# 9.3average Valueap Calculus

The average of a set of data is typically defined as the sum of the values divided by the number of data points. But what if you have infinitely many data points? What is the average value of a function? Read on to find out! Average Value of Functions Suppose f is a continuous function defined over an interval. In particular f(x) exists at every one of the infinitely-many points x between. Evia Family of Implantable Pulse Generators Technical Manual Evia Implantable Pulse Generators Evia DR X-Ray identification Evia DR-T X-Ray identification Radiopaque Identification A radiopaque identification code is visible on standard x-ray, and identifies the pulse generator: Evia DR, DR-T, SR, and SR-T SF C AUTION Because of the numerous available 3.2-mm configurations (e.g., the IS-1.

The average of some finite set of values is a familiar concept. If,for example, the class scores on a quiz are 10, 9, 10, 8, 7, 5, 7, 6,3, 2, 7, 8, then the average score is the sum of these numbers dividedby the size of the class:$$ hbox{average score} = {10+ 9+ 10+ 8+ 7+ 5+ 7+ 6+ 3+ 2+ 7+ 8over 12}={82over 12}approx 6.83.$$Suppose that between $t=0$ and $t=1$ the speed of an object is$sin(pi t)$. What is the average speed of the object over that time?The question sounds as if it must make sense, yet we can't merely addup some number of speeds and divide, since the speed is changingcontinuously over the time interval.

To make sense of 'average' in this context, we fall back on the ideaof approximation. Consider the speed of the object at tenth of asecond intervals: $sin 0$, $sin(0.1pi)$, $sin(0.2pi)$,$sin(0.3pi)$,…, $sin(0.9pi)$. The average speed 'should' befairly close to the average of these ten speeds:$$ {1over 10}sum_{i=0}^9 sin(pi i/10)approx {1over 10}6.3=0.63.$$Of course, if we compute more speeds at more times, the average ofthese speeds should be closer to the 'real' average. If we take theaverage of $n$ speeds at evenly spaced times, we get:$${1over n}sum_{i=0}^{n-1} sin(pi i/n).$$Here the individual times are $ds t_i=i/n$, so rewriting slightly we have$${1over n}sum_{i=0}^{n-1} sin(pi t_i).$$This is almost the sort of sum that we know turns into an integral;what's apparently missing is $Delta t$—but in fact, $Delta t=1/n$,the length of each subinterval. So rewriting again:$$ sum_{i=0}^{n-1} sin(pi t_i){1over n}= sum_{i=0}^{n-1} sin(pi t_i)Delta t.$$Now this has exactly the right form, so that in the limit we get$$ hbox{average speed} = int_0^1 sin(pi t),dt= left.-{cos(pi t)overpi}right _0^1= -{cos(pi)over pi}+{cos(0)overpi}={2overpi}approx 0.6366approx 0.64.$$

It's not entirely obvious from this one simple example how to computesuch an average in general. Let's look at a somewhat more complicatedcase. Suppose that the velocity of an object is $ds 16t^2+5$ feet per second. What is the average velocity between $t=1$ and $t=3$? Again weset up an approximation to the average:$${1over n}sum_{i=0}^{n-1} 16t_i^2+5,$$where the values $ds t_i$ are evenly spaced times between 1 and 3. Once again we are 'missing' $Delta t$, andthis time $1/n$ is not the correct value. What is $Delta t$ ingeneral? It is the length of a subinterval; in this case we take theinterval $[1,3]$ and divide it into $n$ subintervals, so eachhas length $(3-1)/n=2/n=Delta t$. Now with the usual 'multiply anddivide by the same thing' trick we can rewrite the sum:$$ {1over n}sum_{i=0}^{n-1} 16t_i^2+5= {1over 3-1}sum_{i=0}^{n-1} (16t_i^2+5){3-1over n}= {1over 2}sum_{i=0}^{n-1} (16t_i^2+5){2over n}= {1over 2}sum_{i=0}^{n-1} (16t_i^2+5)Delta t.$$In the limit this becomes$${1over 2}int_1^3 16t^2+5,dt={1over 2}{446over 3}={223over 3}.$$Does this seem reasonable? Let's picture it: infigure 9.4.1 is the velocity function togetherwith the horizontal line $y=223/3approx 74.3$. Certainly the height of thehorizontal line looks at least plausible for the average height of thecurve.

Here's another way to interpret 'average' that may make ourcomputation appear even more reasonable. The object of our examplegoes a certain distance between $t=1$ and $t=3$. If instead the objectwere to travel at the average speed over the same time, it should gothe same distance. At an average speed of $223/3$ feet per second fortwo seconds the object would go $446/3$ feet. How far does it actuallygo? We know how to compute this:$$int_1^3 v(t),dt = int_1^3 16t^2+5,dt={446over 3}.$$So now we see that another interpretation of the calculation$${1over 2}int_1^3 16t^2+5,dt={1over 2}{446over 3}={223over 3}$$is: total distance traveled divided by the time in transit, namely,the usual interpretation of average speed.

In the case of speed, or more properly velocity, we can alwaysinterpret 'average' as total (net) distance divided by time. But inthe case of a different sort of quantity this interpretation does notobviously apply, while the approximation approach always does. We mightinterpret the same problem geometrically: what is the average heightof $16x^2+5$ on the interval $[1,3]$? We approximate this in exactlythe same way, by adding up many sample heights and dividing by thenumber of samples. In the limit we get the same result:$$ lim_{ntoinfty}{1over n}sum_{i=0}^{n-1} 16x_i^2+5= {1over 2}int_1^3 16x^2+5,dx={1over 2}{446over 3}={223over 3}.$$We can interpret this result in a slightly different way. The areaunder $y=16x^2+5$ above $[1,3]$ is$$int_1^3 16t^2+5,dt={446over 3}.$$ The area under $y=223/3$ over the same interval $[1,3]$ is simply thearea of a rectangle that is 2 by $223/3$ with area $446/3$. So theaverage height of a function is the height of the horizontal line thatproduces the same area over the given interval.

## Exercises 9.4

**Ex 9.4.1**Find the average height of $cos x$ over the intervals$[0,pi/2]$, $[-pi/2,pi/2]$, and $[0,2pi]$.(answer)

**Ex 9.4.2**Find the average height of $ds x^2$ over the interval$[-2,2]$.(answer)

**Ex 9.4.3**Find the average height of $ds 1/x^2$ over the interval$[1,A]$.(answer)

**Ex 9.4.4**Find the average height of $ds sqrt{1-x^2}$ over the interval$[-1,1]$.(answer)

## 9 3 Average Valueap Calculus Formula

**Ex 9.4.5**An object moves with velocity $ds v(t)=-t^2+1$ feet per secondbetween $t=0$ and $t=2$. Find the average velocity and the averagespeed of the object between $t=0$ and $t=2$.(answer)

## 9 3 Average Valueap Calculus Equation

**Ex 9.4.6**The observation deck on the 102nd floor of the Empire State Buildingis 1,224 feet above the ground. If a steel ball is dropped from theobservation deck its velocity at time $t$ is approximately $v(t)=-32t$feet per second. Find the average speed between the time it is droppedand the time it hits the ground, and find its speed when it hits theground.(answer)