## Calculus AB and Calculus BC

## CHAPTER 8 Further Applications of Integration

**Concepts and Skills**

In this chapter, we will review many ways that definite integrals can be used to solve a variety of problems, notably distance traveled by an object in motion along a line. We’ll see that in a variety of settings accumulated change can be expressed as a Riemann sum whose limit becomes an integral of the rate of change.

For BC students, we’ll expand our discussion of motion to include objects in motion in a plane along a parametrically defined curve.

### A. MOTION ALONG A STRAIGHT LINE

If the motion of a particle *P* along a straight line is given by the equation *s* = *F*(*t*), where *s* is the distance at time *t* of *P* from a fixed point on the line, then the velocity and acceleration of *P* at time *t* are given respectively by

This topic was discussed as an application of differentiation. Here we will apply integration to find velocity from acceleration and distance from velocity.

**Distance**

If we know that particle *P* has velocity *v*(*t*), where *v* is a continuous function, then the *distance* traveled by the particle during the time interval from *t* = *a* to *t* = *b* is the definite integral of its speed:

If *v*(*t*) 0 for all *t* on [*a*, *b*] (i.e., *P* moves only in the positive direction), then (1) is equivalent to similarly, if *v*(*t*) 0 on [*a*, *b*] (*P* moves only in the negative direction), then (1) yields If *v*(*t*) changes sign on [*a*, *b*] (i.e., the direction of motion changes), then (1) gives the total distance traveled. Suppose, for example, that the situation is as follows:

Then the total distance traveled during the time interval from *t* = *a* to *t* = *b* is exactly

The *displacement* or *net change* in the particle’s position from *t* = *a* to *t* = *b* is equal, by the Fundamental Theorem of Calculus (FTC), to

**EXAMPLE 1**

If a body moves along a straight line with velocity *v* = *t*^{3} + 3*t*^{2}, find the distance traveled between *t* = 1 and *t* = 4.

**SOLUTION:**

Note that *v* > 0 for all *t* on [1, 4].

**EXAMPLE 2**

A particle moves along the *x*-axis so that its velocity at time *t* is given by *v*(*t*) = 6*t*^{2} − 18*t* + 12.

(a) Find the total distance covered between *t* = 0 and *t* = 4.

(b) Find the displacement of the particle from *t* = 0 to *t* = 4.

**SOLUTIONS:**

**(a)** Since *v*(*t*) = 6*t*^{2} − 18*t* + 12 = 6(*t* − 1)(*t* − 2), we see that:

Thus, the total distance covered between *t* = 0 and *t* = 4 is

When we replace *v*(*t*) by 6*t*^{2} − 18*t* + 12 in (2) and evaluate, we obtain 34 units for the total distance covered between *t* = 0 and *t* = 4. This can also be verified on your calculator by evaluating

This example is the same as Example 26, in which the required distance is computed by another method.

**(b)** To find the *displacement* of the particle from *t* = 0 to *t* = 4, we use the FTC, evaluating

This is the net change in position from *t* = 0 to *t* = 4, sometimes referred to as “position shift.” Here it indicates the particle ended up 32 units to the right of its starting point.

**EXAMPLE 3**

The acceleration of an object moving on a line is given at time *t* by *a* = sin *t*; when *t* = 0 the object is at rest. Find the distance *s* it travels from *t* = 0 to

**SOLUTION:** Since it follows that

Also, *v*(0) = 0 yields *C* = 1. Thus *v*(*t*) = 1 − cos *t*; and since cos *t* 1 for all *t* we see that *v*(*t*) 0 for all *t*. Thus, the distance traveled is