survivorboi Posts: 431, Reputation: 9 Full Member #1 Sep 1, 2009, 03:27 PM
very difficult math equation
Prove it!
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 ebaines Posts: 12,131, Reputation: 1307 Expert #2 Sep 2, 2009, 07:21 AM
Originally Posted by survivorboi
Prove it!
Where did you get this equation from? I think I can show graphically that the sum of the two integrals is precisely 9, so I think the inequality (less than sign) on the left is in error.
 galactus Posts: 2,271, Reputation: 282 Ultra Member #3 Sep 2, 2009, 08:32 AM
Running them through my TI, I see the integrals add to 9.00004286557

Which is between 9 and 9.001

Try using $m(b-a)\leq \int_{a}^{b}f(x)dx\leq M(b-a)$

Suppose f is integrable on [a,b] and $m\leq f(x)\leq M \;\ \forall x \;\ \in [a,b], \;\ m>0, \;\ M>0$
 ebaines Posts: 12,131, Reputation: 1307 Expert #4 Sep 2, 2009, 01:17 PM

OK, here's a method - I'm going to do a bit of "hand waving" here but I trust it'll be reasonably clear...

First, note that the two integrals are functions that are inverses of each other. That is, if

$
y = (x^4+1)^{1/4}
$

then

$
x = (y^4-1)^ {1/4}
$

Hence we know that the area between the lower curve and the x axis is equal to the area between the y axis and upper curve y. These areas are shown as A and A' in the figure. This also means that the areas between each curve and the diagonal are equal -- I have marked these areas as B and B' in the figure. Similarly, the two areas marked C and C' are identical. Finally, there is a little tail out at the end of the upper curve, labelled D.

Now, the integral for the lower curve from x = 1 to x = 3 is equal to the area A. This area is also equal to the area under the diagonal minus B minus C. Since the area under the diagonal is simply a triangle, we know its area is 1/2*3*3 = 4.5.

The integral for the upper curve is equal to the area under the diagonal, plus the areas B' plus C' + D.

Add the two integral together and you have (4.5 - B - C)+(4.5 + B'+ C' +D). Or in other words, the sum of the integrals is simply 9 +D.

The area D is slightly smaller than a triangle, and with a little work you can show that the length of the legs of the triangle are approximately 0.0092 and 0.0093, so the area D is 1/2* 0.0092 * 0.0093 = .000043. Hence the total area is greater than 9, and less than 9.000043. QED.
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 jcaron2 Posts: 986, Reputation: 204 Senior Member #5 Sep 2, 2009, 01:59 PM
Originally Posted by ebaines
The area D is slightly smaller than a triangle, and with a little work you can show that the length of the legs of the triangle are approximately 0.0092 and 0.0093, so the area D is 1/2* 0.0092 * 0.0093 = .000043. Hence the total area is greater than 9, and less than 9.000043. QED.
Brilliant! And so simple now that you point it out. Great job!
 galactus Posts: 2,271, Reputation: 282 Ultra Member #6 Sep 3, 2009, 04:16 AM
I wish they would adjust the 'rate this answer'. Every time I try to give feedback, it tells me I can't. I am just going to quit altogether. :(
I tried ebaines. Nice proof.
 survivorboi Posts: 431, Reputation: 9 Full Member #7 Sep 13, 2009, 07:51 AM
Originally Posted by ebaines
Where did you get this equation from? I think I can show graphically that the sum of the two integrals is precisely 9, so I think the inequality (less than sign) on the left is in error.
It was one of the Math olympiad questions in 1990's something/
 survivorboi Posts: 431, Reputation: 9 Full Member #8 Sep 13, 2009, 07:53 AM
Originally Posted by ebaines
Where did you get this equation from? I think I can show graphically that the sum of the two integrals is precisely 9, so I think the inequality (less than sign) on the left is in error.
It's not an equation is it?
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #9 Sep 13, 2009, 08:14 AM

How could I miss this thread! I think its more appropriate name would be inequality.
 marijaISKRA Posts: 1, Reputation: 1 New Member #10 Jan 25, 2011, 03:37 AM
A little les than 9 ;D

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