# Integral Sandwich

Generalise this inequality involving integrals.

## Problem

(i) Suppose that $f(0)=0$ and that, for $x\neq 0$, $$0 \leq {f(x)\over x} \leq 1$$ Show that $$-{1\over 2} \leq \int_{-1}^1 f(x)\,dx \leq {1\over 2}$$ (ii) Suppose that $f(0)=0$ and that, for $x\neq 0$, $$0 \leq {f(x)\over x^2} \leq 1$$ Show that $$0 \leq \int_{-1}^1 f(x)\,dx \leq {2\over 3}$$ (iii) Generalize (i) and (ii) to the case where $f(0)=0$ and, for $x\neq 0$, $$0 \leq {f(x)\over x^n} \leq 1$$ where $n$ is a positive integer.

## Getting Started

You might find it useful to sketch a graph. The graph of $f(x)$ can
take any shape but the inequalities tell you it is sandwiched
between the axes and some other graph, so you can deduce something
about the area under the graph of $f(x)$ and hence about the
integral.

## Student Solutions

Thank you for your solutions to Daniel (no school given), to Shaun from Nottingham High School and to Andrei from Tudor Vianu National College, Bucharest, Romania.

(a) Here we have

$$

\begin{cases}

0 \leq f(x) \leq x &\text{if}\; x > 0\\

0 \geq f(x) \geq x &\text{if} \; x < 0

\end{cases}\;.

$$

Thus $$0 \leq \int_0^1f(x)\,dx \leq \int_0^1x\,dx = {1\over 2}\;,$$ and $$0 \geq \int_{-1}^0f(x)\,dx \geq \int_{-1}^0x\,dx = {-1\over 2}\;,$$ and so $$-{1\over 2} \leq \int_{-1}^1 f(x)\,dx \leq {1\over 2}\;.$$ (b) Here we have $0 \leq f(x) \leq x^2$ for all $x$, so that $$0 = \int_{-1}^1 0 \,dx \leq \int_{-1}^1 f(x)\,dx \leq \int_{-1}^1 x^2\,dx = {2\over 3}\;.$$ (c) In general, consider $f(0)=0$ and, for $x \neq 0$ $$0 \leq {f(x)\over x^n} \leq 1\;.$$ I will now use this constraint to establish upper and lower bounds to the definite integral of $f(x)$ from -1 to 1. First I will break this down into two constraints:

(A) $f(x)/x^n \geq 0$

(B) $f(x)/x^n \leq=1.$

First note that (A) tells us that $f(x)$ has the same sign as $x^n$.

Now we will consider 2 possibilities, $n$ being even and odd:

#### 1. n is even

If n is even then $x^n \geq 0$ for all $x$ thus $0 \leq f(x)\leq x^n$. Now the definite integral is minimized when $f(x)=0$ for all $x$ thus the definite integral is greater than or equal to 0.

It is also maximized when $f(x)=x^n$ and thus the integral is less than or equal to $2/(n+1)$ thus the definite integral is bounded by 0 and $2/(n+1)$.

#### 2. $n$ is odd.

Now since $x^n$ shares the sign of $x$ then so does $f(x)$ so for $x\geq 0$, $f(x)\geq 0$ and $f(x)\leq x^n$. For $x\leq 0$, $f(x)\leq 0$ and $f(x)\geq x^n$.

Now if we want to minimize the integral we will have $f(x)=0$ for $x \geq 0$ and $f(x)=x^n$ for x < 0 and thus we get $-1/(n+1)$ for the lower bound.

In order to maximize the integral we will do the opposite and have $f(x)=x^n$ for $x\geq 0$ and $f(x)=0$ for x < 0 and thus we get $1/(n+1)$ for the upper bound. So in closing...

If $n$ is even then the integral is within the interval $[0,2/(n+1)]$ and if $n$ is odd then the integral is within the interval $[-1/(n+1),1/(n+1)]$.

## Teachers' Resources

This is an example where you have to consider the cases of $n$ even
and $n$ odd separately.

The general cases are proved exactly as for parts (i) and (ii) and you may find it easier to consider even values of $n$ first.

The general cases are proved exactly as for parts (i) and (ii) and you may find it easier to consider even values of $n$ first.