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
^{1/4}<br />
)
then
^ {1/4}<br />
)
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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