## 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.

### Work

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 ith 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.