First Application of Definite Integral

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 i^th 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

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

We know the average of n numbers, $a_{1}, a_{2}, \ldots, a_{n}$, is their sum divided by $n: average (mean) =  \frac{1}{n} \sum_{k=1}^{n} a_{k}$.

Finding the average of a function on an interval, an infinite number of values, requires an integral.

To find a Riemann sum approximation of the average value of $f$ on the interval $[a,b]$, we can partition $[a,b]$ into $n$ equally long subintervals of length $\Delta x=(b-a) / n$, pick a value $\mathrm{c}_{\mathrm{i}}$ of $x$ in each subinterval, and find the average of the numbers $\mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right)$. Then

$\text { average of } \mathrm{f} \approx \frac{\mathrm{f}\left(\mathrm{c}_{1}\right)+\mathrm{f}\left(\mathrm{c}_{2}\right)+\ldots+\mathrm{f}\left(\mathrm{c}_{\mathrm{n}}\right)}{\mathrm{n}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right)=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right) \cdot \frac{1}{\mathrm{n}}$

This last sum is not a Riemann sum since it does not have the form $\sum \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right) \cdot \Delta \mathrm{x}_{\mathrm{i}}$ , but it can be manipulated into one:

$\sum_{i=1}^{n} f\left(c_{i}\right) \cdot \frac{1}{n}=\sum_{i=1}^{n} f\left(c_{i}\right) \cdot \frac{b-a}{n} \cdot \frac{1}{b-a}=\frac{1}{b-a} \sum_{i=1}^{n} f\left(c_{i}\right) \cdot \frac{b-a}{n}=\frac{1}{b-a} \sum_{i=1}^{n} f\left(c_{i}\right) \cdot \Delta x$

Then $\{\text { average of } f\} \approx \frac{1}{b-a} \sum_{k=1}^{n} f\left(c_{i}\right) \cdot \Delta x \longrightarrow \frac{1}{b-a} \int_{a}^{b} f(x) d x=\{\text { average of } f\}$

as the number of points $n$ gets larger and the mesh , $(\mathrm{b}-\mathrm{a}) / \mathrm{n}$, approaches 0.

Definition: Average (Mean) Value of a Function

For an integrable function $f$ on the interval $[a,b]$,

the average value of $f$ on $[a,b]$ is $\frac{1}{b-a} \int{a}^{b} f(x) d x$

The average value of a positive $f$ has a nice geometric interpretation. Imagine that the area under $f$ (Fig.6a) is a liquid that can "leak" through the graph to form a rectangle with the same area (Fig. 6b). If the height of the rectangle is $H$, then the area of the rectangle is$\mathrm{H} \cdot(\mathrm{b}-\mathrm{a})$. We know the area of the rectangle is the same as the area under $f$ so

$\mathrm{H} \cdot(\mathrm{b}-\mathrm{a})=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{dx}$ then

$\mathrm{H}=\frac{1}{\mathrm{~b}-\mathrm{a}} \int{a}^{b} \mathrm{f}(x) \mathrm{d} \mathrm{x}$

The average value of positive $f$ is the height $H$ of the rectangle whose area is the same as the area under $f$.

Example 4: Find the average value of $\mathrm{f}(x)=\sin (x)$ on the interval $[0, \pi]$. (Fig. 7)

Solution: Average value = $\frac{1}{\pi-0} \int_{0}^{\pi} \sin (x) \mathrm{dx}$

$=\left.\frac{1}{\pi}(-\cos (x))\right|_{0} ^{\pi}=\frac{1}{\pi}\{-(-1)-(-1)\}=\frac{2}{\pi} \approx 0.6366$

A rectangle with height $2 / \pi \approx 0.64$ on the interval $[0, \pi]$ encloses the same area as one arch of the sine curve. The average value of $\sin (x)$ on the interval $[0,2 \pi]$ is 0 since $\frac{1}{2 \pi} \int_{0}^{2 \pi} \sin (x) \mathrm{dx}=0$.

Practice 2: During a 9 hour work day, the production rate at time $t$ hours was $\mathrm{r}(t)=5+\sqrt{t}$ cars per hour. Find the average hourly production rate. 

Function averages, involving means and more complicated averages, are used to "smooth" data so that underlying patterns are more obvious and to remove high frequency "noise" from signals. In these situations, the original function $f$ is replaced by some "average of $f$". If f is rather jagged time data, then the ten year average of $f$ is the integral $g(x)=\frac{1}{10} \int_{x-5}^{x+5} f(t) d t$. An average of $f$ over 5 units on each side of $x$. For example, Fig. 9 shows the graphs of a Monthly Average (rather “noisy” data) of surface temperature data, an Annual Average (still rather “jagged), and a Five Year Average (a much smoother function). Typically the average function reveals the pattern much more clearly than the original data. This use of a “moving average” value of “noisy” data (weather information, stock prices) is a very common.

The amount of work done on an object is the force applied to the object times the distance the object is moved while the force is applied (Fig. 10) or, more succinctly, $\text { work }=\text { (force) } \cdot(\text { distance })$.

If you lift a 3 pound book 2 feet, then the force is 3 pounds, the weight of the book, and the distance moved is 2 feet, so you have done $(3 \text { pounds }) \cdot(2 \text { feet })=6$ foot- pounds of work. When the applied force and the distance are both constants, then calculating work is simply a matter of multiplying.

Practice 3: How much work is done lifting a 10 pound object from the ground to the top of a 30 foot building (assume the cable is weightless).

If either the force or the distance is variable, then integration is needed.

Example 5: How much work is done lifting a 10 pound object from the ground to the top of a 30 foot building if the cable weighs 2 pounds per foot. (Fig. 11)

Solution: This is more difficult than the Practice problem. When the object is at ground level, a force of 70 pounds (10 pounds plus the weight of 30 feet of cable) must be applied, but when the object is 29 feet above the ground, only 12 pounds of force are needed. In general, if the object is $x$ feet above the ground (Fig. 12), then $30-x$ feet of cable, weighing $2(30-x)$ pounds, is used so the required force is $f(x)=10+2(30-x)=70-2 x$ pounds.

Let's partition the height of the building into small increments so the force needed in each subinterval does not change much. The force in the i^th subinterval will be approximately $\mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right)$ for some $\mathrm{c}_{\mathrm{i}}$ in the subinterval, and the distance moved will be the length of the subinterval, $\Delta \mathrm{x}_{\mathrm{i}}$. The work done to move the object through the subinterval will be $f\left(c_{i}\right) \cdot \Delta x_{i}$, and the total work will be the sum of the work on each subinterval:

$\text { work } \left.\approx \sum \text { (subinterval work }\right)=\sum \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right) \Delta \mathrm{x}_{\mathrm{i}}=\sum\left\{70-2 \mathrm{c}_{\mathrm{i}}\right\} \Delta \mathrm{x}_{\mathrm{i}}$ (a Riemann sum)

$\longrightarrow \int_{0}^{30}\{70-2 x\} \mathrm{dx}$ as the mesh approaches 0.

We have approximated the solution to an applied problem by a Riemann sum, and obtained an exact solution by taking the limit of the Riemann sum to get a definite integral. Now we just need to evaluate the definite integral:

$\int_{0}^{30}(70-2 x) \mathrm{dx}=70 x-\left.x^{2}\right|_{0} ^{30}=\left\{70 \cdot 30-(30)^{2}\right\}-\left\{70 \cdot 0-(0)^{2}\right\}=(2100-900)-(0)=1200 foot–pounds$.

Practice 4: Suppose the building in Example 5 is 50 feet tall and the cable weighs 3 pounds per foot. (a) How much force is needed when the 10 pound object is $x$ feet above the ground? (b) Write an integral for the work done raising the object from the ground to a height of 10 feet. From a height of 10 feet to a height of 20 feet.

In the previous Example and Practice problem, the force was variable and the distance was $\Delta \mathrm{x}$. In later sections we will examine situations where the force is constant and the distance changes.

Practice 1: Using geometry (Fig. 17): $\mathrm{A}=\frac{1}{2}(2)(1)=1 \text { and } \mathrm{B}=\frac{1}{2}(4)(2)=4$ so $total area = A+B=5$.

Using integrals:

$A=\int_{0}^{1}(3-x)-(1+x) d x=\int_{0}^{1}(2-2 x) d x=2 x-\left.x^{2}\right|_{0} ^{1}=(2-1)-(0)=1$

$B=\int_{1}^{3}(1+x)-(3-x) \mathrm{dx}=\int_{1}^{3}(2 x-2) d x=x^{2}-\left.2 x\right|_{1} ^{3}=(9-6)-(1-2)=4$

The single integral $\int_{0}^{3}(1+x)-(3-x) d x$ is not correct: $\int_{0}^{3}(1+x)-(3-x) d x=3$.

Practice 2: $Average value = \frac{1}{\mathrm{~b}-\mathrm{a}} \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{dx}=\frac{1}{9-0} \int_{0}^{9} 5+\sqrt{t} \mathrm{dt}$

$=\frac{1}{9} \int_{0}^{9} 5+\mathrm{t}^{1 / 2} \mathrm{dt}=\left.\frac{1}{9}\left(5 \mathrm{t}+\frac{2}{3} \mathrm{t}^{3 / 2}\right)\right|_{0} ^{9}$

$=\frac{1}{9}\left(45+\frac{2}{3} 9^{3 / 2}\right)-\frac{1}{9}(0)=\frac{1}{9}(45+18)=7$ cars per hour.

Practice 3: $\text { Work }=(\text { force }) \cdot(\text { distance })=(10 \text { pounds }) \cdot(30 \text { feet })=\mathbf{3 0 0}  foot-pounds$.

Practice 4: (a) force = (force for cable) + (force for object)

= (length of cable)(density of cable) + 10 pounds

$=(50-x \text { feet })(3 \text { pounds } / \text { foot })+10 \text { pounds }=160-3 \mathbf{x}$ pounds.

(b) "from the ground to a height of 10 feet:"

$\text { Work } \approx \sum(\text { subinterval work })=\sum \mathrm{f}\left(\mathrm{c}_{\mathrm{i}}\right) \Delta \mathrm{x}_{\mathrm{i}}=\sum\left\{160-3 \mathrm{c}_{\mathrm{i}}\right\} \Delta \mathrm{x}_{\mathrm{i}}$ (a Riemann sum)

$\longrightarrow \int_{0}^{10}\{160-3 x\} d x$ as the mesh approaches 0.

"From a height of 10 feet to a height of 20 feet:" work = $\int_{10}^{20}\{160-3 x\} d x$.