5.4 Integration formulas and the net change theorem (Page 2/8)

Calculus volume 1 Page 2 / 8

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. — 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} 4 0 d t + \int_{4}^{5} −30 d t \\ = 80 - 30 \\ = 50 . \end{matrix}

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} 4 0 d t + \int_{4}^{5} 30 d t \\ = 80 + 30 \\ = 110 . \end{matrix}

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

To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.

Finding net displacement

Given a velocity function $v (t) = 3 t - 5$ (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{array}{l} \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{array}

The net displacement is $- \frac{3}{2}$ m ( [link] ).

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. — The graph shows velocity versus time for a particle moving with a linear velocity function.

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Finding the total distance traveled

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

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}

The two subintervals are $[0, \frac{5}{3}]$ and $[\frac{5}{3}, 3] .$ To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval $[0, \frac{5}{3}],$ we have $| v (t) | = − v (t)$ over that interval. Over $[\frac{5}{3}, 3],$ 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}

So, the total distance traveled is $\frac{14}{6}$ m.

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Find the net displacement and total distance traveled in meters given the velocity function $f (t) = \frac{1}{2} e^{t} - 2$ over the interval $[0, 2] .$

Net displacement: $\frac{e^{2} - 9}{2} \approx - 0.8055 m;$ total distance traveled: $4 ln 4 - 7.5 + \frac{e^{2}}{2} \approx 1.740$ m

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Applying the net change theorem

The net change theorem can be applied to the flow and consumption of fluids, as shown in [link] .

How many gallons of gasoline are consumed?

If the motor on a motorboat is started at $t = 0$ and the boat consumes gasoline at the rate of $5 - t^{3}$ gal/hr, how much gasoline is used in the first 2 hours?

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 $[0, 2] .$ We have

\begin{matrix} \int_{0}^{2} (5 - t^{3}) d t & = (5 t - \frac{t^{4}}{4}) |_{0}^{2} \\ = [5 (2) - \frac{{(2)}^{4}}{4}] - 0 \\ = 10 - \frac{16}{4} \\ = 6 . \end{matrix}

Thus, the motorboat uses 6 gal of gas in 2 hours.

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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