, amost all English mathematicians went to Italian universities to get grounded in techniques.

I think that we are living at a point in mathematical history where a huge paradigm shift is occurring. Unless one wants to be a computer programmer, I have real doubts that valued techniques, such as doing obscure integrals, will have any value in future mathematics education. Other topics are just so much more importent. - Dec 4th 2007, 05:09 PMshilz222
I got the answer. It's $\displaystyle 1 $. All you had to do was multiply the integral by $\displaystyle 2 $. Then divide by $\displaystyle 2 $.

So $\displaystyle 2I = \int_{2}^{4} \ dx $, $\displaystyle I = 1 $. - Dec 4th 2007, 05:17 PMThePerfectHacker
I hate calculators, they ruin the mathematical abilities of students. Look at China they still use abacus and they got one of the best math education programs. Russia also, and I am sure they do not have those fancy calculators like the TI (in fact I do not own one) and they have also one of the best math programs.

But eventhough I hate calculators I use them when I have to do complicated calculations. I have no problem with using calculators when doing calculations I can do. But when I am learning something new I would never ever use them. They blind the persons abilities. I am happy of my College because it banned calculators on exams, with what I completely agree.

Thus, I see nothing to be a waste of time when a person does integral computations by hand. Or when he does arithmetical calculations like proving the 100th digit in exansion of sqrt(2) is a 5, yes eventhough a computer can easily do that a person who can do that by hand certainly knows how numbers work well.

But I do feel that in the future people will be worse at math. I heard stories of people passing differecial eqautions courses all because their calculator can do symbolic algebra.