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
ì
í
î
0 £ f(x) £ x
if x > 0;
0 ³ f(x) ³ x
if x < 0.
Thus
0 £ ó
õ 		1

0 
 f(x) dx £ ó
õ 		1

0 
 x dx = 1
2
 ,
and
0 ³ ó
õ 		0

-1 
 f(x) dx ³ ó
õ 		0

-1 
 x dx = -1
2
 ,
and so
- 1
2
 £ ó
õ 		1

-1 
 f(x) dx £ 1
2
 .
(b)     Here we have 0 £ f(x) £ x2 for all x, so that
0 = ó
õ 		1

-1 
 0  dx £ ó
õ 		1

-1 
 f(x) dx £ ó
õ 		1

-1 
 x2 dx = 2
3
 .
(c)     In general, consider f(0)=0 and, for x ¹ 0
0 £ f(x)
xn
 £ 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)/xn ³ 0
(B) f(x)/xn £ = 1.

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

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

1. n is even.
If n is even then xn ³ 0 for all x thus 0 £ f(x) £ xn. 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)=xn 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 xn shares the sign of x then so does f(x) so for x ³ 0, f(x) ³ 0 and f(x) £ xn. For x £ 0, f(x) £ 0 and f(x) ³ xn.

Now if we want to minimize the integral we will have f(x)=0 for x ³ 0 and f(x)=xn 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)=xn for x ³ 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)].