Problem source

The notation ${\lfloor } x \rfloor$ denotes the greatest integer less than or equal to the real number $x$. Thus, for example, $\lfloor \pi\rfloor =3\,$, $\lfloor 18\rfloor =18\,$ and $\lfloor-4.2\rfloor = -5\,$.
\begin{questionparts}
\item Two curves are given by $y= x^2+3x-1$ and $y=x^2 +3\lfloor  x\rfloor -1\,$.
Sketch the curves, for $1\le x \le 3\,$, on the same axes.
Find the area between the two curves for $1\le x \le n$, where $n$ is a positive integer.
\item Two curves are given by $y= x^2+3x-1$ and $y=\lfloor x\rfloor ^2+3\lfloor x\rfloor -1\,$.
Sketch the curves, for $1\le x \le 3\,$, on the same axes.
Show that  the area  between the two curves for $1\le x \le n$, where $n$ is  a positive integer, is 
\[
\tfrac 16 (n-1)(3n+11)\,.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$

\begin{center}
    \begin{tikzpicture}
    \def\a{-0.8};
    \def\functionf(#1){(#1)^2+3*(#1)-1};
    \def\xl{-0.2};
    \def\xu{3.5};
    \def\yl{-2};
    \def\yu{20};
    
    % 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[blue, smooth, thick, domain=1:3, samples=101]
            plot({\x}, {\functionf(\x)}); 
        \draw[red, smooth, thick, domain=1:2, samples=101]
            plot({\x}, {\x*\x+3-1}); 
        \draw[red, smooth, thick, domain=2:3, samples=101]
            plot({\x}, {\x*\x+6-1}); 
        
        \node[blue, above, rotate=52] at (2.5, {\functionf(2.5)}) {\tiny $y=x^2+3x-1$};

        \node[red, below, rotate=40] at (2.5, {\functionf(2.5)-1.5}) {\tiny $y=x^2+3\lfloor x\rfloor-1$};
        
        % \filldraw (0.5, 0) circle (1.5pt) node[below right]  {$a$};
        % \draw[blue, thick] ({\b*\t}, {-\t}) -- ({-\b*\t}, {\t});
        \filldraw[red] (1, {\functionf(1)}) circle (1.5pt);
        \filldraw[red] (2, {\functionf(2)}) circle (1.5pt);
        \filldraw[red] (3, {\functionf(3)}) circle (1.5pt);
        \draw[red] (2, {\functionf(2)-3}) circle (1.5pt);
        \draw[red] (3, {\functionf(3)-3}) circle (1.5pt);

        % \filldraw (0, 4) circle (1.5pt) node[right] {$4$};
        \filldraw (1, 0) circle (1.5pt) node[below] {$1$};
        \filldraw (3, 0) circle (1.5pt) node[below] {$3$};

    \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}

The difference between the curves is $3x - 3\lfloor x \rfloor$, which has area $\frac32$ for each step. Therefore the area between the curves from $1 \leq x \leq n$ is $\frac32 (n-1)$

\item $\,$

\begin{center}
    \begin{tikzpicture}
    \def\a{-0.8};
    \def\functionf(#1){(#1)^2+3*(#1)-1};
    \def\xl{-0.2};
    \def\xu{3.5};
    \def\yl{-2};
    \def\yu{20};
    
    % 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[blue, smooth, thick, domain=1:3, samples=101]
            plot({\x}, {\functionf(\x)}); 
        \draw[red, smooth, thick, domain=1:2, samples=101]
            plot({\x}, {1+3-1}); 
        \draw[red, smooth, thick, domain=2:3, samples=101]
            plot({\x}, {4+6-1}); 
        
        \node[blue, above, rotate=52] at (2.5, {\functionf(2.5)}) {\tiny $y=x^2+3x-1$};

        \node[red, below] at (2.5, {9}) {\tiny $y=\lfloor x\rfloor ^2+3\lfloor x\rfloor-1$};
        
        % \filldraw (0.5, 0) circle (1.5pt) node[below right]  {$a$};
        % \draw[blue, thick] ({\b*\t}, {-\t}) -- ({-\b*\t}, {\t});
        \filldraw[red] (1, {\functionf(1)}) circle (1.5pt);
        \filldraw[red] (2, {\functionf(2)}) circle (1.5pt);
        \filldraw[red] (3, {\functionf(3)}) circle (1.5pt);
        \draw[red] (2, {3}) circle (1.5pt);
        \draw[red] (3, {9}) circle (1.5pt);

        % \filldraw (0, 4) circle (1.5pt) node[right] {$4$};
        \filldraw (1, 0) circle (1.5pt) node[below] {$1$};
        \filldraw (3, 0) circle (1.5pt) node[below] {$3$};

    \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}

The area between the curves is $x^2 - \lfloor x \rfloor ^2 + 3(x - \lfloor x \rfloor)$. 

Looking at 
\begin{align*}
&& A &= \int_1^n \left ( x^2 - \lfloor x \rfloor ^2 \right )\d x \\
&&&= \frac{n^3-1^3}{3} - \sum_{k=1}^{n-1} k^2 \\
&&&= \frac{(n-1)(n^2+n+1)}{3} - \frac{(n-1)n(2n-1)}{6} \\
&&&= \frac{(n-1) \left (2n^2+2n+2-2n^2+n \right)}{6} \\
&&&= \frac{(n-1)(3n+2)}{6}
\end{align*}

Therefore the total area is $\frac{(n-1)(3n+2)}{6}+\frac32(n-1) = \frac{(n-1)}{6}\left ( 3n+2+9\right) =\frac{(n-1)(3n+11)}{6}$
\end{questionparts}