SUNY Geneseo Department of Mathematics

Area between Curves

Wednesday, December 4

Math 221 06
Fall 2019
Prof. Doug Baldwin

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Misc

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Final Exam

Wednesday, December 11, 8:00 - 11:20 AM, in our South Hall classroom.

Comprehensive, but concentrates on material since the second hour exam (problem sets 9 through 13, e.g., optimization, L’Hospital’s rule, summations, integrals, the Fundamental Theorem, areas, volumes, etc.)

Designed for about 2 hours, you’ll have 3 hours 20 minutes.

Rules and format otherwise similar to hour exams, especially open-references rules.

I’ll try to post some sample questions.

Donuts and cider.

Review Sessions

3-hour SI next Monday (Dec. 9), 3:00 - 6:00, Fraser 104.

1 hour on study day (Tuesday, Dec. 10), 12:00 - 1:00, in our regular classroom. Bring questions or topics you want to go over.

Last Day for Grading

Wednesday, December 11 (i.e., final exam day).

Please turn in any extra credit you want counted in your grade by then.

Please also complete any late grading appointments by then.

Questions?

Area between Curves

We looked at the basic idea last class: if f(x) ≥ g(x) for a ≤ x ≤ b, then the area between the graphs of f(x) and g(x) in the interval a ≤ x ≤ b is the integral from a to b of f(x) - g(x):

Graphs of functions f and g, area between highlighted and equal to integral from a to b of f minus g

Section 6.1.

One Detail

Find the area between the graphs of y = sin x and y = cos x between x = 0 and x = π.

Graphs of sine x and cosine x

The problem here is that neither function is consistently bigger than the other. The graphs cross at x = π/2. So much as happened when finding the area between a graph and the x axis if the function was sometimes negative, break the integration into several integrals over partial intervals, and switch the order of the subtraction in order to always subtract the smaller function from the larger.

Area is integral from 0 to pi over 4 of cosine minus sine plus integral from pi over 4 to pi of sine minus cosine

Key Idea

You always need to subtract the smaller function from the larger, so split the interval if necessary.

Another way to think of this is that you always integrate | f(x) - g(x) |.

Next

Volumes by slicing.

For example, what is the volume of this pyramid lying along the x axis:

Pyramid on its side with apex at origin and base at x equals 2; width and height equal x over 2

If you split the tapered pyramid into lots of thin slices, you can think of each slice as having a constant height and width, and calculate its volume by multiplying length times width times height:

Pyramid with slice of height and width one half x and length delta x

Now approximate the volume of the whole pyramid by adding the volumes of all the slices:

Volume is approximately the sum over n slices of the quantity 1 half x squared times delta x

This is an approximation, but it gets more accurate as the slices get thinner. So in the limit, as you have an infinite number of infinitely thin slices, you get the exact volume.

Volume is the limit as number of slices goes to infinity of sum over slices of slice volumes

And this limit is just a limit of a Riemann sum, i.e., a definite integral!

Volume is a limit of a sum which is an integral

This example gives you a sense of how volume can be described by a Riemann sum, and thus an integral. Tomorrow we’ll look at it a little more generally, and do some more examples.

Read section 6.2.

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