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

absci2010
Messages
10
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top