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Weighted Average

Pure mathematics is, in its way, the poetry of logical ideas, Albert Einstein.

Recall

Antiderivatives are fundamental concepts in calculus. They are the inverse operation of derivatives.

Given a function f(x), an antiderivative, also known as indefinite integral, F, is the function that can be differentiated to obtain the original function, that is, F’ = f, e.g., 3x2 -1 is the antiderivative of x3 -x +7 because $\frac{d}{dx} (x^3-x+7) = 3x^2 -1$. Symbolically, we write F(x) = $\int f(x)dx$.

The process of finding antiderivatives is called integration.

The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F'(x) = f(x)) between the ends of the interval, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$

Definition. The average value of a continuous function f(x) over the interval [a, b] is AVG = $\frac{y_1+y_2+···+y_n}{n}$ where y1, y2, ···, yn represent the function values at specific points within the interval [a, b] and n is the total number of points where the function is evaluated within the interval [a, b].

Since integrals are essentially a sum of all of the possible f(x) values, AVG = $\frac{1}{b-a}\int_{a}^{b} f(x)dx$.

Said in a more mathematical way, $\lim_{n \to ∞}\frac{y_1+y_2+···+y_n}{n}$ = $\frac{1}{b-a}\int_{a}^{b} f(x)dx$ where $\int_{a}^{b} f(x)dx$ represents the signed area under the curve of f(x) between x = a and x = b, $\frac{1}{b-a}$ is a scaling factor.

Average Value Theorem. If f is a continuous function on [a, b], then its average or mean value on [a, b] is given by favg = $\frac{\int_{a}^{b} f(x)dx}{b-a}$

Weighted Average

A weighted average is a method of computing an average where some elements contribute more than others. In other words, some elements or data points have different weights or importance assigned to them. The weights reflect the relative significance or contribution of each element to the overall average.

The weighted average gives more importance to certain values, elements, or data points in the dataset based on their assigned weights. This type of average is commonly used in various fields and situations where different elements contribute differently to the overall average.

Weighted averages are calculated by multiplying each value by its assigned weight, summing up these products, and then dividing by the sum of all the weights.

The weighted average of a list of elements x1, ···, xn with corresponding weights w1, ···, wn is $\frac{w_1x_1+w_2x_2 + ··· + w_nx_n}{w_1 + ··· + w_n}$. Note that if the weights are all just 1, that is, all elements or data points have the same importance, then the weighted average is just a plain average.

Exercises

To calculate the average time spent exercising over a 30-day month, we need to find the total time spent exercising and then divide it by the total number of days.

Average_Time = $\frac{20·7+60·5+45·14+0·4}{7+5+14+4}≈35.67$, the weighted average time you spent working out for this month was approximately 35.67 minutes per day.

The continuous analogue of a weighted average can be obtained as an integral. Let f be a function defined on an interval [a, b], with weight w(x), a non-negative function on [a, b]. Then, the formula for calculating a weighted average of f on the interval [a, b] is $\frac{\int_{a}^{b} f(x)w(x)dx}{\int_{a}^{b} w(x)dx}$, e.g., f(x) = c ⇒ $\frac{\int_{a}^{b} f(x)w(x)dx}{\int_{a}^{b} w(x)dx} = \frac{c\int_{a}^{b} w(x)dx}{\int_{a}^{b} w(x)dx} = c$.

Solved exercises

For simplicity sake, let’s define T = 100 -8y. We need to slice the cauldron in disks because at every level, the temperature is indeed constant.

Recall our previous example of the witches’ cauldron with a initial and final temperature with 0° C and 100 -8y respectively. The question is how much energy do we need? We will use the method of disks because the temperature is constant on horizontal levels. Notice that energy equals the product of volume -πx2dy- and temperature -100-8y-, and we are adding up from 0 (bottom) to 5 (top) -Figure 1.c.- .

 

$\int_{0}^{5} (100-8y)(πx^2)dy = \int_{0}^{5} (100-8y)(πy)dy = \int_{0}^{5} 100πy -8πy^2dy = (50πy^2 -\frac{8}{3}πy^3)\bigg|_{0}^{5} = \frac{2750}{3}π ≈ 47.3333π$ °C·m3.

$\int_{0}^{5} πydy = π\frac{y^2}{2}\bigg|_{0}^{5} = π\frac{25}{2}.$

Therefore, the weighted temperature average equals $\frac{\int_{0}^{5} (100-8y)(πx^2)dy}{\int_{0}^{5} πydy} = \frac{2750}{3}π : π\frac{25}{2} = \frac{5500}{75} ≈ 73.3°$.

 

The area of the disks (radius r = x) is πr2 =[y = 2x2 ⇒ $x = \sqrt{\frac{y}{2}}$]

Therefore, the weighted temperature average equals $\frac{\int_{0}^{4} (90-10y)(π\frac{y}{2})dy}{\int_{0}^{4} π\frac{y}{2}dy} = \frac{\int_{0}^{4} (45y-5y^2)dy}{\int_{0}^{4} \frac{y}{2}dy} = \frac{\frac{45y^2}{2}-\frac{5y^3}{3}}{\frac{y^2}{4}}\bigg|_{0}^{4} = \frac{\frac{45·4^2}{2}-\frac{5·4^3}{3}}{\frac{4^2}{4}} = \frac{360-\frac{320}{3}}{4} = \frac{\frac{760}{3}}{4} = \frac{760}{12} ≈ 63.3°$

 

The area of the disks (radius r = x) is πr2 =[y = x + 1 ⇒ x = y -1]

Let’s suppose that the temperature is uniformly decreasing (in a straight line), and recall that the equation of a line passing through two points is $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$ or $y-y_2=\frac{y_2-y_1}{x_2-x_1}(x-x_2)$ ⇒ $T-87=\frac{23-87}{4-1}(y-1) = \frac{-64}{3}(y-1) ↭ 3(T-87) = -64(y-1) ↭ 3T -261 = -64y + 64 ↭ 3T = 325 - 64y$

Average Temperature = $\frac{\int_{1}^{4} (\frac{325 - 64y}{3})π(y-1)^2dy}{\int_{1}^{4} π(y-1)^2dy} = \frac{\int_{1}^{4} (\frac{325 - 64y}{3})(y^2-2y+1)dy}{\int_{1}^{4} (y-1)^2dy} = \frac{\int_{1}^{4} (\frac{325}{3}y^2-\frac{650}{3}y+\frac{325}{3}-\frac{64}{3}y^3+\frac{128}{3}y^2-\frac{64}{3}y)dy}{\int_{1}^{4} (y-1)^2dy} = \frac{\int_{1}^{4} (\frac{-64}{3}y^3+\frac{453}{3}y^2-\frac{714}{3}y+\frac{325}{3})dy}{\int_{1}^{4} (y-1)^2dy} = \frac{\frac{-64}{12}y^4+\frac{453}{9}y^3-\frac{714}{6}y^2+\frac{325}{3}y}{\frac{1}{3}(y-1)^3}\bigg|_{1}^{4} = \frac{385.33-34.33}{9}=39°C$

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
  10. Chau Tu.
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