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# Definite Integrals for Future Value

In this notebook, we concern ourselves with the determination of the future value of a quantity of interest (such as the amount of water in a reservoir, the amount of pollution in a lake, the number of indivuals in a population, etc.).

In Sections 1-3, we return to a very familiar concept, namely the calculation of <b>net or cumulative change</b> of a quantity by integrating the known rate of change of that quantity with respect to time.  This readily leads to the determination of the future value.

In Sections 4-6, we introduce a new concept, namely the interaction of a renewal function (representing the rate at which the quantity increases) and a survival function (representing the proportion of a quantity that survives for a certain period of time).  We use the "slicing and integrating method" to develop a formula for future value. 


### 1. An Old Idea: Interpreting a Definite Integral in Terms of Net Change or Cumulative Change

From the Fundamental Theorem of Calculus, we know that 
$$\int_a^b f(x) \ dx = F(b) - F(a),$$
where $F$ is any antiderivative of $f(x)$.

A common first application of the Fundamental Theorem of Calculus is in the determination of displacement of an object with given velocity function.  Suppose the position of an object as a function of time is given by $s(t)$.  The velocity of the object as a function of time is $v(t) = s'(t)$, that is, $v$ is an antiderivative of $s$.  Then 
$$\int_a^b v(t) \ dt = s(b) - s(a).$$
We interpret the latter as the displacement of the object from $t=a$ to $t=b$.

If we let $a=0$ and $b=T$, then it immediately follows that 
$$s(T) = s(0) + \int_0^T v(t) \ dt.$$

In words, we can interpret the latter as "the position at future time $T$ is the current position (at time 0) plus the net or cumulative change in position".

### 2. Generalization: Future Value from Net Change or Cumulative Change

We can generalize the above ideas to other quantities, such as the amount of water in a reservoir, the amount of pollution in a lake, the number of indivuals in a population, etc.

Suppose a quantity $Q$ changes over time at a known rate $Q'(t)$. Then the net (cumulative) change in Q between $t=a$ and $t=b$ is $$Q(b) - Q(a) = \int_a^b Q'(t) \ dt,$$
and the <b>future value</b> of $Q$ at time $T$ is $$Q(T) = Q(0) + \int_0^T Q'(t) \ dt.$$

### 3. Example of Future Value from Net Change or Cumulative Change

<font size="1">[Credit: W. Briggs and L. Cochran, Calculus: Early Transcendentals, 2011, Pearson Educationc Inc. ]</font>

A culture of cells in a lab has a population of 100 cells when nutrients are added at time $t=0$.  Suppose the population $N(t)$ increases at a rate given by $\displaystyle N'(t) = 90 e^{-0.1t}$ cells/hr.  

Then the population at future time $T$ is 
\begin{align}
N(T) 
& = N(0) + \int_0^T N'(t) \ dt \\
& = 100 + \int_0^T 90 e^{-0.1t} \ dt \\
& = 100 + \left. \left( \frac{90}{-0.1} e^{-0.1t} \right) \right|_0^T \\
& = 1000 - 900 e^{-0.1T}.
\end{align}

Note that this population increases, but at a decreasing rate.  As $t \to \infty$, the population size approaches 1000 cells.

### 4. A New Idea: Future Value from Survival and Renewal

For some quantities of interest, the effective rate of change is not known, but there is information about other processes (which we will refer to as survival and renewal) affecting change that allow us to determine the future value of the quantity.

Renewal refers to the rate at which the quantity Q increases.  Let $R(t)$ be the renewal function. 
- If $Q$ represents the size of a population, then $R(t)$ represents the rate at which new members are added to the population at time $t$.  
- If $Q$ represents the amount of a drug in the body, then $R(t)$ represents the amount of drug administered at time $t$.  

Survival refers to the proportion of the quantity $Q$ that survives or remains.  Let $S(t)$ be the survival function.
- If $Q$ represents the size of a population, then $S(t)$ represents the proportion of the population that survives at least $t$ time units from now.
- If $Q$ represents the amount of a drug in the body, then $S(t)$ represents the proportion of the drug that remains in the body at least $t$ time units from now.  
>For example, if $S(10) = 0.7$, then 70% of the current quantity $Q$ survives/remains after $10$ time units.

As before in Section 2, we are interested in determining $Q(T)$, the value of $Q$ at future time $T$.  We need to do this from our knowledge of $Q(0)$ and the renewal and survival functions, $R(t)$ and $S(t)$, respectively.

> <font color=red><b>It is key to realize that the renewal and survival processes referred to here are not independent; they interact and work simultaneously.</b>  In general, more of a small quantity added late in time will survive/remain at future time $T$ than a small quantity added earlier in time.</font>

### 5. Slicing Time and Integrating to determine Future Value from Survival and Renewal

Let's determine first what happens to the initial quantity $Q(0)$.  
- The <i>proportion</i> of $Q(0)$ that survives/remains at time $T$ is $S(T)$.  
- The <i>amount</i> of $Q(0)$ that survives/remains at time $T$ is the product, that is, 
$$S(T) \cdot Q(0).$$

Now focus on what happens during the time interval $[0,T]$.  We slice time, focus on a small infinitessimal time period $dt$ at generic time $t$, as illustrated below, and integrate.



- The small quantity added during this infinitessimal time period $dt$ is "rate of renewal times time", or $R(t) \ dt.$
- The length of the time interval $[t,T]$ is $T-t$.  Thus, the <i>proportion</i> of the small quantity added that survives/remains at time $T$ is $S(T-t).$
- The <i>amount</i> of the small quantity that is added at time $t$ AND that survives/remains at time $T$ is the product of the above, that is, $$dQ = S(T-t) \cdot R(t) \ dt.$$
- The total quantity that is added during the time period from $t=0$ to $t=T$ AND that survives/remains at time $T$ is $$\int dQ = \int_0^T S(T-t) \cdot R(t) \ dt.$$

Putting everything together, we have that $Q(T)$ is the initial amount that survives/remains at time $T$ PLUS the amount that has been added during the time interval $[0,T]$ and survives/remains at time $T$, namely
\begin{align}
Q(T) 
& = S(T) \cdot Q(0) + \int dQ \\ 
& = S(T) \cdot Q(0) + \int_0^T S(T-t) \cdot R(t) \ dt.
\end{align}

### 6. Example of Future Value from Survival and Renewal

<font size="1">[Credit: J. Stewart and T. Day, Biocalculus: Calculus for the Life Sciences, 2015, Cengage Learning.]</font>

Consider a lake with trout.  There currently are 5600 trout in the lake.  The trout are reproducing at the rate $\displaystyle R(t) = 720 e^{0.1t}$ trout/year.  However, polution is killing many of the trout; the proportion that survive after $t$ years is given by $\displaystyle S(t) = e^{-0.2t}$.  How many trout will be in the lake 10 years from now?

Let $P(t)$ be the number of trout in the lake at time $t$, measured in years.  

We are given $P(0) = 5600$ trout, and $T = 10$ years.  Then 
\begin{align}
P(10) 
& = S(10) \cdot P(0) + \int_0^{10} S(10-t) \cdot R(t) \ dt \\ 
& = e^{-0.2(10)} \cdot 5600 + \int_0^{10} e^{-0.2(10-t)} \cdot 720 e^{0.1t} \ dt \\
& = 5600 e^{-2} + 720 \int_0^{10} e^{-2+0.2t+0.1t} \ dt \\
& = 5600 e^{-2} + 720 e^{-2} \int_0^{10} e^{0.3t} \ dt \\ 
& = 5600 e^{-2} + 720 e^{-2} \left. \left( \frac{ e^{0.3t} }{0.3} \right) \right|_0^{10} \\
& = 5600 e^{-2} + \frac{ 720 e^{-2} }{0.3} \left( e^{3} - 1 \right) \\ 
& = 5600 e^{-2} + 2400 \left( e - e^{-2} \right) \\ 
& \approx 6956.95.
\end{align}
Thus, we expect the number of trout in the lake 10 years from now to be about 6957.

### 7. Summary

<b>Future Value from Net Change or Cumulative Change:</b>

Suppose a quantity $Q$ changes over time at a known rate $Q'(t)$. 
- The net (cumulative) change in Q between $t=a$ and $t=b$ is $$Q(b) - Q(a) = \int_a^b Q'(t) \ dt.$$
- The <b>future value</b> of $Q$ at time $T$ is $$Q(T) = Q(0) + \int_0^T Q'(t) \ dt.$$

<b>Future Value from Survival and Renewal:</b>

Suppose a quantity $Q(t)$ has initial value $Q(0)$, renewal function $R(t)$ that represents the rate at which the quantity increases, and survival function $S(t)$ that represents the proportion of the quantity that survives/remains after $t$ years.
- The <b>future value</b> of $Q$ at time $T$ is  
$$Q(T) = S(T) \cdot Q(0) + \int_0^T S(T-t) \cdot R(t) \ dt.$$

### 8. Further Study

Please refer to Section 6.3 in the textbook for additional treatment of this topic.

### 9. Don't Forget

Don't forget to return to eClass to complete the pre-class quiz.

<font color=blue>In the pre-class quiz feedback question, please let me know if this notebook would benefit from one or two videos.  If so, which section would you like to see become a video?  The derivation of Future Value in Section 5, or the example in Section 6, or both?  Thank you!</font>


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