### Area L

By sketching a graph of a continuous increasing function, can you prove a useful result about integrals?

### Integral Equation

Solve this integral equation.

### Integral Inequality

An inequality involving integrals of squares of functions.

# Integral Sandwich

##### Stage: 5 Challenge Level:

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 $[-1/(n+1),1/(n+1)]$.