Prove it!

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.

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

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.

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!

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.

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/

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?

How could I miss this thread! I think its more appropriate name would be inequality.

A little les than 9 ;D