Calculus 220, section 6.5 Applications of the Definite IntegCalculus 220, section 6.5 Applications of the Definite Integral notes by Tim Pilachowski So far in this class, we have spent chapters 1 through 5 beginning with a function that represents an amount. The derivative (= slope of the curve) gives us a rate of change. In chapter 6, we are beginning with a rate of change and use the integral (= area under the curve) to determine an amount. In applications, it is first important to determine the question—to identify what sort of answer you are looking for: Is it an amount or is it a rate of change? Next identify the nature of the information you have: Is it an amount or is it a rate of change? Example A: For a retail company, the monthly marginal cost when x units have been purchased is 50 – 0.08x dollars. Their monthly fixed costs are $700. a) Determine the cost for producing x units. b) What is the cost of raising monthly purchases from 10 to 15 units? For part a) we want to know a cost, the amount of money needed to purchase items. The cost to purchase x units is the Cost function C(x). The information give is a rate of change—the marginal cost is , the derivative of the Cost function. “Monthly fixed costs are $700” is an amount when x = 0, i.e. C(0) = 700. We can find C(x) from using the integral. Answer: ()xC′()xC′()7005004.02++−= xxxC For part b) we want to know a cost, an amount. The information involves a rate of change: raising the number of units from 10 to 15. The numbers 10 and 15 set definite boundaries for us. We need a definite integral. Answer: $245 Example B: An object falls from a high-altitude balloon with a speed of 32t feet per second. How far does the object fall during the first 4 seconds? “How far” asks for a distance, which is an amount. The velocity is a rate of change of distance with respect to time: v(t) in in time changedistancein change is secondfeet. “The first four seconds” gives us boundaries of t = 0 and t = 4. Answer: downward 256 feetExample C: The rate at which gasoline was consumed in the U.S. was approximately r(t) = 0.075t + 1.7 billion gallons per year from 1964 (t = 0) to 1976. Find the number of gallons of gas consumed from 1964 to 1976. The equation give is a rate of change of consumption. The “number of gallons” asked for is an amount. “From 1964 to 1976” gives us boundaries: t = 0 to t = 12. We need a definite integral. Answer: 25.8 billion gallons All three examples above are essentially anti-derivatives: we had a rate of change and used the information given to determine an amount. There are, however, other uses of integrals. Consider the average of a group of numbers: nxxxxxn++++==K321average . The average value of a function is calculated in a similar manner: ()()()()nxfxfxfxfyn++++==K321function a of average . However, a continuous function consists not of discrete values which can be added together, but an infinite number of values. A little algebraic manipulation will help us find a way to calculate this infinite sum. First we “raise to higher terms” by multiplying numerator and denominator by the same number. We choose the number (b – a). ()()()()ababnxfxfxfxfyn−−∗++++==K321function a of average Rearranging using the commutative and associative properties: () ( ) () ( )[]nabxfxfxfxfabyn−∗++++∗−== K3211function a of average The latter part of the formula should look familiar: it’s the Riemann sum for the area under the curve. Thus, () ( ) () ( )[]()∫−=∆∗++++∗−==→∆banxdxxfabxxfxfxfxfaby1lim1function a of average3210K Example D (Example F from Lecture 6.3 revisited): Blood flows through an artery fastest at the center and slowest next to the artery wall. The rate of blood flow in an artery with radius 0.2 cm is described by the function where v is in centimeters per second and x = distance from the center in centimeters. What is the average velocity of the blood flow through the artery? ()299040 xxv −=Another application using integrals comes from the business world. The text does a consumer surplus problem as Example 4. The explanation below is designed to add to your understanding of why the process works the way it does. Let p = price and x = the number of units sold. A demand curve illustrates the reality that as price of goods increases, fewer people will buy, and the demand gdown. The function p = f(x) will have a negative slope. A predictable number of items x = A will be sold if the price is set at a particular level B = f(A). The area of the rectangle in the picture has area oes BA∗ = number sold times price = revenue generated by sales. Note that the top edge of the rectangle is the horizontal line function p = f(A) = B. The area under the curve above the rectangle represents the money saved by the people who were willing to pay a higher price: the consumer’s surplus. We already know that area under a curve can be found by integrating. Thus consumer’s surplus = area under demand curve – area of rectangle = [area under p = f(x)] – [area under p = f(A) = B] = ()[]∫−AdxBxf0 Example E: Let p = 15 – 0.5x represent a demand function where p is price in dollars and x is the number of units sold. Let the level of demand = 7. a) What was the price? b) How much revenue was generated? c) What was the consumer’s surplus? Answers: $11.50; $80.50; $12.25 You’ll need Example 5 and the formula at the top of the page following it in the text to do some of the assigned homework. It is similar to the continuous compounding done in section 5.2, but instead of one deposit there are a series of deposits (a continuous income stream) that increase the balance along with the interest. If you think of each new deposit as an trPeA = calculation where A = accumulated balance, the sum of balances from each deposit forms a Riemann sum, i.e. an integral calculation. From your knowledge of basic geometry, you are probably already familiar with the formula for volume of a rectangular prism: length times width times height. If we think of that prism as laying on its side, the formula looks more like (volume) = (area of cross-section) times (length). cross-section is rectangular A = width * heightlength cross-section is circular A = 2r∗π length The volume of a cylinder is the same: area of cross section times length.Example F: A solid is formed in the region between the functions f(x) =