The area under a graph.

Theorem 8.5.1   Let $ f$ be a function defined on an interval $ [a,b]$ . Suppose that: Then the area of the region defined by $ a \leq x \leq b$ and $ 0 \leq y \leq f(x)$ is equal to $ \int_a^b f(x) \; dx$ .

Figure 1: The area under the graph of a function.
\begin{figure}\mbox{\epsfig{file=Area01.eps,height=4cm}}\end{figure}

Example 8.5.2   The area $ A$ of the plane region bounded by the $ x-$ axis, the $ y-$ axis, the parabola whose equation is $ y=x^2+1$ and the line whose equation is $ x=2$ is given by:

$\displaystyle A= \int_0^2 (x^2+1) \; dx = \left[ \frac 13 x^3 +x \right]_0^2 = \frac {14}{3}.$    

Figure 2: The area under a parabola.
\begin{figure}\mbox{\epsfig{file=area2.eps,height=4cm}}\end{figure}

Proposition 8.5.3   Let $ \mathcal{C}_1$ and $ \mathcal{C}_2$ be the respective graphs of the functions $ f_1$ and $ f_2$ , both defined and continuous on the interval $ [a,b]$ , where $ a<b$ . Suppose that $ \forall x \in [a,b], \; f_1(x) \leq f_2(x)$ . Then the area of the plane region bounded by the graphs $ \mathcal{C}_1$ and $ \mathcal{C}_2$ , and by the lines whose equations are $ x=a$ and $ x=b$ is given by:

$\displaystyle A=\int_a^b [f_2(x) - f_1(x) ] \; dx.$    

Example 8.5.4   Take $ f_1(x)=\cos x$ and $ f_2(x)= \cos x$ , for $ \frac {\pi}{4} \leq x \leq \frac {5 \pi}{4}$ .
Figure 3: The area between two curves.
\begin{figure}\mbox{\epsfig{file=areabetween.eps,height=4cm}}\end{figure}
The two graphs intersect at $ A$ and $ B$ , as shown on Figure 3. Moreover, the graph $ \mathcal{C}_1$ is under $ \mathcal{C}_2$ for $ \frac {\pi}{4} \leq x \leq \frac {5 \pi}{4}$ . Thus:

$\displaystyle A=\int_{\pi / 4}^{5 \pi / 4} [ \sin x - \cos x ] \; dx = 2 \sqrt{2}.$    

Noah Dana-Picard 2007-12-28