Query 9

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course Mth 272

July 11, 2012

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

009. `query 9

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Question: `q 5.5.1 (previously 5.5.6 (was 5.5.6) ) area between (x-1)^3 and x-1 from 0 to 2

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Your solution:

I graphed both equations then took the area in between the intersection of the lines and got approximately 1.5.

confidence rating #$&*: 2

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Given Solution:

`a The graphs cross at (0, -1), (1,0), and (2, 1) as we easily find by solving the equation (x-1)^3 = (x - 1).

Since (1, 0) lies between the endpoints of our interval we have to be careful about which function lies above which, and we'll have to split the calculation into two separate intervals.

• The x-1 graph lies above the (x-1)^3 graph from 0 to 1, so the area on this interval will be the integral of (x-1) - (x-1)^3 between these limits.

• The (x-1)^3 graph lies above the (x-1) graph from 1 to 2, so the area on this interval will be the integral of (x-1)^3 - (x-1) between these limits.

Each integral is equal to .25, so the total area is the sum .25 + .25 = .5 of these areas. **

DER

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Self-critique (if necessary): I’m not sure if I’m supposed to be substituting a value in for x in these equation or what. I’m not sure how I am supposed to use the integral of the expression if there’s no definite numbers to solve for x.

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On this problem you have to integrate the difference between the lower and the higher function, i.e., between the function whose graph lies lower and the function whose graph lies higher.

The complication here is that on the given interval from 0 to 2, the graphs of the functions cross, which reverses the order in which the functions are subtracted.

We therefore need to answer the question in terms of two integrals, one for each of the two intervals.

The graphs intersect where (x - 1)^3 = x - 1. Solving this equation for x, we obtain solutions x = 1 and x = 2.

Between x = 0 and x = 1, the x - 1 function is the lower, so the integral for that part of the interval is

integral ( ( (x - 1) ^3 - (x - 1)) dx, x from 0 to 1)

Between x = 1 and x = 2, the (x-1)^3 graph is lower, so for that part of the interval the integral is

integral ( ( (x - 1) - (x - 1)^3 ) dx, x from 1 to 2).

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Self-critique Rating: 3

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Question: `q 5.5.2 (previously 5.5.10 (was 5.5.10) ) describe the region for integral of (1-x^2) - (x^2-1) from -1 to 1

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Your solution:

2-2x^2

Integral: 2x-(2x^3)/3

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Good.

If you evaluate this from x = -1 to x = 1 you will get 8/3, which is the correct area.

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Therefore, after graphing the equation between -1 and 1 appears to be a diagonal line connecting 2 parabolas.

confidence rating #$&*: 2

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Given Solution:

`a*& The graph of 1 - x^2 is an upside down parabola with vertex at (0, 1), intercepting the x axis at x = -1 and x = 1.

The graph of x^2 - 1 is a rightside up parabols with vertex at (0, -1), intercepting the x axis at x = -1 and x = 1.

The region between the graphs is close to a circle passing thru (-1,0), (0,1), (1,0) and (0, -1), but the region is not exactly circular since it is formed by two parabolas. The graphs aren't vertical at (1,0) and (-1,0), for example, and a circle would be. The parabolas curve in such a way as to stay inside the circular region, so the region between the parabolas will have a bit less area than the circle.

The integrand (1-x^2) - (x^2-1) can be simplified to 2 - 2 x^2. An antiderivative would be 2x - 2/3 x^3. Evaluating this at -1 and 1 we obtain integral 8/3. The area of the region is 8/3 = 2.67 approx..

Note that the area of the circle described above would be pi = 3.14, approx., a bit bigger than the area of the region between the parabolas. *&*&

STUDENT ERROR: The graph is a circular region centered on the point (0,0)

INSTRUCTOR COMMENT:

The region is not exactly circular, (for example the graphs aren't vertical at (1,0) and (-1,0), for example), but it's fairly close to the circle. **

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Self-critique (if necessary): When you say fairly close to a circle are you referring to the parabola?

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Self-critique Rating: 3

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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Check my notes. If you still have questions, be sure to ask.

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