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{matrix} 0 \leq f (x) \leq x & if x > 0; \\ 0 \geq f (x) \geq x & if x < 0 . \end{matrix}

Thus

0 \leq \int_{0}^{1} f (x) dx \leq \int_{0}^{1} x dx = \frac{1}{2},

and

0 \geq \int_{- 1}^{0} f (x) dx \geq \int_{- 1}^{0} x dx = \frac{- 1}{2},

and so

- \frac{1}{2} \leq \int_{- 1}^{1} f (x) dx \leq \frac{1}{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 = \frac{2}{3} .

f (0) = 0

and, for

x \neq 0

0 \leq \frac{f (x)}{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 $[- 1 / (n + 1), 1 / (n + 1)]$ .