Integration Formulas and the Net Change Theorem: Learn It 2

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Integration Formulas and the Net Change Theorem: Learn It 2

The Net Change Theorem

The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity.

The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.

net change theorem

The new value of a changing quantity equals the initial value plus the integral of the rate of change:

\begin{matrix} F (b) = F (a) + \int_{a}^{b} F' (x) d x or \int_{a}^{b} F' (x) d x = F (b) - F (a) . \end{matrix}

$\begin{array}{}\\ \\ F(b)=F(a)+{\displaystyle\int }_{a}^{b}F\text{'}(x)dx\hfill \\ \hfill \text{or}\hfill \\ {\displaystyle\int }_{a}^{b}F\text{'}(x)dx=F(b)-F(a).\hfill \end{array}$

Subtracting $F(a)$ from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.

The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral.

To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement. We looked at a simple example of this in The Definite Integral section.

Suppose a car is moving due north (the positive direction) at $40$ mph between $2$ p.m. and $4$ p.m., then the car moves south at $30$ mph between $4$ p.m. and $5$ p.m. We can graph this motion as shown in the figure below.

A graph with the x axis marked as t and the y axis marked normally. The lines y=40 and y=-30 are drawn over [2,4] and [4,5], respectively.The areas between the lines and the x axis are shaded. — Figure 1. The graph shows speed versus time for the given motion of a car.

Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by

\begin{matrix} \int_{2}^{5} v (t) d t & = \int_{2}^{4} 40 d t + \int_{4}^{5} - 30 d t = 80 - 30 = 50. \end{matrix}

$\begin{array}{cc}{\displaystyle\int }_{2}^{5}v(t)dt\hfill & ={\int }_{2}^{4}40dt+{\displaystyle\int }_{4}^{5}-30dt\hfill \\ & =80-30\hfill \\ & =50.\hfill \end{array}$

Thus, at $5$ p.m. the car is $50$ mi north of its starting position. The total distance traveled is given by

\begin{matrix} \int_{2}^{5} | v (t) | d t & = \int_{2}^{4} 40 d t + \int_{4}^{5} 30 d t = 80 + 30 = 110. \end{matrix}

$\begin{array}{} {\displaystyle\int }_{2}^{5}|v(t)|dt\hfill & ={\int }_{2}^{4}40dt+{\displaystyle\int }_{4}^{5}30dt\hfill \\ & =80+30\hfill \\ & =110.\hfill \end{array}$

Therefore, between $2$ p.m. and $5$ p.m., the car traveled a total of $110$ mi.

To summarize, net displacement can include both positive and negative values, accounting for both forward and backward distances. To find the net displacement, integrate the velocity function over the given interval.

Total distance traveled, however, is always positive. To find the total distance traveled by an object, regardless of direction, integrate the absolute value of the velocity function.

Given a velocity function $v(t)=3t-5$ (in meters per second) for a particle in motion from time $t=0$ to time $t=3,$ find the net displacement of the particle.

Applying the net change theorem, we have

\begin{matrix} \int_{0}^{3} (3 t - 5) d t & = \frac{3 t^{2}}{2} - 5 t |_{0}^{3} = [\frac{3 {(3)}^{2}}{2} - 5 (3)] - 0 = \frac{27}{2} - 15 = \frac{27}{2} - \frac{30}{2} = - \frac{3}{2} . \end{matrix}

$\begin{array}{ll}{\int }_{0}^{3}(3t-5)dt\hfill & =\frac{3{t}^{2}}{2}-5t{|}_{0}^{3}\hfill \\ \\ & =\left[\frac{3{(3)}^{2}}{2}-5(3)\right]-0\hfill \\ & =\frac{27}{2}-15\hfill \\ & =\frac{27}{2}-\frac{30}{2}\hfill \\ & =-\frac{3}{2}.\hfill \end{array}$

The net displacement is $-\frac{3}{2}$ m.

A graph of the line v(t) = 3t – 5, which goes through points (0, -5) and (5/3, 0). The area over the line and under the x axis in the interval [0, 5/3] is shaded. The area under the line and above the x axis in the interval [5/3, 3] is shaded. — Figure 2. The graph shows velocity versus time for a particle moving with a linear velocity function.

Use the previous example to find the total distance traveled by a particle according to the velocity function $v(t)=3t-5$ m/sec over a time interval $\left[0,3\right].$

The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the absolute value of the velocity function to find the total distance traveled.

To continue with the example, use two integrals to find the total distance. First, find the $t$ -intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for $t$ .

Thus,

\begin{matrix} 3 t - 5 & = & 0 3 t & = & 5 t & = & \frac{5}{3} . \end{matrix}

$\begin{array}{ccc}3t-5\hfill & =\hfill & 0\hfill \\ \hfill 3t& =\hfill & 5\hfill \\ \hfill t& =\hfill & \frac{5}{3}.\hfill \end{array}$

The two subintervals are $\left[0,\frac{5}{3}\right]$ and $\left[\frac{5}{3},3\right].$

To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval $\left[0,\frac{5}{3}\right],$ we have $|v(t)|=\text{−}v(t)$ over that interval. Over $\left[\frac{5}{3},3\right],$ the function is positive, so $|v(t)|=v(t).$

Thus, we have

\begin{matrix} \int_{0}^{3} | v (t) | d t & = \int_{0}^{5 / 3} - v (t) d t + \int_{5 / 3}^{3} v (t) d t = \int_{0}^{5 / 3} 5 - 3 t d t + \int_{5 / 3}^{3} 3 t - 5 d t = {(5 t - \frac{3 t^{2}}{2}) |}_{0}^{5 / 3} + {(\frac{3 t^{2}}{2} - 5 t) |}_{5 / 3}^{3} = [5 (\frac{5}{3}) - \frac{3 {(5 / 3)}^{2}}{2}] - 0 + [\frac{27}{2} - 15] - [\frac{3 {(5 / 3)}^{2}}{2} - \frac{25}{3}] = \frac{25}{3} - \frac{25}{6} + \frac{27}{2} - 15 - \frac{25}{6} + \frac{25}{3} = \frac{41}{6} . \end{matrix}

$\begin{array}{} {\int }_{0}^{3}|v(t)|dt\hfill & ={\int }_{0}^{5\text{/}3}\text{−}v(t)dt+{\int }_{5\text{/}3}^{3}v(t)dt\hfill \\ \\ & ={\int }_{0}^{5\text{/}3}5-3tdt+{\int }_{5\text{/}3}^{3}3t-5dt\hfill \\ & ={(5t-\frac{3{t}^{2}}{2})|}_{0}^{5\text{/}3}+{(\frac{3{t}^{2}}{2}-5t)|}_{5\text{/}3}^{3}\hfill \\ & =\left[5(\frac{5}{3})-\frac{3{(5\text{/}3)}^{2}}{2}\right]-0+\left[\frac{27}{2}-15\right]-\left[\frac{3{(5\text{/}3)}^{2}}{2}-\frac{25}{3}\right]\hfill \\ & =\frac{25}{3}-\frac{25}{6}+\frac{27}{2}-15-\frac{25}{6}+\frac{25}{3}\hfill \\ & =\frac{41}{6}.\hfill \end{array}$

So, the total distance traveled is $\frac{41}{6}$ m.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.4 Integration Formulas and the Net Change Theorem” here (opens in new window).

Applying the Net Change Theorem

The net change theorem can be applied to the flow and consumption of fluids, as shown in the example below.

If the motor on a motorboat is started at $t=0$ and the boat consumes gasoline at a rate which can be modeled for the first two hours as $5-\frac{t^{3}}{100}$ gal/hr for the first hour, how much gasoline is used in the first hour?

Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval $\left[0,1\right].$ We have

\begin{matrix} \int_{0}^{1} (5 - \frac{t^{3}}{100}) d t & = (5 t - \frac{t^{4}}{400}) |_{0}^{1} = [5 (1) - \frac{{(1)}^{4}}{400}] - 0 = 5 - \frac{1}{400} = 4.9975. \end{matrix}

$\begin{array}{cc}{\displaystyle\int _{0}^{1}}\left(5-\dfrac{{t}^{3}}{100}\right)dt\hfill & =\left(5t-\dfrac{{t}^{4}}{400}\right){\displaystyle |_{0}^{1}}\hfill \\ \\ \\ & =\left[5(1)-\dfrac{{(1)}^{4}}{400}\right]-0\hfill \\ & =5-\frac{1}{400}\hfill \\ & =4.9975.\hfill \end{array}$

Thus, the motorboat uses $4.9975$ gal of gas in $1$ hour.

Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function $v(t)=-10t+15.$ In other words, the wind speed is given by

$v(t)=\bigg\{\begin{array}{lll}20t+5\hfill & \text{ for }\hfill & 0\le t\le \frac{1}{2}\hfill \\ -10t+15\hfill & \text{ for }\hfill & \frac{1}{2}\le t\le 1.\hfill \end{array}$

Under these conditions, how far from his starting point is Andrew after $1$ hour?

$17.5$ mi