If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 2011 Calculus AB free response #2 (a & b)

Approximating rate of change and total area under a curve. Trapezoidal sums to approximate integrals. Created by Sal Khan.

## Want to join the conversation?

• At : Why is this expression the average temperature?
• Recall the definition of a Definite Integral

You're basically doing a summation of H(t)*dt
which can be read as a summation of Temperature*change in time

You really only need to focus on the units

You're basically adding up a bunch of degrees Celsius*minutes
which then leaves you with [total] degrees Celsius*minutes

But then you divide by 10 [total] [minutes]

What do you get if you divide [total] degrees Celsius*minutes by [total] minutes?
you get [average] degrees Celsius, or in another word: [average] Temperature

The Definite Integral expression yields [average] Temperature

-- Edit to clarify:

The Definite Integral itself yields [total] degrees Celsius*minutes
It is implied that the 1/10 outside the integral is 1/10minutes.

The "minutes" unit cross cancels, so the whole expression leaves just Temperature

And yes.. degrees Celsius*minutes makes no sense
Length*Width can at least be understood as Area
or Area*Height can at least be understood as Volume
degrees Celsius/minutes can be understood as rate of change
but degrees Celsius*minutes..... no intuitive sense
• I am confused when he does the derivative for f(3.5) at . My textbook tells me that you use two secant lines(that are close to the point whose slope you are trying to estimate) to approximate the derivative, is there something I am missing here?
• You can do that if you have a formula, but if you only have a table of values to draw from, you might be limited on how many slopes you can calculate. Sal is using a version of the difference quotient that uses an input value on either side of the input you are looking for. This is sometimes called the "balanced difference quotient" and is a bit more efficient than the standard formulation. The formula for the balanced difference quotient is
lim h->0 of (f(x+h)-f(x-h))/(2h).
• for (a), could you not say that the graph is probably an exponential decay graph? Then you could substitute in a couple of the points and do a little bit of algebra to find the approximate equation of the curve, then take the derivative at that point? I understand it would take longer, but it would give an even better approximation, would it not?
• It would be possible to extrapolate the equation of the curve, but this is one of many questions from a timed test, and the instructions say to use trapezoidal sum. Defining the curve itself algebraically would cost a lot of a limited resource to find a more precise answer than the test is asking for.
• How fast are you expected to do each of these free response questions on the AP test?
• There have been minor changes in recent years. There are two calculator questions in a 30 minute span of time, and then 4 non calculator questions in a 60 minute chunk. However, you can go back and complete the calculator questions (without the use of your calculator) if you finish early in the non calculator section.
• Why was part b more complicated than it needed to be? it would had been easier if we added all the degrees that the sample gave us and divided by five (The answer would then be 53).
• Well, it is true that that approximation is pretty close to the one obtained by integration over the interval. I can think of a couple of reasons not to use that shortcut.

First, the question asks us to use the trapezoidal sum method and to use the defined subintervals that we are given in the table. Usually the rubrics don't take kindly to not following directions on the test, and your method would not demonstrate mastery of the specified technique. Mathematical intuition might get to the ballpark of the required approximation, but probably won't give you points on this part of the test, which is kinda why people take it.

Second, your method would be further off if the curve were different, right? It happens that this curve is not that far off a linear function. But what if the data points were as follow:
0 66
2 60
5 44
9 42
10 41
You'd start to see more divergence and that is why the technique was invented.
• How do you suggest to complete this problems in a timely manner when stressed and in a hurry during the exam?! Honestly, I would like input.
(1 vote)
• Why does the area of a trapezoid A = 1/2H(B1 +B2) not used?
(1 vote)
• It is used.
He is saying (1/2)(b1+b2)*h
Where h is the length of the bottom, and the (1/2)(b1+b2) he calls the "average of the heights"
So when he finds the average of 66 and 60, he is finding (66+60)/2, then he multiplies that by the "height" which is 2
• For problem b part 1. Is it ok if you say: The average temperature in °C/min? that was my guess. Thank you
(1 vote)
• Ruben,
The average temperature is expressed (in this case) in °C and is a scalar, and would not be per unit time. The time units cancel: [H(t) (°C) x dt (min)] / [10 (min)]
I hope this helps.