Integral as Net Change Chapter 8.1

The Accumulation Function The function determined by FTC1 is sometimes called an accumulation function 𝑔 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡 Suppose that 𝑓 is a function that is a derivative of a function 𝑔; that is, 𝑔 ′ 𝑥 =𝑓(𝑥) As we “accumulate” net area from a to some arbitrary value x, we are observing the net amount of change

The Accumulation Function Consider the well-known distance formula: 𝑑=𝑟𝑡 This formula applies only when the rate is constant; that is, at constant velocity Suppose an object is moving horizontally at velocity 𝑣 that is a non- constant function of time We want to determine a process by which we can model the displacement of the object; we cannot directly use 𝑑=𝑟𝑡

The Accumulation Function To do this, we partition the t-axis in the same way we did for determining area

The Accumulation Function The previous slide shows that, if we choose a small enough value of Δ𝑡, then the particle will be displaced an amount that is approximately equal to 𝑣 𝑐 𝑘 Δ𝑡 (this is 𝑑=𝑟𝑡 in disguise) Now, if we add 𝑣 𝑐 1 Δt+𝑣 𝑐 2 Δt+…+𝑣 𝑐 𝑛 Δt we have a Riemann sum 𝑘=1 𝑛 𝑣 𝑐 𝑘 Δt

The Accumulation Function

The Accumulation Function If we now allow the number of partitions to approach infinity, we have lim 𝑛→∞ 𝑘=1 𝑛 𝑣 𝑐 𝑘 Δ𝑡 = 𝑎 𝑏 𝑣(𝑡) 𝑑𝑡 Note that the difference here is that we came up with a Riemann sum without having to resort to thinking about area under the curve Also, it accounts for why a function of the form 𝑔 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡 is sometimes called the accumulation function

The Accumulation Function In this chapter, you will learn how to model with integrals using the method just demonstrated To model with integrals Approximate what you want to find as a Riemann sum of values of a continuous function multiplied by interval lengths. If 𝑓(𝑥) is a function and [𝑎,𝑏] the interval, and you partition the interval into subintervals of length Δ𝑥, the approximating sums will have the form 𝑎 𝑓 𝑐 𝑘 Δ𝑥 with 𝑐 𝑘 a point in the kth subinterval. Write a definite integral, here 𝑎 𝑏 𝑓(𝑥) 𝑑𝑥, to express the limit of these sums as the norms of the partition go to zero. Evaluate the integral numerically or with an antiderivative.

Linear Motion Revisited Remember the difference between these terms Displacement is the net distance during an interval of time [𝑎.𝑏]: displacement= 𝑎 𝑏 𝑣(𝑡) 𝑑𝑡=𝑠 𝑏 −𝑠(𝑎) Position is a number on a number line (a frame of reference); it is relative to some initial position, 𝑠(0) at 𝑡=0 position=𝑠 𝑡 =𝑠 0 + 0 𝑡 𝑓(𝑥) 𝑑𝑥 Speed is the magnitude of the velocity speed=|𝑣 𝑡 | Total distance is the distance traveled without regard to direction total distance= 𝑎 𝑏 |𝑣 𝑡 | 𝑑𝑡

Example 1: Interpreting a Velocity Function The figure below shows the velocity 𝑑𝑠 𝑑𝑡 =𝑣 𝑡 = 𝑡 2 − 8 𝑡+1 2 of a particle moving along a horizontal s-axis for 0≤𝑡≤5. Describe the motion.

Example 1: Interpreting a Velocity Function

Example 1: Interpreting a Velocity Function The graph of v starts with 𝑣(0)=−8, which we interpret as saying that the particle has an initial velocity of 8 cm/sec to the left. It slows to a halt about 𝑡=1.25 sec, after which it moves to the right (𝑣>0) with increasing speed, reaching a velocity of 𝑣 5 ≈24.8 cm/sec at the end.

Example 2: Finding Position from Displacement Suppose the initial position of the particle in Example 1 is 𝑠 0 =9. (a) What is the particle’s displacement over the interval 0≤𝑡≤5? (b) What is the particle’s position at 𝑡=1 second? (c) What is the particle’s position at 𝑡=5 seconds?

Example 2: Finding Position from Displacement Suppose the initial position of the particle in Example 1 is 𝑠 0 =9. What is the particle’s displacement over the interval 0≤𝑡≤5? The displacement is 0 5 𝑡 2 − 8 𝑡+1 2 𝑑𝑡=35 centimeters How should we interpret this? “The particle is 35 centimeters to the right of its initial position.”

Example 2: Finding Position from Displacement Suppose the initial position of the particle in Example 1 is 𝑠 0 =9. (b) What is the particle’s position at 𝑡=1 second? The displacement over 0≤𝑡≤1 is 0 1 ( 𝑡 2 − 8 𝑡+1 2 𝑑𝑡=− 11 3 centimeters The position is 9+ 0 1 ( 𝑡 2 − 8 𝑡+1 2 𝑑𝑡=9− 11 3 = 16 3 centimeters The particle moved to the left from position 9 so its position at 𝑡=1 is 9− 11 3 = 16 3

Example 2: Finding Position from Displacement Suppose the initial position of the particle in Example 1 is 𝑠 0 =9. (c) What is the particle’s position at 𝑡=5 second? The displacement over 0≤𝑡≤5 is 0 5 𝑡 2 − 8 𝑡+1 2 𝑑𝑡=35 centimeters The position is 9+ 0 1 ( 𝑡 2 − 8 𝑡+1 2 𝑑𝑡=9+35=44 centimeters The particle moved (net movement) to the left from position 9 so its position at 𝑡= 5 is 9+35=44

Example 3: Calculating Total Distance Traveled Find the total distance traveled by the particle in Example 1.

Example 3: Calculating Total Distance Traveled Find the total distance traveled by the particle in Example 1. To evaluate this numerically (i.e., with a calculator): 0 5 𝑡 2 − 8 𝑡+1 2 𝑑𝑡≈42.587 centimeters To evaluate this analytically, separate the integrals at the zeros of the function and use negative for those parts for which the function is negative − 0 1.25 𝑡 2 − 8 𝑡+1 2 𝑑𝑡+ 1.25 5 𝑡 2 − 8 𝑡+1 2 𝑑𝑡

Example 4: Modeling the Effects of Acceleration A car moving with initial velocity of 5 mph accelerates at the rate of 𝑎 𝑡 =2.4𝑡 mph per second for 8 seconds. How fast is the car going when the 8 seconds are up? How far did the car travel during those 8 seconds?

Example 4: Modeling the Effects of Acceleration How fast is the car going when the 8 seconds are up? Note that, if acceleration is constant, then the velocity at any given time t is given by 𝑣=𝑎𝑡. To model this, we imagine that the interval [0,8] is partitioned so that, if the width of a partition is Δ𝑡 is a small enough value, then velocity at each partition k is approximately 𝑎 𝑐 𝑘 Δt (equivalent to 𝑣=𝑎𝑡), where 𝑐 𝑘 is a value in the kth interval. We can add these values using a Riemann sum 𝑘=1 𝑛 𝑎 𝑐 𝑘 Δ𝑡 Note that the units here are mph sec ×sec=mph

Example 4: Modeling the Effects of Acceleration How fast is the car going when the 8 seconds are up? If we now allow the number of partitions 𝑛 to approach infinity lim 𝑛→∞ 𝑘=1 𝑛 𝑎 𝑐 𝑘 Δ𝑡 By definition, this is the same as 0 8 𝑎(𝑡) 𝑑𝑡

Example 4: Modeling the Effects of Acceleration How fast is the car going when the 8 seconds are up? Finally, we evaluate the definite integral 5+ 0 8 𝑎(𝑡) 𝑑𝑡=5+ 0 8 2.4𝑡 𝑑𝑡 =5+ 1.2 𝑡 2 0 8 =1.2 8 2 −1.2 0 2 =5+76.8 mph=81.8 mph

Example 4: Modeling the Effects of Acceleration How far did the car travel during those 8 seconds? We must solve the differential equation 𝑑𝑣 𝑑𝑡 =2.4𝑡 given that 𝑣 0 =5 𝑣 𝑡 =5+ 0 𝑡 2.4𝑥 𝑑𝑥=1.2 𝑡 2 +5 To find the total distance, we solve the differential equation 𝑑𝑠 𝑑𝑡 = 1.2 𝑡 2 +5 total distance= 0 8 1.2𝑡 2 +5 𝑑𝑡= 0 8 1.2 𝑡 2 +5 𝑑𝑡 Since velocity is positive over this interval, the absolute value sign is not needed.

Example 4: Modeling the Effects of Acceleration How far did the car travel during those 8 seconds? total distance= 0 8 1.2 𝑡 2 +5 𝑑𝑡= 0.4 𝑡 3 +5𝑡 0 8 = 0.4 8 3 +5 8 −0.4 0 3 −5 0 =244.8 mph×sec Since the interval units are in seconds, we must convert the units to get a total distance of 0.068 miles traveled during the 8 seconds.

Example 5: Potato Consumption From 1970 to 1980, the rate of potato consumption in a particular country was 𝐶 𝑡 =2.2+ 1.1 𝑡 millions of bushels per year, with t being years since the beginning of 1970. How many bushels were consumed from the beginning of 1972 to the end of 1973?

Example 5: Potato Consumption The simple case, as with velocity and acceleration, is an equation of the form 𝑇=𝐶𝑡 where T is total consumption, t is time in years, and C is consumption rate as a constant. Since 𝐶 𝑡 =2.2+ 1.1 𝑡 varies, we can imagine partitioning the interval [2,4] into small enough values of Δ𝑡 so that the consumption rate over the small interval is almost constant. For the kth subinterval, the amount consumed would be approximately 𝐶 𝑐 𝑘 Δ𝑡. If we sum all of these approximations over the entire interval, then allow the widths of the partitions approach zero, we have lim | 𝑃 |→0 𝑘=1 𝑛 𝐶 𝑐 𝑘 Δ𝑡

Example 5: Potato Consumption Therefore, lim 𝑃 →0 𝑘=1 𝑛 𝐶 𝑐 𝑘 Δ𝑡 = 2 4 2.2+ 1.1 𝑡 𝑑𝑡 Evaluating this definite integral: 2 4 (2.2+ 1.1 𝑡 ) 𝑑𝑡= 2.2𝑡+ 1.1 𝑡 ln 1.1 2 4 ≈7.066 million bu.

Example 6: Finding Gallons Pumped from Rate Data A pump connected to a generator operates at a varying rate, depending on how much power is being drawn from the generator to operate other machinery. The rate (gallons per minute) at which the pump operates is recorded at 5-minute intervals for one hour as shown in the table (next slide). How many gallons were pumped during that hour?

Example 6: Finding Gallons Pumped from Rate Data

Example 6: Finding Gallons Pumped from Rate Data Note that the units for the data are gallons per minute: a rate of change. From this we are to find total number of gallons pumped (approximately). If the rate of pumping were constant (call it r), then gallons, g, would be 𝑔=𝑟𝑡. Since the rate is not constant (and is a function of time), it can be represented as 𝑅(𝑡). If we take the rate as approximately constant over each 5 minute interval, then we can find the approximate value of the total by adding up the approximations using a trapezoidal approximation: 0 60 𝑅(𝑡) 𝑑𝑡≈ 5 58+60 2 + 5 60+65 2 +…+ 5 60+63 2 + 5 63+63 2 ≈ 5 2 58+2 60 +2 65 +…+2 63 +63 =3582.5 gallons

Work In your physics class you learned that work is the product of force times distance 𝑊=𝐹𝑑 As with the distance formula, this applies only if the force is constant If the force varies with time (i.e., 𝐹(𝑡) varies), then we turn first to an approximation by creating a Riemann sum, then writing this as an integral and evaluating For the next example you also need to know Hooke’s Law for springs: the force that it takes to stretch or compress a spring x units from its natural (unstressed) length is a constant times x, or 𝐹=𝑘𝑥

Example 7: A Bit of Work It takes a force of 10 N to stretch a spring 2 m beyond its natural length. How much work is done in stretching the spring 4 m from its natural length?

Example 7: A Bit of Work We must first determine the value of the force constant, k, of the spring. If it takes 10 N to stretch the spring 2 m beyond its natural length, then 10=2𝑘⟹𝑘=5 By Hooke’s law we have 𝐹 𝑥 =5𝑥. Note that this means that the force is not constant. Also note that the units for the constant are Newtons per meter. If we graph 𝐹 𝑥 =5𝑥, the horizontal axis is distance x in meters and the vertical axis is force in Newtons.

Example 7: A Bit of Work

Example 7: A Bit of Work Now, we create a Riemann sum by portioning the interval [0,4] into subintervals of width Δ𝑥. As the width approaches zero, the product 𝐹 𝑥 𝑘 Δ𝑥, where 𝑥 𝑘 is a value of x in the kth interval, is approximately equal to W. If we add up the values for each subinterval, we get a Riemann sum 𝑊≈ 𝐹 𝑥 𝑘 Δ𝑥 = 5 𝑥 𝑘 Δ𝑥 Finally, if we allow Δ𝑥 to approach zero the Riemann sum can be written as the integral 𝑊= 0 4 5𝑥 𝑑𝑥= 5 𝑥 2 2 0 4 = 5 4 2 2 =40 N−m