AP Calculus Review Average Value of Functions

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! Defining the average value of a continuous function is not as easy as finding the average of a finite set of data. Average Value of Functions Suppose f is a continuous function defined over an interval [a, b]. In particular f(x) exists at every one of the infinitely-many points x between and including a and b. So, if youre looking for the average value of f on that interval, it wont do any good to try adding up those infinitely-many data points. Instead, the way to tame the infinity is to use calculus. Specifically, we define the average value of a function f as the following definite integral. The Theory Behind the Formula But where does the integral formula for average come from? The key is sampling. If you included enough of the function values, say a thousand, a million, or even more, then that should approximate the average of all infinitely-many points! Lets illustrate with the following example. Estimate the average value of the function f(x) = √(x) + 1 over the interval [1, 3]. Since were just estimating, lets pick four sample points. (The more sample points you pick, the better your estimated average will be.) Divide the interval [1, 3] into four equal subintervals, and lets agree to choose the midpoint of each subinterval. Then plug those midpoints into f to find the sample values. MidpointsHeight: f(x) = (x) + 1 1.25(1.25) + 1 = 2.12 1.75(1.75) + 1 = 2.32 2.25(2.25) + 1 = 2.50 2.75(2.75) + 1 = 2.66 Finally, use the familiar old averaging formula. Add up the data and divide by the number of data points: (2.12 + 2.32 + 2.50 + 2.66)/4 = 2.4 So the (approximate) average of the function is 2.4. Taking it to the Limit However, what weve just done will not give us an exact answer because weve essentially ignored most of the function! What about f(1.3) or f(2.95234)? No matter how many sample points we include, there will always be some missing Unless we can use the magic of calculus to catch them all. The sampling process should remind you of a Riemann Sum. For a quick reminder, feel free to check out AP Calculus Review: Riemann Sums. Midpoint Riemann sum. In the limit as the number of sample points goes to ∞, the Riemann sum becomes a definite integral. Let n be any whole number and xk* stand for the various sample x-values. Then the estimated average is the sum: Next, allow n → ∞ using a limit. We also need to get Δx into the act somehow. The trick is to multiply and divide by (b a). Remember, Δx = (b a)/n. Examples Ok, now that youve seen the theory, lets use the formula in practice! Problem 1 Find the exact average value of f(x) = √(x) + 1 over the interval [1, 3]. Solution Above, we only estimated the average to be 2.4. Now well use the integral formula to determine the average value precisely. (Its interesting to compare our estimate with the exact value above. √(3) + 2/3 ≈ 2.39871747, which means that our estimate of 2.4 was actually pretty good!) Problem 2 The amount of energy associated with a certain chemical reaction is given by E = x ln x, where 1 ≠¤ x ≠¤ e, and x represents the amount of one of the reactants. Find the average energy of the reaction over the range of possible levels of reactant. Solution This problem seems more like chemistry than math! Chemistry can be fun too! But what does this have to do with calculus However, the keyword average tells us that mathematics plays a major role in this problem. In fact, they are simply asking for the average value of f(x) = x ln x, over the interval [1, e]. First set up the integral formula with a = 1 and b = e. Then work out the integration, which involves Integration By Parts in this case. Thus the average energy of the reaction is (e2 + 1)/[ 4(e 1) ], or roughly 1.22. Mean Value Theorem for Integrals Averages are also called means. So you may use the same formula to find the mean value of a function. There is also an important result in calculus that relates the mean value to a particular function value on the given interval. The Mean Value Theorem for Integrals (MVTI). If f is continuous on a closed interval [a, b], then there is at least one point x = c in that interval such that the mean value of the function is equal to f(c). That is, (Caution: There is also a Mean Value Theorem for Derivatives. Its important not to confuse the two.) Problem 3 Let f(x) = 6x2 8x + 1. Determine the value of c at which the mean value of f on [-1, 1] is the same as f(c). Solution According to the Mean Value Theorem for Integrals, there must be at least one such value c. Lets set up the formula and find it! At this point, we will need to solve a quadratic equation. Dont forget your Quadratic Formula! 6c2 8c 2 = 0 c = (2  ± √(7))/3 It seems as though there may be two answers. However, only one lies within the given interval [-1, 1]. c = (2 + √(7))/3 ≈ 1.549, not in the interval. c = (2 √(7))/3 ≈ -0.215, in the interval. Therefore, the only value that satisfies the MVTI is c = (2 √(7))/3. Summary Although average value and the Mean Value Theorem for Integrals are specialized topics and only show up in a few problems on any given AP Calculus test, they are important concepts to master. For one thing, they illustrate how integral calculus can be used in applications. Moreover, working out the average value of a function is no more difficult than computing a definite integral. So now when you see these kinds of problems on the AP Calculus Exam, you can rise to the challenge!