8.6 Integrals and Area

8.51   Theorem. Let $f$ be a piecewise monotonic function from an interval $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$. Then

$$\int_a^b f=A_a^b f=\alpha(S_a^b f).$$

Proof: We already know this result for monotonic functions, and from this the result follows easily for piecewise monotonic functions. $\diamondsuit$

Remark Theorem 8.51 is in fact true for all integrable functions from $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$, but the proof is rather technical. Since we will never need the result for functions that are not piecewise monotonic, I will not bother to make an assumption out of it.

8.52   Theorem. Let $a,b\in\mbox{{\bf R}}$ and let $f\colon [a,b]\to\mbox{{\bf R}}$ be a piecewise monotonic function. Then the graph of $f$ is a zero-area set.

Proof: We will show that the theorem holds when $f$ is monotonic on $[a,b]$. It then follows easily that the theorem holds when $f$ is piecewise monotonic on $[a,b]$.

Suppose $f$ is increasing on $[a,b]$. Let $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ and let $P=\{x_0,x_1,\cdots ,x_n\}$ be the regular partition of $[a,b]$ into $n$ equal subintervals.

Then

$$x_i-x_{i-1}={{b-a}\over n} \mbox{ for } 1\leq i\leq n$$

and

$$\mbox{graph}(f)\subset\bigcup_{i=1}^n B\Big(x_{i-1},x_i\colon f(x_{i-1}),f(x_i)\Big).$$

Hence

$$\begin{eqnarray*} 0\leq\alpha\Big(\mbox{graph}(f)\Big) &\leq& \alpha \Big( \big... ...(x_i)-f(x_{i-1})\Big)\cr &=& {{b-a}\over n}\Big( f(b)-f(a)\Big). \end{eqnarray*}$$

Now $${\Big\{ {{b-a}\over n}\Big( f(b)-f(a)\Big)\Big\} \to 0}$$, so it follows from the squeezing rule that the constant sequence $\Big\{ \alpha \Big(\mbox{graph}(f)\Big)\Big\}$ converges to $0$, and hence

$$\alpha\Big(\mbox{graph}(f)\Big)=0.\mbox{ $\diamondsuit$}$$

Remark: Theorem 8.52 is actually valid for all integrable functions on $[a,b]$.

8.53   Theorem (Area between graphs.) Let $f,g$ be piecewise monotonic functions on an interval $[a,b]$ such that $g(x)\leq f(x)$ for all $x\in [a,b]$. Let

$$S=\{(x,y)\colon a\leq x\leq b \mbox{ {\rm and} } g(x)\leq y\leq f(x)\}.$$

Then

$$\mbox{{\rm area}}(S)=\int_a^b f(x)-g(x)\; dx.$$

Proof: Let $M$ be a lower bound for $g$, so that

$$0\leq g(x)-M\leq f(x)-M \mbox{ for all } x\in [a,b].$$

Let

$$F(x)=f(x)-M,\qquad G(x)=g(x)-M$$

for all $x\in [a,b]$, and let

$$T=\{(x,y)\colon a\leq x\leq b \mbox{ and } G(x)\leq y\leq F(x)\}.$$

Then

$$\begin{eqnarray*} (x,y)\in T &\iff& a\leq x\leq b \mbox{ and } G(x)\leq y\leq F(... ... y+M\leq f(x)\\ &\iff& (x,y+M)\in S\\ &\iff& (x,y)+(0,M)\in S. \end{eqnarray*}$$

It follows from translation invariance of area that

$$\mbox{area}(S)=\mbox{area}(T).$$

Let

$$\begin{eqnarray*} R &=& \{ (x,y)\colon a\leq x\leq b \mbox{ and } 0\leq y\leq F(... ...y)\colon a\leq x\leq b \mbox{ and } 0\leq y\leq G(x)\} =S_a^b G. \end{eqnarray*}$$

Then $V\cup T=R$, and

$$V\cap T=\{(x,y)\colon a\leq x\leq b \mbox{ and } y=G(x)\}=\mbox{graph}(G).$$

It follows from theorem 8.52 that $V$ and $T$ are almost disjoint, so

$$\mbox{\rm area}(R)=\mbox{area}(V\cup T)=\mbox{area}(V)+\mbox{area}(T),$$

and thus

$$\mbox{\rm area}(T)=\mbox{area}(R)-\mbox{area}(V).$$

By theorem 8.51 we have

$$\mbox{area}(R)=\mbox{area}(S_a^b F)=\int_a^b F(x)\; dx$$

and

$$\mbox{area}(V)=\mbox{area}(S_a^b G)=\int_a^b G(x)\; dx.$$

Thus

$$\begin{eqnarray*} \mbox{area}(S) &=& \mbox{\rm area}(T) = \mbox{\rm area}(R) - \... ...Bigg)\; dx \\ &=&\int_a^b f(x)-g(x)\; dx.\mbox{ $\diamondsuit$} \end{eqnarray*}$$

Remark: Theorem 8.53 is valid for all integrable functions $f$ and $g$. This follows from our proof and the fact that theorems 8.51 and 8.52 are both valid for all integrable functions.

8.54   Example. We will find the area of the set $R$ in the figure, which is bounded by the graphs of $f$ and $g$ where

$$f(x) = {1\over 2} x^2$$

and

$$g(x) = x^3 - 3x^2 + 3x.$$

Now

$$\begin{eqnarray*} g(x) - f(x) &=& x^3 - 3x^2 + 3x - {1\over 2}x^2 = x^3 - {7 \ov... ...+ 3x\\ &=& x(x^2 - {7\over 2} x + 3) = x(x-2)(x - {3\over 2}). \end{eqnarray*}$$

Hence

$$\big( g(x) = f(x) \big) \hspace{1ex}\Longleftrightarrow\hspac... ...space{1ex}\left( x \in \left\{0,{3\over 2},2 \right\} \right).$$

It follows that the points $\mbox{{\bf p}}$ and $\mbox{{\bf q}}$ in the figure are

$$\mbox{{\bf p}}= ({3\over 2}, f({3\over 2})) = ({3\over 2},{9\over 8}) \mbox{ and } \mbox{{\bf q}}= (2,f(2)) = (2,2).$$

Also, since $x(x-2) \leq 0$ for all $x \in [0,2]$,

$$g(x) - f(x) \geq 0 \hspace{1ex}\Longleftrightarrow\hspace{1ex... ...0 \hspace{1ex}\Longleftrightarrow\hspace{1ex}x \leq {3\over 2}.$$

(This is clear from the picture, assuming that the picture is accurate.) Thus

$$\begin{eqnarray*} \mbox{\rm area}(R) &=& \int_0^{3\over 2} (g - f) + \int_{3\ove... ...} \over 3\right) -3\left({2^2 -{({3\over 2})}^2 \over 2}\right). \end{eqnarray*}$$

We have now found the area, but the answer is not in a very informative form. It is not clear whether the number we have found is positive. It would be reasonable to use a calculator to simplify the result, but my experience with calculators is that I am more likely to make an error entering this into my calculator than I am to make an error by doing the calculation myself, so I will continue. I notice that three terms in the answer are repeated twice, so I have

$$\begin{eqnarray*} \mbox{\rm area}(R) &=& 2\left( {81 \over 64} - {63 \over 16} +... ...{21 \over 32 } +{1\over 12} = {63 + 8 \over 96 } = {71\over 96}. \end{eqnarray*}$$

Thus the area is about $0.7$ From the sketch I expect the area to be a little bit smaller than $1$, so the answer is plausible.

8.55   Exercise. The curve whose equation is

$$y^2+2xy+2x^2=4$$ (8.56)

is shown in the figure. Find the area enclosed by the curve.

(The set whose area we want to find is bounded by the graphs of the two functions. You can find the functions by considering equation (8.56) as a quadratic equation in $y$ and solving for $y$ as a function of $x$.) A

8.57   Exercise. A Find the areas of the two sets shaded in the figures below:

8.58   Exercise. A Find the area of the shaded region.