If your solution to 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.

015. `query 15

Question: `q query problem 5.2.39 6th; 5.2.24 was 5.2.23 integral of x^2+1 from 0 to 6. What is the value of your integral, and what were your left-hand and right-hand estimates?

Your solution:

Confidence Assessment:

Given Solution:

 ** An antiderivative of x^2 + 1 is x^3 / 3 + x. The change in the antiderivative from x = 0 to x = 6 is equal to the integral:

Integral = 6^3 / 3 + 6 - (0^3 / 3 + 0) = 216/3 + 6 = 234/3 = 78.

The values of the function at 0, 2, 4, 6 are 1, 5, 17 and 37.

The left-hand sum would therefore be 2 * 1 + 2 * 5 + 2 * 17 = 46.

The right-hand sum would be 2 * 5 + 2 * 17 + 2 * 37 = 118.

The sums differ by ( f(b) - f(a) ) * `dx = (37 - 1) * 2 = 72.**

From the shape of the graph which estimate would you expect to be low and which high, and what property of the graph makes you think so?

** The graph is increasing so the left-hand sum should be the lesser.

the graph is concave upward and increasing, and for each interval the average value of the function will therefore be closer to the left-hand estimate than the right. This is corroborated by the fact that 46 is closer to 78 than is 118. **

Self-critique (if necessary):

Self-critique Rating:

Question: `q query problem 5.2.31 6th; 5.2.32 f(x) piecewise linear from (-4,1) to (-2,-1) to (1,1) to (3,0) then like parabola with vertex at (4,-1) to (5,0). What is your estimate of the integral of the function from -3 to 4 and how did you get this result?

Your solution:

Confidence Assessment:

Given Solution:

 ** We use the area interpretation of the integral. This graph can be broken into a trapezoidal graph and since all lines are straight the areas can be calculated exactly.

From x=-3 to x=0 the area between the graph and the x axis is 2 (divide the region into two triangles and a rectangle and find total area, or treat it as a trapezoid). The area is below the x axis so the integral from x = -3 to x = 0 is -2.

From x = 0 to x = 3 the area is 2, above the x axis, so the integral on this interval is 2.

If the graph from x=3 to x=4 was a straight line from (3,0) to (4,-1) the area over this interval would be .5 and the integral would be -.5. Since the graph curves down below this straight line the area will be more like .6 and the integral closer to -.6.

The total integral from x=-3 to x=4 would therefore be about -2 + 2 - .6 = -.6. **

**** revised solution ... ****

It appears that the solution given above doesn't match the problem.

The graph as described would pass through (-3, 0), (-2, -1), (-1/2, 0), (1, 1), (3, 0) and (4, -1).

You should sketch the graph and check to see that these points are consistent with the given description. The regions between the graph and the x axis will all be triangles, except for the last, which will have a curved boundary.

Except for the last interval the graph would consist of broken straight lines.

I believe the areas of the resulting triangular regions would be, respectively, 1/2, 3/4, 3/4 and 1. The last region would encompass more area than just the triangle defined by (3, 0) and (4, -1), so would be greater than 1/2. Let's call it 2/3.

Taking account of whether each region is above or below the x axis, we have the following contributions to the integral:

-1/2, -3/4, 3/4, 1 and -2/3.

This would result in a total integral of -1/6.

Self-critique (if necessary):

Self-critique Rating:

Question: `q query problem 5.3.10 was 5.3.20 ave value of e^t over [0,10]. What is the average value of the function over the given interval?

Your solution:

Confidence Assessment:

Given Solution:

 ** An antiderivative of f(t) = e^t is F(t) = e^t. Using the Fundamental Theorem you get the definite integral by calculating F(10) - F(0). This gives you e^10 - e^0 = 22025, approx.

The average value of the function is the integral divided by the length of the interval. The length of the interval is 10 and 22025 / 10 is 2202.5. **

What would be the height of a rectangle sitting on the t axis from t = 0 to t = 10 whose area is the same as that under the y = e^t graph from t=0 to t=10?

** The height of the rectangle would be the average value of the function. That would have to be the case so that height * length would equal the integral. **

Self-critique (if necessary):

Self-critique Rating:

Question: `q query problem 5.3.38, was 5.3.26 v vs. t polynomial (0,0) to (1/6,-10) thru (2/6,0) to (.7,30) to (1,0). When is the cyclist furthest from her starting point?

Your solution:

Confidence Assessment:

Given Solution:

 ** The change in position of the cyclist from the start to any instant will be the integral of the v vs. t graph from the t = 0 to that instant.

From t = 0 to t = 1/3 we see that v is negative, so the cyclist is getting further from her starting point. The average velocity over this interval appears to be about -6 mph, so the integral over this segment of the graph is about -6 * 1/3 = -2, indicating a displacement of -2 miles. At t = 1/3 she is therefore 3 miles from the lake.

For the remaining 2/3 of an hour the cyclist is moving in the positive direction, away from the lake. Her average velocity appears to be a bit greater than 15 mph--say about 18 mph--so the integral over this segment is about 18 * 2/3 = 12. She gets about 12 miles further from the lake, ending up about 15 miles away.

The key to interpretation of the integral of a graphed function is the meaning of a rectangle of the Riemann Sum. In this case any rectangle in the Riemann sum has dimensions of velocity (altitude) vs. time interval (width), so each tiny rectangle represents a product of velocity and time interval, giving you a displacement. The integral, represented by the total area under the curve, therefore represents the displacement. **

STUDENT QUESTIONS

I know that you can find the change in position will be the integral multiplied by the change in time
How did you find the average velocity over the intervals? From the points listed above?
Integrals are really difficult for me to understand

INSTRUCTOR RESPONSES

<h3>@&
The change in position is the area beneath the v vs. t graph. This area represents the integral. You don't multiply the integral by anything to get the change in position, since the integral gives you the change in position.

The integral is equal to the average vertical coordinate of the graph (i.e., the average velocity) multiplied by the time interval.
*@</h3>


<h3>@&
The average velocity of the cylist on an interval is the average vertical coordinate of the graph.

If the graph between two points is a straight line segment, then the average velocity on that interval would be represented by the vertical coordinate of the midpoint of the segment.

If the graph curves above the straight line connecting the points at the left and right ends of the interval, then the average velocity will be somewhat higher than the midpoint of the straight line connecting those points.


*@</h3>



<h3>@&
The definite integral from point a to point b can be understood as the area under the graph between a and b, where 'area' in this context is negative when it's below the x axis.

So to answer this question you could look at the area from, say, 0 to .1, then 0 to .2, then 0 to .3, etc.., asking yourself to which point the total area would be the greatest.

Then you would think of moving the right endpoint of the interval continuously, until you arrive at the point where the total area is greatest.

Once you look at the graph in this way, the explanation in the given solution might become clearer.

You are of course welcome to ask additional questions about this problem.
*@</h3>
 

Self-critique (if necessary):

Self-critique Rating:

Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment.