## First Application of Definite Integral

Read this section to see how some applied problems can be reformulated as integration problems. Work through practice problems 1-4.

### Area between f and g

We have already used integrals to find the area between the graph of a function and the horizontal axis. Integrals can also be used to find the area between two graphs. If $f(x) \geq g(x)$ for all $x$ in [a,b], then we can approximate the area between $f$ and $g$ by partitioning the interval [a,b] and forming a Riemann sum (Fig. 2). The height of each rectangle is $\mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right)-\mathrm{g}\left(\mathrm{c}_{\mathrm{i}}\right)$ so the area of the ith rectangle is $\text { (height) } \cdot(\text { base })=\left\{\mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right)-\mathrm{g}\left(\mathrm{c}_{\mathrm{i}}\right)\right\} \cdot \Delta \mathrm{x}_{\mathrm{i}} \text {. }$ . This approximation of the total area is

$\text { area } \approx \sum_{i=1}^{n}\left\{f\left(c_{i}\right)-g\left(c_{i}\right)\right\} \cdot \Delta x_{i}$, a Riemann sum.

The limit of this Riemann sum, as the mesh of the partitions approaches 0, is the definite integral $\int_{\mathrm{a}}^{\mathrm{b}}\{\mathrm{f}(x)-\mathrm{g}(x)\} \mathrm{dx}$.

We will sometimes use an arrow to indicate "the limit of the Riemann sum as the mesh of the partitions approaches zero," and will write

$\sum_{\mathrm{i}=1}^{\mathrm{n}}\left\{\mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right)-\mathrm{g}\left(\mathrm{c}_{\mathrm{i}}\right)\right\} \cdot \Delta \mathrm{x}_{\mathrm{i}} \longrightarrow \int_{\mathrm{a}}^{\mathrm{b}}\{\mathrm{f}(x)-\mathrm{g}(x)\} \mathrm{dx}$

If $\mathbf{f}(x) \geq \mathrm{g}(x)$ on the interval [a,b],

then \begin{aligned} &\left\{\begin{array}{l} \text { area bounded by the graphs } \\ \text { of } \mathrm{f} \text { and } \mathrm{g} \text { and vertical } \\ \text { lines at } x=\mathrm{a} \text { and } x=\mathrm{b} \end{array}\right\}=\int_{\mathrm{a}}^{\mathrm{b}}\{\mathrm{f}(x)-\mathrm{g}(x)\} \mathrm{dx} . \end{aligned}

.

Example 1: Find the area bounded between the graphs of $f(x)=x$ and $g(x)=3$ for $1 \leq x \leq 4$ (Fig. 3).

Solution: It is clear from the figure that the area between $f$ and g is $2.5$ square inches. Using the theorem, area between $f$ and $g$ for $1 \leq x \leq 3$ is

$\int_{1}^{3}\{3-x\} \mathrm{dx}=3 x-\left.\frac{x^{2}}{2}\right|_{1} ^{3}=\left(\frac{9}{2}\right)-\left(\frac{5}{2}\right)=2$, and area between f and g for $3 \leq x \leq 4$ is $\int_{3}^{4}\{x-3\} \mathrm{dx}=\frac{x^{2}}{2}-\left.3 x\right|_{3} ^{4}=\left(\frac{-8}{2}\right)-\left(\frac{-9}{2}\right)=\frac{1}{2}$.

The two integrals also tell us that the total area between $f$ and $g$ is 2.5 square inches.

The single integral $\int_{1}^{4}\{3-x\} \mathrm{dx}=1.5$ hich is not the area we want in this problem. The value of the integral is 1.5, and the value of the area is 2.5.

Practice 1: Use integrals and the graphs of $\mathrm{f}(x)=1+x$ and $g(x)=3-x$ to determine the area between the graphs of $f$ and $g$ for $0 \leq x \leq 3$.

Example 2: Two objects start from the same location and travel along the same path with velocities $\mathrm{v}_{\mathrm{A}}(t)=t+3$ and $\mathrm{v}_{\mathrm{B}}(t)=t^{2}-4 t+3$ meters per second (Fig. 4). ow far ahead is $A$ after 3 seconds? After 5 seconds?

Solution: Since $\mathrm{v}_{\mathrm{A}}(t) \geq \mathrm{v}_{\mathrm{B}}^{(t)}$, the "area" between the graphs of $\mathrm{v}_{\mathrm{A}}$ and $\mathrm{v}_{\mathrm{B}}$ represents the distance between the objects.

After 3 seconds, the distance apart $\int_{0}^{3} \mathrm{v}_{\mathrm{A}}(t)-\mathrm{v}_{\mathrm{b}}(t) \mathrm{dt}=\int_{0}^{3}(t+3)-\left(t^{2}-4 t+3\right) \mathrm{dt}$

$=\int_{0}^{3} 5 t-t^{2} \mathrm{dt}=\frac{5}{2} t^{2}-\left.\frac{t^{3}}{3}\right|_{0} ^{3}=\left(\frac{5}{2} \cdot 9-\frac{27}{3}\right)-\left(\frac{5}{2} \cdot 0-\frac{0}{3}\right)=13 \frac{1}{2}$ meters.

After 5 seconds, the distance apart = $\int_{0}^{5} \mathrm{v}_{\mathrm{A}}(t)-\mathrm{v}_{\mathrm{b}}(t) \mathrm{dt}=\frac{5}{2} t^{2}-\left.\frac{t^{3}}{3}\right|_{0} ^{5}=20 \frac{5}{6} \text { meters. }$

If $f(x) \geq g(x)$, we can use the simpler argument that the area of region $A$ is $\int_{a}^{b} f(x) d x$ and the area of region $B$ is $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{g}(x) \mathrm{dx}$, so the area of region C, the area between $f$ and $g$, is $area of C = (\text { area of } \mathrm{A})-(\text { area of } \mathrm{B})=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{dx}-\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{g}(x) \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x)-\mathrm{g}(x) \mathrm{dx}$.

If the same function is not always greater, then we need to be very careful and find the intervals where $\mathrm{f} \geq \mathrm{g}$ and the intervals where $\mathrm{g} \geq \mathrm{f}$.

Example 3: Find the area of the shaded region in Fig. 5.

Solution: For $0 \leq x \leq 5, f(x) \geq g(x)$ so the area of $A$ is \begin{aligned} &\int_{0}^{5} \mathrm{f}(x)-\mathrm{g}(x) \mathrm{dx}=\int_{0}^{5}(x+3)-\left(x^{2}-4 x+3\right) \mathrm{dx} \\ & \end{aligned}$=\int_{0}^{5} 5 x-x^{2} \mathrm{dx}=\frac{5}{2} x^{2}-\left.\frac{x^{3}}{3}\right|_{0} ^{5}=20 \frac{5}{6}$

For $5 \leq x \leq 7, \mathrm{~g}(x) \geq \mathrm{f}(x)$ so the area of $B$ is

\begin{aligned} &\int_{5}^{7} \mathrm{~g}(x)-\mathrm{f}(x) \mathrm{dx}=\int_{5}^{7}\left(x^{2}-4 x+3\right)-(x+3) \mathrm{dx}=\int_{5}^{7} x^{2}-5 x \mathrm{dx}=\frac{x^{3}}{3}-\left.\frac{5}{2} x^{2}\right|_{5} ^{7}=12 \frac{4}{6} \end{aligned}

Altogether, the total area between $f$ and $g$ for $0 \leq x \leq 7$ is

\begin{aligned} &\int_{0}^{5} \mathrm{f}(x)-\mathrm{g}(x) \mathrm{dx}+\int_{5}^{7} \mathrm{~g}(x)-\mathrm{f}(x) \mathrm{dx}=20 \frac{5}{6}+12 \frac{4}{6}=33 \frac{1}{2} . \end{aligned}