Calculus

Integration of Functions

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Net Change Theorem

  • Let \(R\left( t \right)\) represent the rate at which water flows into a tank. The rate can be measured in cubic meters per second or, for example, in gallons per minute.

    Then the definite integral \(\int\limits_a^b {R\left( t \right)dt} \) expresses the total change in volume from time \(a\) to time \(b.\)

    Instead of a water flow, we can consider any other quantity. In general case, if \(f\left( t \right)\) is the rate of change of some quantity, then the integral \(\int\limits_a^b {f\left( t \right)dt} \) represents the net change in that quantity over the time interval \(\left[ {a,b} \right].\)

    This leads us to the Net Change Theorem, which states that if a quantity changes and is represented by a differentiable function, the final value equals the initial value plus the integral of the rate of change of that quantity:

    \[F\left( b \right) = F\left( a \right) + \int\limits_a^b {f\left( t \right)dt} .\]

    The Net Change Theorem can be applied to various problems involving rate of change (such as finding volume, area, population, velocity, distance, cost, etc.)


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    The engine on a boat starts at \(t = 0\) and consumes fuel at the rate of\[c\left( t \right) = 5 – {e^{ – t}}\]litres per hour. How much fuel does it consume for the first 2 hours?

    Example 2

    Suppose a fish population in a lake is increasing with a rate of\[f\left( t \right) = 10 + 5{t^{\frac{3}{2}}}\] thousands of fish per year, where \(t\) is the number of years from now. How much will the fish population increase in \(4\) years?

    Example 3

    A tank has a capacity of \(V = 1000\,\text{litres}.\) Water is pumped into the tank at the rate of\[R\left( t \right) = 60 – t\]litres per minute, where time \(t\) is measured in minutes. How long will it take to fill the tank to its full capacity.

    Example 4

    Suppose that \(t\) months from now the population of a town will be growing at the rate of\[g\left( t \right) = 50 + 10\sqrt t\]people per month. The initial population is 5000. What will be the population in \(3\) years?

    Example 5

    An initially empty swimming pool is being filled by water. The rate at which the height of water changes is given by the equation\[\frac{{dH}}{{dt}} = 2 – \sqrt t ,\]where \(H\) is measured in meters, and \(t\) is measured in hours. Calculate the maximum height of water.

    Example 6

    The growth rate of a vegetable on a field linearly depends on temperature according to the law\[g\left( T \right) = 2\left( {T – 10} \right),\]where \({g\left( T \right)}\) is measured in kg per hour and \(T \ge 15^{\circ}C.\) The hourly temperature varies as follows:\[T\left( t \right) = 20 + 5\sin \frac{{\pi t}}{{12}}.\]Find the total growth of the vegetables (in kg) for the \(12-\text{hour}\) period.

    Example 7

    The relative output power of solar panels decreases with increasing temperature. Suppose that the relative power output is determined by the function\[P\left( T \right) = 100 + \frac{1}{2}\left( {25 – T} \right),\]where \(P\left( T \right)\) is measured in percent and temperature \(T\) is measured in \({^{\circ}C}.\) The temperature varies during a month according to the law\[T\left( t \right) = 15 + 10\sin \frac{{\pi t}}{{30}},\]where \(t\) is measured in days. What is the average power output of the solar panel?

    Example 8

    A radioactive material decays according to the law\[N\left( t \right) = {N_0}{e^{ – \lambda t}},\]where \(N\left( t \right)\) is the amount of the radioactive material at time \(t,\) \({N_0}\) is the initial amount at time \(t = 0,\) \(\lambda\) is the rate of decay. Calculate the cumulative exposure to radiation for all time and for the period of \(T = n{T_{1/2}},\) where \({T_{1/2}}\) is the half life time of the radioactive element and \(n\) is an arbitrary positive number.

    Example 9

    Oil leaks from a tank at a rate of\[r\left( t \right) = {r_0}{e^{ – \lambda t}},\]where \(\lambda = 1\) cubic meters per hour. During the \(1\text{st}\) hour, \({R_1} = 3\,{\text{m}^3}\) of oil has leaked out. How much oil will leak out during the next hour?

    Example 10

    Suppose a hormone production rate during a day is given by the cyclic function\[p\left( t \right) = 1 + \sin \frac{{\pi t}}{{12}}\]for \(0 \le t \le 24.\) The removal rate of the hormone is described by the linear function\[q\left( t \right) = \frac{t}{{12}}.\]Find the highest hormone level over the period.

    Example 1.

    The engine on a boat starts at \(t = 0\) and consumes fuel at the rate of\[c\left( t \right) = 5 – {e^{ – t}}\]litres per hour. How much fuel does it consume for the first 2 hours?

    Solution.

    To estimate the fuel consumption, we use the net change theorem. This yields:

    \[{C = \int\limits_0^2 {c\left( t \right)dt} }={ \int\limits_0^2 {\left( {5 – {e^{ – t}}} \right)dt} }={ \left. {\left( {5t + {e^{ – t}}} \right)} \right|_0^2 }={ 10 + {e^{ – 2}} – 1 }={ 9 + \frac{1}{{{e^2}}} }\approx{ 9.14\,\text{}}\]

    Example 2.

    Suppose a fish population in a lake is increasing with a rate of\[f\left( t \right) = 10 + 5{t^{\frac{3}{2}}}\] thousands of fish per year, where \(t\) is the number of years from now. How much will the fish population increase in \(4\) years?

    Solution.

    The integral of the rate of change from \(t = 0\) to \(t = 4\) is the net change of the fish population:

    \[{F = \int\limits_0^4 {f\left( t \right)dt} }={ \int\limits_0^4 {\left( {10 + 5{t^{\frac{3}{2}}}} \right)dt} }={ \left. {\left( {10t + 2{t^{\frac{5}{2}}}} \right)} \right|_0^4 }={ 40 + 2 \cdot 32 }{= 104.}\]

    Hence, the fish population will increase by \(F = 104\) thousands.

    Example 3.

    A tank has a capacity of \(V = 1000\,\text{litres}.\) Water is pumped into the tank at the rate of\[R\left( t \right) = 60 – t\]litres per minute, where time \(t\) is measured in minutes. How long will it take to fill the tank to its full capacity.

    Solution.

    Let \(T\) be the time required to fill the tank. Using integration, we have

    \[{V = \int\limits_0^T {R\left( t \right)dt} }={ \int\limits_0^T {\left( {60 – t} \right)dt} }={ \left. {\left( {60t – \frac{{{t^2}}}{2}} \right)} \right|_0^T }={ 60T – \frac{{{T^2}}}{2}.}\]

    To find the time \(T,\) we must solve the quadratic equation

    \[{\frac{{{T^2}}}{2} – 60T + V = 0,\;\text{ or }\;}\kern0pt{{T^2} – 120T + 2000 = 0.}\]

    The roots of the equation are \({T_{1,2}} = 100, 20\,\text{min}.\) The first root \({T_1} = 100\,\text{min}\) does not make sense as the rate of the water flow becomes negative for \(T \gt 60\,\text{min}.\)

    Thus, the answer is \(T = 20\,\text{min}.\)

    Example 4.

    Suppose that \(t\) months from now the population of a town will be growing at the rate of\[g\left( t \right) = 50 + 10\sqrt t\]people per month. The initial population is 5000. What will be the population in \(3\) years?

    Solution.

    By the net change theorem,

    \[G = {G_0} + \int\limits_0^T {g\left( t \right)dt} ,\]

    where \({G_0}\) and \(G\) are the initial and final populations, respectively.

    By integrating from \(t = 0\) to \(T = 36\) months, we get the total increase in the number of people:

    \[{\int\limits_0^T {g\left( t \right)dt} }={ \int\limits_0^{36} {\left( {50 + 10\sqrt t } \right)dt} }={ \left. {\left( {50t + \frac{{20{t^{\frac{3}{2}}}}}{3}} \right)} \right|_0^{36} }={ 1800 + 1440 }={ 3240.}\]

    Thus, the final population in \(3\) years is expected to be

    \[G = 5000 + 3240 = 8240.\]

    Example 5.

    An initially empty swimming pool is being filled by water. The rate at which the height of water changes is given by the equation\[\frac{{dH}}{{dt}} = 2 – \sqrt t ,\]where \(H\) is measured in meters, and \(t\) is measured in hours. Calculate the maximum height of water.

    Solution.

    Determine the time \(T\) it takes to fill the pool.

    \[{\frac{{dH}}{{dt}} = 0,}\;\; \Rightarrow {2 – \sqrt T = 0,}\;\; \Rightarrow {T = 4\,\text{hours}}.\]

    Using the net change theorem we find the maximum height:

    \[{{H_{\max }} = \int\limits_0^T {\frac{{dH}}{{dt}}dt} }={ \int\limits_0^4 {\left( {2 – \sqrt t } \right)dt} }={ \left. {\left( {2t – \frac{{2{t^{\frac{3}{2}}}}}{3}} \right)} \right|_0^4 }={ 8 – \frac{2}{3} \cdot 8 }\approx{ 2.67\,\text{meters}}\]

    Example 6.

    The growth rate of a vegetable on a field linearly depends on temperature according to the law\[g\left( T \right) = 2\left( {T – 10} \right),\]where \({g\left( T \right)}\) is measured in kg per hour and \(T \ge 15^{\circ}C.\) The hourly temperature varies as follows:\[T\left( t \right) = 20 + 5\sin \frac{{\pi t}}{{12}}.\]Find the total growth of the vegetables (in kg) for the \(12-\text{hour}\) period.

    Solution.

    Calculating total growth of vegetables depending on hourly temperature
    Figure 1.

    By substituting \(T\left( t \right)\) into the equation for \(g\left( T \right)\) we obtain the growth rate as an explicit function of time \(t:\)

    \[{g\left( t \right) = 2\left( {T\left( t \right) – 10} \right) }={ 2\left( {20 + 5\sin \frac{{\pi t}}{{12}} – 10} \right) }={ 20 + 10\sin \frac{{\pi t}}{{12}}.}\]

    Integrating over the \(12-\text{hour}\) period yields:

    \[{G = \int\limits_0^{12} {g\left( t \right)dt} }={ \int\limits_0^{12} {\left( {20 + 10\sin \frac{{\pi t}}{{12}}} \right)dt} }={ \left. {\left( {20t – \frac{{120}}{\pi }\cos \frac{{\pi t}}{{12}}} \right)} \right|_0^{12} }={ 480 – \frac{{120}}{\pi }\left( {\cos \pi – \cos 0} \right) }={ 480 + \frac{{240}}{\pi } }\approx{ 556.4\,\text{kg}}\]

    Example 7.

    The relative output power of solar panels decreases with increasing temperature. Suppose that the relative power output is determined by the function\[P\left( T \right) = 100 + \frac{1}{2}\left( {25 – T} \right),\]where \(P\left( T \right)\) is measured in percent and temperature \(T\) is measured in \({^{\circ}C}.\) The temperature varies during a month according to the law\[T\left( t \right) = 15 + 10\sin \frac{{\pi t}}{{30}},\]where \(t\) is measured in days. What is the average power output of the solar panel?

    Solution.

    Calculating the average power of solar panels
    Figure 2.

    First we write the relative power output as a function of time \(t:\)

    \[{P\left( t \right) }={ 100 + \frac{1}{2}\left( {25 – T\left( t \right)} \right) }={ 100 + \frac{1}{2}\left( {25 – 15 – 10\sin \frac{{\pi t}}{{30}}} \right) }={ 105 – 5\sin \frac{{\pi t}}{{30}}.}\]

    Integrating the power over the period of \(1\) month gives us the total energy produced for the period. The average power value is equal to the energy divided by the duration of the time interval. Hence,

    \[{\bar P = \frac{1}{{30}}\int\limits_0^{30} {P\left( t \right)dt} }={ \frac{1}{{30}}\int\limits_0^{30} {\left( {105 – 5\sin \frac{{\pi t}}{{30}}} \right)dt} }={ \frac{1}{{30}}\left. {\left( {105t + \frac{{150}}{\pi }\cos \frac{{\pi t}}{{30}}} \right)} \right|_0^{30} }={ 105 + \frac{5}{\pi }\left( {\cos \pi – \cos 0} \right) }={ 105 – \frac{{10}}{\pi } }\approx{ 101.8\% }\]

    Example 8.

    A radioactive material decays according to the law\[N\left( t \right) = {N_0}{e^{ – \lambda t}},\]where \(N\left( t \right)\) is the amount of the radioactive material at time \(t,\) \({N_0}\) is the initial amount at time \(t = 0,\) \(\lambda\) is the rate of decay. Calculate the cumulative exposure to radiation for all time and for the period of \(T = n{T_{1/2}},\) where \({T_{1/2}}\) is the half life time of the radioactive element and \(n\) is an arbitrary positive number.

    Solution.

    We assume that the absorbed dose of radiation \(R\left( t \right)\) is proportional to the amount of radioactive material:

    \[{R\left( t \right) \sim N\left( t \right),}\;\; \Rightarrow {R\left( t \right) = {R_0}{e^{ – \lambda t}},}\]

    where \({R_0}\) is the initial rate of radiation absorption.

    The total amount of absorbed dose is given by the improper integral:

    \[{{R_\infty } = \int\limits_0^\infty {R\left( t \right)dt} }={ {R_0}\int\limits_0^\infty {{e^{ – \lambda t}}dt} }={ {R_0}\lim \limits_{b \to \infty } \int\limits_0^b {{e^{ – \lambda t}}dt} }={ {R_0}\lim \limits_{b \to \infty } \left. {\left( { – \frac{1}{\lambda }{e^{ – \lambda t}}} \right)} \right|_0^b }={ \frac{{{R_0}}}{\lambda }\lim \limits_{b \to \infty } \left( {1 – {e^{ – \lambda b}}} \right) }={ \frac{{{R_0}}}{\lambda }.}\]

    Let’s now determine the amount of absorbed dose of radiation for the period of \({nT_{1/2}}.\) The half life time \({T_{1/2}}\) is defined from the condition:

    \[{N\left( {{T_{1/2}}} \right) = \frac{{{N_0}}}{2} }={ {N_0}{e^{ – \lambda {T_{1/2}}}}.}\]

    It follows from here that

    \[{T_{1/2}} = \frac{{\ln 2}}{\lambda }.\]

    Integrating from \(t = 0\) to \(t = nT_{1/2}\) yields

    \[{{R_n} = \int\limits_0^{n{T_{1/2}}} {R\left( t \right)dt} }={ {R_0}\int\limits_0^{\frac{{n\ln 2}}{\lambda }} {{e^{ – \lambda t}}dt} }={ {R_0}\left. {\left( { – \frac{1}{\lambda }{e^{ – \lambda t}}} \right)} \right|_0^{\frac{{n\ln 2}}{\lambda }} }={ \frac{{{R_0}}}{\lambda }\left( {1 – {e^{ – n\ln 2}}} \right) }={ \frac{{{R_0}}}{\lambda }\left( {1 – \frac{1}{{{e^{\ln {2^n}}}}}} \right) }={ \frac{{{R_0}}}{\lambda }\left( {1 – \frac{1}{{{2^n}}}} \right).}\]

    Thus, at the moment equal to half life time, the cumulative absorbed dose is \(\large{\frac{1}{2}}\normalsize\) of the total amount. At time \(2{T_{1/2}},\) the absorbed dose is already \(\large{\frac{3}{4}}\normalsize\) of the total amount, etc.

    Amount of absorbed radiation dose depending on time
    Figure 3.

    This pattern is typical for any processes described by the exponential decay function.

    Example 9.

    Oil leaks from a tank at a rate of\[r\left( t \right) = {r_0}{e^{ – \lambda t}},\]where \(\lambda = 1\) cubic meters per hour. During the \(1\text{st}\) hour, \({R_1} = 3\,{\text{m}^3}\) of oil has leaked out. How much oil will leak out during the next hour?

    Solution.

    Using the net change theorem, we can write \({R_1}\) in the form

    \[{{R_1} = \int\limits_0^1 {r\left( t \right)dt} }={ \int\limits_0^1 {{r_0}{e^{ – \lambda t}}dt} }={ \left. { – \frac{{{r_0}}}{\lambda }{e^{ – \lambda t}}} \right|_0^1 }={ \frac{{{r_0}}}{\lambda }\left( {1 – {e^{ – \lambda }}} \right).}\]

    Similarly, we can express the volume of leaked oil for the \(2\text{nd}\) hour \({R_2}:\)

    \[{{R_2} = \int\limits_1^2 {r\left( t \right)dt} }={ \int\limits_1^2 {{r_0}{e^{ – \lambda t}}dt} }={ \left. { – \frac{{{r_0}}}{\lambda }{e^{ – \lambda t}}} \right|_1^2 }={ \frac{{{r_0}}}{\lambda }\left( {{e^{ – \lambda }} – {e^{ – 2\lambda }}} \right).}\]

    It then follows that

    \[\require{cancel}{\frac{{{R_2}}}{{{R_1}}} }={ \frac{{\cancel{\frac{{{r_0}}}{\lambda }}\left( {{e^{ – \lambda }} – {e^{ – 2\lambda }}} \right)}}{{\cancel{\frac{{{r_0}}}{\lambda }}\left( {1 – {e^{ – \lambda }}} \right)}} }={ \frac{{{e^{ – \lambda }} – {e^{ – 2\lambda }}}}{{1 – {e^{ – \lambda }}}} }={ \frac{{{e^{ – \lambda }}\cancel{\left( {1 – {e^{ – \lambda }}} \right)}}}{\cancel{1 – {e^{ – \lambda }}}} }={ {e^{ – \lambda }}.}\]

    Therefore the value of \({R_2}\) is equal to

    \[{{R_2} = {R_1}{e^{ – \lambda }} }={ 3 \cdot {e^{ – 1}} }={ \frac{3}{e} }\approx{ 1.10\,{\text{m}^3}}\]

    Example 10.

    Suppose a hormone production rate during a day is given by the cyclic function\[p\left( t \right) = 1 + \sin \frac{{\pi t}}{{12}}\]for \(0 \le t \le 24.\) The removal rate of the hormone is described by the linear function\[q\left( t \right) = \frac{t}{{12}}.\]Find the highest hormone level over the period.

    Solution.

    Serotonin molecule
    Figure 4.

    We can make sure that the net change in hormone level over the \(24-\)hour interval is equal to zero:

    \[\require{cancel}{P – Q }={ \int\limits_0^{24} {\left[ {p\left( t \right) – q\left( t \right)} \right]dt} }={ \int\limits_0^{24} {\left[ {1 + \sin \frac{{\pi t}}{{12}} – \frac{t}{{12}}} \right]dt} }={ \left. {\left[ {t – \frac{{12}}{\pi }\cos \frac{{\pi t}}{{12}} – \frac{{{t^2}}}{{24}}} \right]} \right|_0^{24} }={ \cancel{24} – \frac{{12}}{\pi }\left( {\cancel{\cos 2\pi} – \cancel{\cos 0}} \right) – \cancel{24} }={ 0,}\]

    where \(P\) and \(Q\) denote the total amount of the hormone, respectively, produced and removed for the period.

    Now let’s derive the expression for the hormone level \(H\) at an arbitrary time \(T,\) where \({0 \le T \le 24}.\) Using the net change theorem, we have

    \[{H\left( T \right)} = {P(T) – Q(T) }={ \int\limits_0^T {\left[ {p\left( t \right) – q\left( t \right)} \right]dt} }={ \int\limits_0^T {\left[ {1 + \sin \frac{{\pi t}}{{12}} – \frac{t}{{12}}} \right]dt} }={ \left. {\left[ {t – \frac{{12}}{\pi }\cos \frac{{\pi t}}{{12}} – \frac{{{t^2}}}{{24}}} \right]} \right|_0^T }={ T – \frac{{12}}{\pi }\left( {\cos \frac{{\pi T}}{{12}} – \cos 0} \right) – \frac{{{T^2}}}{{24}} }={ T – \frac{{{T^2}}}{{24}} – \frac{{12}}{\pi }\cos \frac{{\pi T}}{{12}} + \frac{{12}}{\pi }.}\]

    The hormone level \(H\) is maximal when

    \[{ p\left( t \right) – q\left( t \right) }={ 1 + \sin \frac{{\pi t}}{{12}} – \frac{t}{{12}} }={ 0.}\]

    Although this equation is nonlinear, we can notice that \(t = 12\) is its solution. The total rate \(p\left( t \right) – q\left( t \right)\) changes sign from positive to negative at this point. Therefore the hormone level \(H\) reaches the maximum value at \(t = 12.\)

    The maximum concentration of the hormone is equal to

    \[{{H_{\max }} = P(12) – Q(12) }={ 12 – \frac{{{{12}^2}}}{{24}} – \frac{{12}}{\pi }\cos \pi + \frac{{12}}{\pi } }={ 6 + \frac{{24}}{\pi } }\approx{ 13.6\,\text{units}}\]