# Tricky integral inequality question

## Homework Statement

Prove the following inequality:

$$\frac{1}{6}\leq\int_{R}\frac{1}{y^{2}+x+1}\chi_{B}(x,y)dxdy\leq\frac{1}{2}$$

where B={(x,y)|0$$\leq (x)\leq (y)\leq1$$} and R=[0,1]x[0,1]
EDIT: The B region should be 0 less than or equal to x less than or equal to y less than or equal to 1.

## Homework Equations

I understand the $$\chi_{B}$$ to be the characteristic function, ie takes value 1 if x is in B, and zero else.

## The Attempt at a Solution

I've bashed my head against a wall for ages with this one. It keeps coming out wrong and different every time.
I don't know if whether to solve it is to brute force integrate on the double integral the function like so:
$$\int^{1}_{0}\int^{y}_{0}\frac{1}{y^{2}+x+1}dxdy$$ (or equivalently the second integral between x and 1, and do dydx, which i believe is the same thing.)

Is this the right thing to do? Or is there a much quicker, shortcut way of doing it?
Many thanks,

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Mark44
Mentor

## Homework Statement

Prove the following inequality:

$$\frac{1}{6}\leq\int_{R}\frac{1}{y^{2}+x+1}\chi_{B}(x,y)dxdy\leq\frac{1}{2}$$

where B={(x,y)|0$$\leq (x)\leq (y)\leq1$$} and R=[0,1]x[0,1]
EDIT: The B region should be 0 less than or equal to x less than or equal to y less than or equal to 1.

## Homework Equations

I understand the $$\chi_{B}$$ to be the characteristic function, ie takes value 1 if x is in B, and zero else.

## The Attempt at a Solution

I've bashed my head against a wall for ages with this one. It keeps coming out wrong and different every time.
I don't know if whether to solve it is to brute force integrate on the double integral the function like so:
$$\int^{1}_{0}\int^{y}_{0}\frac{1}{y^{2}+x+1}dxdy$$ (or equivalently the second integral between x and 1, and do dydx, which i believe is the same thing.)

Is this the right thing to do?
I'm reasonably sure it's not.
Or is there a much quicker, shortcut way of doing it?
Many thanks,
I would start by find the largest value and the smallest value of 1/(y^2 + x + 1) on your region, and finding solids that overestimate and underestimate the volume beneath the surface.

Ah that´s great thanks. So that proves it for when (x,y) are in B.
But when (x,y) are not in B, then we are integrating the function 0 on the interval R=[0,1]x[0,1] with limits of integration 0,1 and x,1 as above, which is 0....
Which is evidently not between 1/6 and 1/2...
So it is only proved for when x,y is in B, and not otherwise...? I'm confused.
It is not true whenever x is greater than y.
?

Mark44
Mentor
Ah that´s great thanks. So that proves it for when (x,y) are in B.
But when (x,y) are not in B, then we are integrating the function 0 on the interval R=[0,1]x[0,1] with limits of integration 0,1 and x,1 as above, which is 0....
Which is evidently not between 1/6 and 1/2...
So it is only proved for when x,y is in B, and not otherwise...?
You don't need to deal with points that aren't in the B region. Do you understand what this region looks like? It is a triangle.
I'm confused.
It is not true whenever x is greater than y.
?
It doesn't matter about points in [0, 1] X [0, 1] for which x >= y. All you need to be concerned with are the points in B.

The solid whose volume you are estimating (with a lower bound and an upper bound) has a triangle base, and the upper surface is curved. The function z = 1/(y2 + x + 1) is continuous on B and is decreasing as x and y increase away from (0, 0), so there are a largest function value and a smallest function value. The two triangle-shaped solids whose heights are the smallest z-value and the largest z-value have volumes that are the lower and upper bounds, respectively.