Interpretations of the Definite Integral
Schedule this week
- Today: Finish up "Product Rule" and "Definite Integral Interpretations"
- Wed: quiz, Review/Q&A
- Thurs: Exam 2
But first...
That coffee problem
The quantity (in pounds) of a gourmet ground coffee that is sold by a coffee company at a price of $p$ dollars per pound is $Q=f(p)$
Units: $Q=f(p)$ is in pounds, and the price, $p$ is in dollars/pound.
- What is the meaning of the derivative $f'(8)?
When the price of the coffee is already 8 dollars/lb, $f'(8)$ is the change in the number of pounds of coffee sold when the price per pound is raised from 8 dollars. - What are the units of $f'(8)?
The units are [units of Q]/[units of p]=lbs/(dollars/lb)=(lbs)${}^2$/dollars. - Would $f'(8)$ be positive or negative?
Raising $p$ means raising the price/lb of the coffee. It seems like people would buy more/less if the coffee is more expensive (per pound)?
- Interpretations of the definite integral (.ppt)
- Old Handout (.pdf)
Average value
Given this table of values, how would you come up with an "average temperature for the day"?
| time (hours after midnight) | 6 | 12 | 18 | 24 |
|---|---|---|---|---|
| Temperature (${}^o$F) | 52 | 68 | 75 | 57 |
Probably: $$T_\text{average}=\frac{52+68+75+57}{4}=63.$$
If $T(t)$ is a continuous function of time, then there is a related "right sum" that we could do to approximate the integral from midnight ($t=0$) to midnight ($t=24$). With this table of values, we have values every 6 hours, so we could write the right sum in terms of 4 rectangles, each with a width of 6 hours, as $$\begineq\int_0^{24}T(t)\,dt&\approx \text{RH Sum}=52*6+68*6+75*6+57*6\\ =&(52+68+75+57)*6\\ =&\frac{52+68+75+57}{4}*6*4=T_\text{average}*24. \endeq$$
If instead of 4 equally spaced temperatures, imagine that we had 12. Then we'd add up the 12 temperatures and divide by 12. Like this... $$T_\text{average}=\frac{T_1+T_2+...+T_{12}}{12}.$$ And this would presumably be a better approximation to the "real average" of temperatures throughout the day.
The corresponding right sum would have 12 Temperature values, and each rectangle would have a width of 2 hours. So, now we'd have... $$\begineq \text{RH Sum}=&(T_1+T_2+...+T_{12})*2\\ =&\frac{T_1+T_2+...+T_{12}}{12}*2*12=T_\text{average}*24 \endeq $$
We know that as we take more and more rectangles which are ever narrower, that we'll get closer and closer to the definite integral. So, without further ado, the general result is that: $$\int_a^b f(x)\,dx=f_\text{average}*(b-a).$$ and re-arranging this just a bit:
The average value of the function $f(x)$ on the interval $a\lt x \lt b$ is: $$f_\text{average}=\frac{1}{b-a}\int_a^b f(x)\,dx.$$
