2.3.1 A Little More Calculus: Integrals

Taking the derivative of a function is equivalent to finding the slope of a graph of the function. Similarly, evaluating an integral is equivalent to finding the area under a graph of the function. The graphical method is very important for building intuition about motion but is limited in its practical application. Just as derivatives of standard functions can be evaluated and tabulated, so can integrals.

The integral in Equation 2.11 is called a definite integral because there are two definite boundaries to the area we want to find. These boundaries are called the lower $(t_{i})$ and upper $(t_{f})$ limits of integration. For the important function $u (t) = c t^{n}$ , the essential result from calculus is that

\begin{matrix} \int_{t_{i}}^{t_{f}} u d t = \int_{t_{i}}^{t_{f}} c t^{n} d t = \frac{c t^{n + 1}}{n + 1} |_{t_{i}}^{t_{f}} = \frac{c t_{f}^{n + 1}}{n + 1} - \frac{c t_{i}^{n + 1}}{n + 1} & (n \neq - 1) \end{matrix}

(2.13)

The vertical bar in the third step with subscript $t_{i}$ and superscript $t_{f}$ is a shorthand notation from calculus that means—as seen in the last step—the integral evaluated at the upper limit $t_{f}$ minus the integral evaluated at the lower limit $t_{i}$ . You also need to know that for two functions $u$ and $w$ ,

\int_{t_{i}}^{t_{f}} (u + w) d t = \int_{t_{i}}^{t_{f}} u d t + \int_{t_{i}}^{t_{f}} w d t

(2.14)

That is, the integral of a sum is equal to the sum of the integrals.

EXAMPLE 2.7 Using calculus to find the position

Use calculus to solve Example 2.6.

SOLVE Figure 2.17 is a linear graph. Its “y-intercept” is seen to be 10 m/s and its slope is $- 5 (m / s) / s$ . Thus the velocity can be described by the equation

v_{x} = (10 - 5 t) m / s

where $t$ is in s. We can find the position $x$ at time $t$ by using Equation 2.11:

\begin{array}{l} x & = x_{i} + \int_{0}^{t} v_{x} d t = 30 m + \int_{0}^{t} (10 - 5 t) d t \\ = 30 m + \int_{0}^{t} 10 d t - \int_{0}^{t} 5 t d t \end{array}

We used Equation 2.14 for the integral of a sum to get the final expression. The first integral is a function of the form $u = c t^{n}$ with $c = 10$ and $n = 0;$ the second is of the form $u = c t^{n}$ with $c = 5$ and $n = 1$ . Using Equation 2.13, we have

\int_{0}^{t} 10 d t = 10 t |_{0}^{t} = 10 \cdot t - 10 \cdot 0 = 10 t m

and

\int_{0}^{t} 5 t d t = \frac{5}{2} t^{2} |_{0}^{t} = \frac{5}{2} \cdot t^{2} - \frac{5}{2} \cdot 0^{2} = \frac{5}{2} t^{2} m

Combining the pieces gives

x = (30 + 10 t - \frac{5}{2} t^{2}) m

This is a general result for the position at any time $t$ .

The particle’s turning point occurs at $t = 2 s,$ and its position at that time is

x (at t = 2 s) = 30 + (10) (2) - \frac{5}{2} {(2)}^{2} = 40 m

The time at which the particle reaches the origin is found by setting $x = 0 m :$

30 + 10 t - \frac{5}{2} t^{2} = 0

This quadratic equation has two solutions: $t = - 2 s$ or $t = 6 s$ .

When we solve a quadratic equation, we cannot just arbitrarily select the root we want. Instead, we must decide which is the meaningful root. Here the negative root refers to a time before the problem began, so the meaningful one is the positive root, $t = 6 s$ .

REVIEW The results agree with the answers we found previously from a graphical solution.