(i)     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 ,

so the inequality follows. In general, if n is odd, say n=2m+1, we have

ì í î 0 £ f(x) £ x2m+1 if x > 0; 0 ³ f(x) ³ x2m+1 if x < 0.

Thus

0 £ ó õ 1 0  f(x) dx £ ó õ 1 0  x2m+1 dx = 1 2m+2 ,

and

0 ³ ó õ 0 -1  f(x) dx ³ ó õ 0 -1  x2m+1 dx = -1 2m+2 ,

so that

- 1 2m+2 £ ó õ 1 -1  f(x) dx £ 1 2m+2 .

(ii)     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 .

In general if n is even, say n=2m, we have 0 £ f(x) £ x^2m, so that

0 = ó õ 1 -1  0  dx £ ó õ 1 -1  f(x) dx £ ó õ 1 -1  x2m dx = 2 2m+1 .