{"id":393,"date":"2025-02-13T19:44:34","date_gmt":"2025-02-13T19:44:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-formulas-and-the-net-change-theorem-learn-it-2\/"},"modified":"2025-02-13T19:44:34","modified_gmt":"2025-02-13T19:44:34","slug":"integration-formulas-and-the-net-change-theorem-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-formulas-and-the-net-change-theorem-learn-it-2\/","title":{"raw":"Integration Formulas and the Net Change Theorem: Learn It 2","rendered":"Integration Formulas and the Net Change Theorem: Learn It 2"},"content":{"raw":"\n

The Net Change Theorem<\/h2>\n

The net change theorem<\/strong> considers the integral of a rate of change<\/em><\/span>. 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.<\/p>\n

The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.<\/p>\n

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net change theorem<\/h3>\n

The new value of a changing quantity equals the initial value plus the integral of the rate of change:<\/p>\n

[latex]\\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}[\/latex]<\/div>\n<\/section>\n
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Subtracting [latex]F(a)[\/latex] from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.<\/p>\n<\/section>\n

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.<\/p>\n

To illustrate, let\u2019s apply the net change theorem to a velocity<\/span> function in which the result is displacement<\/span>. We looked at a simple example of this in The Definite Integral section.<\/p>\n

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Suppose a car is moving due north (the positive direction) at [latex]40[\/latex] mph between [latex]2[\/latex] p.m. and [latex]4[\/latex] p.m., then the car moves south at [latex]30[\/latex] mph between [latex]4[\/latex] p.m. and [latex]5[\/latex] p.m. We can graph this motion as shown in the figure below.<\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"286\"]\"A Figure 1. The graph shows speed versus time for the given motion of a car.[\/caption]\n\n\n

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<\/p>\n

[latex]\\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}[\/latex]<\/div>\n

Thus, at [latex]5[\/latex] p.m. the car is [latex]50[\/latex] mi north of its starting position. The total distance traveled is given by<\/p>\n

[latex]\\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}[\/latex]<\/div>\n

Therefore, between [latex]2[\/latex] p.m. and [latex]5[\/latex] p.m., the car traveled a total of [latex]110[\/latex] mi.<\/p>\n<\/section>\n

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.<\/p>\n

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.<\/p>\n

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Given a velocity function [latex]v(t)=3t-5[\/latex] (in meters per second) for a particle in motion from time [latex]t=0[\/latex] to time [latex]t=3,[\/latex] find the net displacement of the particle.<\/p>\n\n\n[reveal-answer q=\"fs-id1170572455638\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1170572455638\"]\n\n\n

Applying the net change theorem, we have<\/p>\n

[latex]\\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}[\/latex]<\/div>\n

 <\/p>\n

The net displacement is [latex]-\\frac{3}{2}[\/latex] m.<\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]\"A Figure 2. The graph shows velocity versus time for a particle moving with a linear velocity function.[\/caption]\n[\/hidden-answer]<\/section>\n

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Use the previous example to find the total distance traveled by a particle according to the velocity function [latex]v(t)=3t-5[\/latex] m\/sec over a time interval [latex]\\left[0,3\\right].[\/latex]<\/p>\n\n\n[reveal-answer q=\"fs-id1170572206288\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1170572206288\"]\n\n\n

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.<\/p>\n

To continue with the example, use two integrals to find the total distance. First, find the [latex]t[\/latex]-intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for [latex]t[\/latex].<\/p>\n

Thus,<\/p>\n

[latex]\\begin{array}{ccc}3t-5\\hfill & =\\hfill & 0\\hfill \\\\ \\hfill 3t& =\\hfill & 5\\hfill \\\\ \\hfill t& =\\hfill & \\frac{5}{3}.\\hfill \\end{array}[\/latex]<\/div>\n

The two subintervals are [latex]\\left[0,\\frac{5}{3}\\right][\/latex] and [latex]\\left[\\frac{5}{3},3\\right].[\/latex]<\/p>\n

To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval [latex]\\left[0,\\frac{5}{3}\\right],[\/latex] we have [latex]|v(t)|=\\text{\u2212}v(t)[\/latex] over that interval. Over [latex]\\left[\\frac{5}{3},3\\right],[\/latex] the function is positive, so [latex]|v(t)|=v(t).[\/latex]<\/p>\n

Thus, we have<\/p>\n

[latex]\\begin{array}{} {\\int }_{0}^{3}|v(t)|dt\\hfill & ={\\int }_{0}^{5\\text{\/}3}\\text{\u2212}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}[\/latex]<\/div>\n

So, the total distance traveled is [latex]\\frac{41}{6}[\/latex] m.<\/p>\n

Watch the following video to see the worked solution to this example.<\/p>\n