Problem source

If   ${\rm f}(t)\ge {\rm g}(t)$ for $a\le t\le b$, explain very briefly why $\displaystyle \int_a^b {\rm f}(t) \d t \ge \int_a^b {\rm g}(t) \d t$.
Prove that if $p>q>0$ and $x\ge1$ then
$$\frac{x^p-1}{ p}\ge\frac{x^q-1}{ q}.$$
Show that this inequality also holds when $p>q>0$ and $0\le x\le1$.
Prove that, if $p>q>0$ and $x\ge0$, then
$$\frac{1}{ p}\left(\frac{x^p}{ p+1}-1\right)\ge
\frac{1}{q}\left(\frac{x^q}{ q+1}-1\right).$$

Solution source

This is just the result that all of the area beneath $g(t)$ is also below $f(t)$

If $p > q > 0, x \geq 1 \Rightarrow x^p \geq x^q$, therefore applying the result we have

\begin{align*}
&& \int_1^x x^p\, \d t & \geq  \int_1^x x^q\, \d t \\
\Rightarrow && \frac{x^p-1}{p} & \geq \frac{x^q-1}{q}
\end{align*}

When $p > q > 0, 0 \leq x \leq 1$ we have $x^p \leq x^q$, ie

\begin{align*}
&& \int_x^1 x^q\, \d t & \geq  \int_{x}^1 x^p\, \d t \\
\Rightarrow && \frac{1-x^q}{q} & \geq \frac{1-x^p}{p} \\
\Rightarrow && \frac{x^p-1}{p} &\geq \frac{x^q-1}{q}
\end{align*}

Now looking at the functions $f(x) = \frac{x^p-1}{p}, g(x) =  \frac{x^q-1}{q}$ and $x \geq 1$ we have

\begin{align*}
&& \int_0^x \frac{t^p-1}{p} \d t & \geq \int_0^x \frac{t^q-1}{q} \d t\\
\Rightarrow &&\frac1p \left[\frac{t^{p+1}}{p+1} - t \right]_0^x &\geq\frac1q \left[\frac{t^{q+1}}{q+1} - t \right]_0^x \\
\Rightarrow &&\frac1p \left(\frac{x^{p+1}}{p+1} -x\right) &\geq\frac1q \left(\frac{x^{q+1}}{q+1} - x \right)\\
\Rightarrow &&\frac1p \left(\frac{x^{p}}{p+1} -1\right) &\geq\frac1q \left(\frac{x^{q}}{q+1} - 1 \right)\\


\end{align*}