Proving Mean Value Inequality for sin(x) on 0≤x≤1 and 0≤y≤1

absci2010
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Homework Statement


Show that 1/2(1-cos1)\leq\int\intsinx/(1+(xy)4)dxdy\leq1 on the area 0\leqx\leq1, 0\leqy\leq1.


Homework Equations


Mean Value Inequality: m*A(D)\leq\int\intf(x,y)dA\leqM*A(D), where m is the minimum and M is the maximum on the interval.


The Attempt at a Solution


A(D)=1
sinx\leq1, sinx/(1+(xy)4)\leq1, so M=1
I tried to find the minimum on the interval, but I can't find any critical points that are in 0\leqx\leq1 and 0\leqy\leq1.
Please help?
 
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I have not worked this question out, but I suspect you need something more than mA(D) as a minimum. When you plug in x = 0 and y = anything in D, you find a minimum value of 0.
 
If there are no critical points, the minimum should occur somewhere on the boundary.
 
Yes, but that's the part I'm confused about. Where does the value (1/2)(1-cos1) come from?
 
absci2010 said:
Yes, but that's the part I'm confused about. Where does the value (1/2)(1-cos1) come from?

If you look hard enough at that quantity, you might be able to guess at a g(x,y), such that

g(x,y)\leq f(x,y), ~ (x,y)\in D,

for which the integration is elementary.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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