(i)     Here we have

${\displaystyle \{\begin{array}{cc}\hfill 0\le f(x)\le x\hfill & \hfill \mathrm{if}x>0;\hfill \\ \multicolumn{1}{c}{0\ge f(x)\ge x}& \hfill \mathrm{if}x<0.\hfill \end{array}}$

Thus

${\displaystyle 0\le {\int }_0^{1}f(x)\hspace{0.5em}\mathrm{dx}\le {\int }_0^{1}x\hspace{0.5em}\mathrm{dx}=\frac{1}{2},}$

and

${\displaystyle 0\ge {\int }_{-1}^{0}f(x)\hspace{0.5em}\mathrm{dx}\ge {\int }_{-1}^{0}x\hspace{0.5em}\mathrm{dx}=\frac{-1}{2},}$

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

${\displaystyle \{\begin{array}{cc}\hfill 0\le f(x)\le x^{2m+1}\hfill & \hfill \mathrm{if}x>0;\hfill \\ \multicolumn{1}{c}{0\ge f(x)\ge x^{2m+1}}& \hfill \mathrm{if}x<0.\hfill \end{array}}$

Thus

${\displaystyle 0\le {\int }_0^{1}f(x)\hspace{0.5em}\mathrm{dx}\le {\int }_0^1{x}^{2m+1}\hspace{0.5em}\mathrm{dx}=\frac{1}{2m+2},}$

and

${\displaystyle 0\ge {\int }_{-1}^{0}f(x)\hspace{0.5em}\mathrm{dx}\ge {\int }_{-1}^0{x}^{2m+1}\hspace{0.5em}\mathrm{dx}=\frac{-1}{2m+2},}$

so that

${\displaystyle -\frac{1}{2m+2}\le {\int }_{-1}^{1}f(x)\hspace{0.5em}\mathrm{dx}\le \frac{1}{2m+2}.}$

(ii)     Here we have $0\le f(x)\le x^2$ for all $x$, so that

${\displaystyle 0={\int }_{-1}^{1}0\hspace{0.5em}\mathrm{dx}\le {\int }_{-1}^{1}f(x)\hspace{0.5em}\mathrm{dx}\le {\int }_{-1}^1{x}^2\hspace{0.5em}\mathrm{dx}=\frac{2}{3}.}$

In general if $n$ is even, say $n=2m$, we have $0\le f(x)\le x^{2m}$, so that

${\displaystyle 0={\int }_{-1}^{1}0\hspace{0.5em}\mathrm{dx}\le {\int }_{-1}^{1}f(x)\hspace{0.5em}\mathrm{dx}\le {\int }_{-1}^1{x}^{2m}\hspace{0.5em}\mathrm{dx}=\frac{2}{2m+1}.}$