Problem source

The function $\mathrm{f}$ satisfies the condition $\mathrm{f}'(x)>0$ for $a\leqslant x\leqslant b$, and $\mathrm{g}$ is the inverse of	$\mathrm{f}.$ By making a suitable change of variable, prove that
	\[
	\int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x=b\beta-a\alpha-\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y,
	\]
	where $\alpha=\mathrm{f}(a)$ and $\beta=\mathrm{f}(b)$. Interpret this formula geometrically, in the case where $\alpha$ and $a$ are both positive. 
	Prove similarly and interpret (for $\alpha>0$ and $a>0$) the formula
	\[
	2\pi\int_{a}^{b}x\mathrm{f}(x)\,\mathrm{d}x=\pi(b^{2}\beta-a^{2}\alpha)-\pi\int_{\alpha}^{\beta}\left[\mathrm{g}(y)\right]^{2}\,\mathrm{d}y.
	\]

Solution source

Let $u = f(x)$ then $\frac{\d u}{\d x} = f'(x)$ and 

\begin{align*}
\int_a^b f(x) \d x &\underbrace{=}_{\text{IBP}} \left [ xf(x) \right]_a^b - \int_a^b x f'(x) \d x \\
&\underbrace{=}_{u = f(x)} b \beta - a \alpha - \int_{u = f(a) = \alpha}^{u = f(b) = \beta} g(u) \d u \\
&= b \beta - a \alpha - \int_{\alpha}^{\beta} g(u) \d u
\end{align*}

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){exp(0.5*#1)+0.05*sin(10*deg(#1))-1};
    \def\xl{-0.1};
    \def\xu{2};
    \def\yl{-0.1};
    \def\yu{2};
    
    % 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
    }

    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    % 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, solid, domain=0:5, samples=100]
            plot ({\x},{\functionf(\x)});

        \draw[thick, red, dashed] (1.5, 0) -- (1.5, {\functionf(1.5)}) -- (0, {\functionf(1.5)});
        \draw[thick, green, dashed] (0.5, 0) -- (0.5, {\functionf(0.5)}) -- (0, {\functionf(0.5)});

        \fill [red, opacity=0.2, domain=0.5:1.5, variable=\x]
          (0.5, 0)
          -- plot ({\x}, {\functionf(\x)})
          -- (1.5, 0)
          -- cycle;
        \fill [blue, opacity=0.2, domain=0.5:1.5, variable=\x]
          (0, {\functionf(0.5)})
          -- plot ({\x}, {\functionf(\x)})
          --  (0, {\functionf(1.5)})
          -- cycle;

        \node [below] at (0.5, 0) {$a$};
        \node [below] at (1.5, 0) {$b$};
        \node [left] at (0, {\functionf(0.5)}) {$\alpha$};
        \node [left] at (0, {\functionf(1.5)}) {$\beta$};

          \fill [green, opacity=0.2]
          (0, 0) 
          -- (0.5, 0) 
          -- (0.5, {\functionf(0.5)}) 
          --  (0, {\functionf(0.5)})
          -- cycle;
    \end{scope}
    
    
    \end{tikzpicture}
\end{center}


\[ \underbrace{\int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x}_{\text{red area}}=\underbrace{b\beta}_{\text{whole area}}-\underbrace{a\alpha}_{\text{area in green}}-\underbrace{\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y}_{\text{area in blue}}, \]

\begin{align*}
2\pi \int_a^b x f(x) \d x &\underbrace{=}_{\text{IBP}}\pi  \left [ x^2 f(x) \right]_a^b - \pi \int_a^b x^2 f'(x) \d x \\
&\underbrace{=}_{x = g(u)} \pi (b^2 \beta - a^2 \alpha) - \pi \int_{u = f(a) = \alpha}^{u = f(b) = \beta} [g(u)]^2 \d u \\
&= \pi(b^2 \beta - a^2 \alpha) - \pi \int_\alpha^\beta [g(u)]^2 \d u
\end{align*}


This is the volume outside the function in the volume of revolution about the $y$ axis between $ \alpha$ and $\beta$.