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

then

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.