Problem source

The numbers $a$ and $b$, where $b > a\ge0$,  are such that 
  \[
  \int_a^b x^2 \d x = \left ( \int_a^b x \d x\right)^{\!\!2}\,.
  \]
  \begin{questionparts}
  \item In the case $a=0$ and $b>0$, find the  value of $b$.
  \item In the case $a=1$, show that $b$ satisfies
    \[
    3b^3 -b^2-7b -7 =0\,.
    \]
    Show further, with the help of a sketch, that there is only one
    (real) value of $b$ that satisfies this equation and that it lies
    between $2$ and $3$.
  \item Show that  $3p^2 + q^2 = 3p^2q$, 
    where $p=b+a$ and $q=b-a$, and express $p^2$ in terms of $q$. 
   Deduce that $1< b-a\le\frac43$.
  \end{questionparts}

Solution source

\begin{questionparts}
\item $\,$
\begin{align*}
&& \int_0^b x^2 \d x &= \left ( \int_0^b x \d x \right)^2 \\
\Rightarrow && \frac{b^3}{3} &= \left ( \frac{b^2}{2} \right)^2 \\
\Rightarrow && b &= \frac{4}{3}
\end{align*}

\item $\,$
\begin{align*}
&& \int_1^b x^2 \d x &= \left ( \int_1^b x \d x \right)^2 \\
\Rightarrow && \frac{b^3}{3} - \frac{1}{3} &= \left ( \frac{b^2}{2} - \frac{1}{2} \right)^2 \\
\Rightarrow && 4(b^3 - 1) &= 3(b^2-1)^2 \\
\Rightarrow && 4(b^3-1) &= 3(b^4-2b^2+1) \\
\Rightarrow && 0 &= 3b^4-4b^3-6b^2+7 \\
&&&= (b-1)(3b^3-b^2-7b-7) \\
\Rightarrow && 0 &= 3b^3-b^2-7b-7
\end{align*}


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){((#1)^3/3 - 1/3)};
    \def\functiong(#1){((#1)^2/2 - 1/2)^2};
    \def\xl{-3.5};
    \def\xu{3.5};
    \def\yl{-1};
    \def\yu{6};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, red, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functiong(\x)});

    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

Let $f(x) = 3x^3-x^2-7x-7$ then $f(2) = 3 \cdot 8 - 4 - 14 - 7 = -1 < 0$, $f(3) = 3 \cdot 27 - 9 - 21 - 7 = 44 > 0$ therefore the root must lie between $2$ and $3$.

\item $,$
\begin{align*}
&& \int_a^b x^2 \d x &= \left ( \int_a^b x \d x \right)^2 \\
\Rightarrow && \frac{b^3}{3} - \frac{a^3}{3} &= \left ( \frac{b^2}{2} - \frac{a^2}{2} \right)^2 \\
\Rightarrow && 4(b^3 - a^3) &= 3(b^2-a^2)^2 \\
\Rightarrow && 4(b^2+ab+a^2) &= 3(b-a)(b+a)^2 \\
\Rightarrow && 4 \left ( \left ( \frac{p+q}{2}\right)^2+\left ( \frac{p+q}{2}\right)\left ( \frac{p-q}{2}\right)+\left ( \frac{p-q}{2}\right)^2\right) &= 3qp^2 \\
\Rightarrow && 3p^2 + q^2 &= 3qp^2 \\
\Rightarrow && 3p^2(q-1) &= q^2 \\
\Rightarrow && p^2 &= \frac{q^2}{3(q-1)} \\
\Rightarrow && 1 &\leq \frac{1}{3(q-1)} \\
\Rightarrow && 3(q-1) &\leq 1 \\
\Rightarrow && q & \leq \frac{4}{3} \\
\end{align*}
\end{questionparts}