The area under a graph.

Theorem 8.5.1   Let $f$ be a function defined on an interval $[a,b]$ . Suppose that:

• $f$ has a primitive on $[a,b]$ ;
• $\forall x \in [a,b], \; f(x) \geq 0$ .

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.

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:

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

Figure 2: The area under a parabola.

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:

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

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.

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:

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