Definite Integrals as Accumulated Change (College Board AP® Calculus AB): Revision Note

For questions involving real world contexts

• Be sure to give your answers in the context of the questions

• This includes using correct units and interpreting results using the language and details of the question

Remember when interpreting rates of change

• A positive rate of change corresponds to an increase in the quantity

• A negative rate of change corresponds to a decrease

A company produces and sells luxury chess sets. The marginal revenue of the company is given by $R^{\text{'}}\left(x\right)=625$, and its marginal cost is given by $C^{\text{'}}\left(x\right)=300+0.02x$. Both revenue $R$ and cost $C$ are measured in dollars, and in both cases $x$ represents number of chess sets sold.

(a) Find the value of $C^{\text{'}}\left(200\right)$ and interpret this in the context of the problem.

(b) Find the change in the cost, $C$, to the company in going from selling 200 to 300 chess sets.

(c) Using $\mathrm{profit}=\mathrm{revenue}-\mathrm{cost}$, find the change in the profit, $P$, made by the company in going from selling 200 to 300 chess sets.

Answer:

(a)

Substitute $x=200$ into $C^{\text{'}}\left(x\right)$

$C^{\text{'}}\left(200\right)=300+0.02\left(200\right)=304$

Interpret this in the context of the question, paying attention to units

The rate of change is positive, so the cost is increasing

$C^{\text{'}}\left(200\right)=304$

At the point when 200 chess sets have been sold, the company's cost is increasing at the rate of $304 per chess set

(b)

This is the integral of $C^{\text{'}}$ between $x=200$ and $x=300$

$\begin{array}{rcl}\mathrm{change}\mathrm{in}\mathrm{cost}& =& {\int }_{200}^{300}\left(300+0.02x\right)dx\\ & =& {\left[300x+0.01x^2\right]}_{200}^{300}\\ & =& 300\left(300\right)+0.01{\left(300\right)}^2-\left(300\left(200\right)+0.01{\left(200\right)}^2\right)\\ & =& 90000+900-\left(60000+400\right)\\ & =& 30500\end{array}$

This is positive, so represents an increase

In going from selling 200 to 300 chess sets, the company's cost increases by $30,500

(c)

$P\left(x\right)=R\left(x\right)-C\left(x\right)$ means that $P^{\text{'}}\left(x\right)=\frac{d}{dx}\left(R\left(x\right)-C\left(x\right)\right)=R^{\text{'}}\left(x\right)-C^{\text{'}}\left(x\right)$

$P^{\text{'}}\left(x\right)=R^{\text{'}}\left(x\right)-C^{\text{'}}\left(x\right)$

$P^{\text{'}}\left(x\right)=625-\left(300+0.02x\right)=325-0.02x$

The change in the profit is the integral of $P^{\text{'}}$ between $x=200$ and $x=300$

$\begin{array}{rcl}\mathrm{change}\mathrm{in}\mathrm{profit}& =& {\int }_{200}^{300}\left(325-0.02x\right)dx\\ & =& {\left[325x-0.01x^2\right]}_{200}^{300}\\ & =& 325\left(300\right)-0.01{\left(300\right)}^2-\left(325\left(200\right)-0.01{\left(200\right)}^2\right)\\ & =& 97500-900-\left(65000-400\right)\\ & =& 32000\end{array}$

This is positive, so represents an increase

In going from selling 200 to 300 chess sets, the company's profit increases by $32,000

When the food bowl in a pet rabbit habitat is filled up, it contains 80 grams of food. The rabbits consume food from the bowl at a rate modeled by

$f\left(t\right)=4+4\cos \left(\frac{\pi }{4}t\right)\mathrm{for}0<t\le 20$

where $f\left(t\right)$ is measured in grams per hour and $t$ is the number of hours after the bowl was filled.

(a) How many grams of food do the rabbits eat during the first four hours after the bowl is filled?

(b) Find $f^{\text{'}}\left(\frac{26}{3}\right)$. Using correct units, explain the meaning of $f^{\text{'}}\left(\frac{26}{3}\right)$ in the context of the problem.

(c) Assuming that no more food is added to the bowl, find an expression in terms of $t$ for $F\left(t\right)$, the amount of food left in the bowl at time $t$.

Answer:

(a)

This will be the definite integral of $f\left(t\right)$ between $t=0$ and $t=4$

$\begin{array}{rcl}{\int }_0^4\left(4+4\cos \left(\frac{\pi }{4}t\right)\right)dt& =& 4{\int }_0^4\left(1+\cos \left(\frac{\pi }{4}t\right)\right)dt\\ & =& 4{\left[t+\frac{4}{\pi }\sin \left(\frac{\pi }{4}t\right)\right]}_0^4\\ & =& 4\left(4+\frac{4}{\pi }\sin \left(\frac{\pi }{4}\left(4\right)\right)-\left(0+\frac{4}{\pi }\sin \left(\frac{\pi }{4}\left(0\right)\right)\right)\right)\\ & =& 4\left(4+\frac{4}{\pi }\sin \left(\pi \right)-\left(0+\frac{4}{\pi }\sin \left(0\right)\right)\right)\\ & =& 4\left(4+0-\left(0+0\right)\right)\\ & =& 16\end{array}$

16 grams

(b)

First differentiate to find $f^{\text{'}}\left(t\right)$

$f^{\text{'}}\left(t\right)=4·\left(-\sin \left(\frac{\pi }{4}t\right)\right)·\frac{\pi }{4}=-\pi \sin \left(\frac{\pi }{4}t\right)$

Substitute in $t=\frac{26}{3}$

$\begin{array}{rcl}{f}^{\text{'}}\left(\frac{26}{3}\right)& =& -\pi \sin \left(\frac{\pi }{4}\left(\frac{26}{3}\right)\right)\\ & =& -\pi \sin \left(\frac{13\pi }{6}\right)\\ & =& -\pi \left(\frac{1}{2}\right)\\ & =& -\frac{\pi }{2}\end{array}$

$f\left(t\right)$ is a rate of change, so $f^{\text{'}}\left(t\right)$ is the rate of change of a rate of change

It is negative, so the rate of change is decreasing

$f^{\text{'}}\left(\frac{26}{3}\right)=-\frac{\pi }{2}$

This means that $\frac{26}{3}$ hours after the food bowl was filled, the rate at which the rabbits are consuming the food is decreasing by $\frac{\pi }{2}$ grams per hour, per hour

(c)

This will be the integral of $f$ between $0$ and $t$, subtracted from the amount of food in the bowl at time $t=0$

Be careful here: $f\left(t\right)$ is the rate at which the rabbits are eating the food, but this means that the rate at which the amount of food in the bowl is changing is $-f\left(t\right)$

$\begin{array}{rcl}F\left(t\right)& =& F\left(0\right)-{\int }_0^{t}f\left(s\right)ds\\ & =& 80-{\int }_0^t\left(4+4\cos \left(\frac{\pi }{4}s\right)\right)ds\\ & =& 80-4{\int }_0^t\left(1+\cos \left(\frac{\pi }{4}s\right)\right)ds\\ & =& 80-4{\left[s+\frac{4}{\pi }\sin \left(\frac{\pi }{4}s\right)\right]}_0^t\\ & =& 80-4\left(t+\frac{4}{\pi }\sin \left(\frac{\pi }{4}\left(t\right)\right)-\left(0+\frac{4}{\pi }\sin \left(\frac{\pi }{4}\left(0\right)\right)\right)\right)\\ & =& 80-4\left(t+\frac{4}{\pi }\sin \left(\frac{\pi }{4}t\right)-\left(0+\frac{4}{\pi }\sin \left(0\right)\right)\right)\\ & =& 80-4\left(t+\frac{4}{\pi }\sin \left(\frac{\pi }{4}t\right)-\left(0+0\right)\right)\\ & =& 80-4t-\frac{16}{\pi }\sin \left(\frac{\pi }{4}t\right)\end{array}$

This could also be solved by finding the indefinite integral $F\left(t\right)=\int f\left(t\right)dt$, and then using $F\left(0\right)=80$ to work out the value of the constant of integration

$F\left(t\right)=\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 80\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 4\end{array}\begin{array}{rcl}& & t\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{16}{\pi }\end{array}\begin{array}{rcl}& & \sin \end{array}\begin{array}{rcl}& & \left(\frac{\pi }{4}t\right)\end{array}$