SUNY Geneseo Department of Mathematics

Area Between Curves

Monday, May 6

Math 221 03
Spring 2019
Prof. Doug Baldwin

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Previous Lecture

Misc

Review Session

Thursday, May 9 (study day)

1:00 - 2:00

In our regular classroom (Sturges 221).

Bring things you want to talk about.

SI

Review on Wednesday, 4:00 - 7:00, Bailey 209.

Final Exam

Thursday, May 16, 12:00 Noon.

In our regular room.

Comprehensive, but emphasizing material since the 2nd hour exam (e.g., Mean Value Theorem, limits at infinity, Riemann sums, definite integrals, substitution, applications of integration; roughly problem sets 7 through 9).

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

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

Practice questions are now available in Canvas.

I’ll bring donuts and cider.

SOFIs

SOFIs are under way.

8 responses (26%) so far. Thank you!

But please keep filling them out. I do read them and apply the feedback where possible to future classes (like the mid-semester feedback).

MLC

Extended hours during study day and finals.

Questions?

Extended Power Rule

How to approach question 5 on the problem set (prove an antiderivative power rule for integrands of the form (x+k)n for constants k).

There are two ways you might approach this: one is to use a u-substitution to find the antiderivative, the other is to differentiate the proposed antiderivative and check that the result equals the original integrand.

Integral of x plus k all to the n maybe equals 1 over n plus 1 times x plus k to the n plus 1

Pursuing the u-substitution approach shows that the proposed antiderivative is correct:

Integral of x plus k to the n becomes integral of u to the n i.e. u to the n plus 1 divided by n plus 1

Here’s the complete analysis, as it looked when we finished it in class:

Integrating x plus k all to the n by substitution

Integration in Mathematica

How to do it? Use the Integrate function. Here is a notebook with examples of both an indefinite integral (antiderivative) and a definite integral.

Area between Curves

Section 6.1.

Example

Friday we started thinking about the area of a (very geometric) pond whose edges are the graphs of y = x2 and y = 4 - x2. We realized that the area would be the integral of the difference of the functions.

Area between right-side-up and upside-down parabolas

Now finish the problem.

The first step is to figure out where the boundary curves intersect, because the x values of the intersections will be the bounds of integration.

Area between parabolas with intersections marked and found to be at x equals plus or minus root 2

Now integrate as you would any other definite integral:

Integrate 4 minus 2 x squared from minus root 2 to plus root 2

Here’s the complete solution, as it looked when we finished it in class:

Finding area between 2 parabolas by integrating their difference

Key Points

Procedure for finding the area between f(x) and g(x), with f(x) ≥ g(x)

Graphs of 2 functions that cross 3 times

Volumes as Integrals of Slices

Section 6.2.

Example

Can you find the volume of this tapered prism as an integral?

Prism tapering from point at origin to triangular face at x equals 1

If not for the taper, you could just multiply the area of the triangular end by the length of the prism. So what if we concentrate on a thin slice of the prism, over which it doesn’t taper much?

Prism with a triangular slice highlighted in middle

Now the approximate volume of the whole prism is the sum of the (approximate) volumes of all the slices:

Sum over n slices of area of triangle times delta x

Look! This is a Riemann sum!

Finally (and not surprisingly, since we have a Riemann sum), we can get the exact volume by taking the limit as we approach an infinite number of infinitely thin slices. That limit, moreover, is an integral.

Limit of sum of areas times delta x is integral of area with respect to x

Here’s the whole analysis, as it looked when we finished it:

Volume as sum of thin slices and thus as an integral

Key Point

Volumes of irregular shapes can sometimes be calculated as sums of volumes of thin slices, which in turn leads to an integration formula for them.

Next

More volume.

Read (or finish reading) section 6.2.

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