query15

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course Mth 173

3/18 10:55 PM

015. `query 15

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Question: `q query problem 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?

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Your solution:

???what is the rule for this antiderivative, i dont remember why/how it works, had to look it up as x^(n+1)/(n+1) which looks somewhat like the derivative of power. f(x) = x^n, f'(x)= nx^(n-1) , but if i reverse that, how does the division come into play

f(x)= x^3/3 + x

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On the first level, take the derivative of x^3 / 3 + x and verify that you do get x^2 + 1.

Then think about why this is so.

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To integrate x^2 + 1 you have to find a function whose derivative is x^2 + 1.

When you take the derivative of a power, the power goes down by one. So to get x^2 you have to take a derivative of a multiple of x^3.

If you take the derivative of x^3 you get 3 x^2, not x^2.

So to get x^2 you would take the derivative of 1/3 x^3.

More generally if you want to integrate x^n you need to use a multiple of x^(n + 1). Since the derivative of x^(n + 1) is (n + 1) x^n, to get x^n you would have to take the derivative of 1 / (n + 1) x^(n + 1).

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so im thinking f(x) * dt, at least i remember reading that in the book

but i also remember yours is f(b) - f(a)

so 6^3/3 + 6 216/3 + 6 72 + 6= 78

78 - 0/3 = 78

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Good.

Think of x^3 / 3 + x representing some quantity. Then x^2 + 1 is its rate of change with respect to x.

It's as if you are given the rate of change function x^2 + 1, asked to find the 'quantity function' (which is this case is x^3 / 3 + x), then to find the change in this quantity as x changes from 0 to 6.

The value of the quantity function x^3 / 3 + x is 0 when x = 0, and 78 when x = 78. So the change in the quantity is 78 - 0 = 78.

In general if you regard the given function as the rate at this the integral function changes, then the definite integral calculates the change in that integral function.

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????the integral is 78? does that mean the area of a trapezoid we would have would be 78, representing the amount of change. and that the slope of the trapezoid is f'(x)?

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If f ' (x) = x^2 + 1, then f(x) = x^3 / 3 + x.

In this case the slope of a given trapezoid on the graph of x^3 / 3 + x will be equal to x^2 + 1 for some value of x in the interval corresponding to the trapezoid.

However the function you are given here is y = x^2 + 1. This is the function for which you would construct the trapezoidal graph, and the area of a series of trapezoidal graphs of this function, on the interval from x = 0 to x = 6, would approach 78.

You would need to use more than one trapezoid over the interval from x = 0 to x = 6. In fact the number of trapezoids will need to approach infinity in order to get the accurate result.

In terms of the trapezoidal approximation graph, if you divide the region into more and more trapezoids by continuing to subdivide the interval between 0 and 6, your total area will get closer and closer to 78, and will in fact approach 78 as a limit.

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x value =0 2 4 6

y value =1, 5, 17, 37

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It's not clear to me what these numbers represent. Ok, actually it is clear. They represent the values of the given function x^2 + 1 at x = 0, 2, 4, 6.

If you were to use these values to define a series of three trapezoids, their total area would be pretty close to 78. It would be worth the time to actually do this.

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???so i guess i just checked between intervals of 2, and what does that say for the y values? arent those rates of change, so am i still working with derivatives? how does the second derivative come into play? all these integral, derivative, differential, all the words make it spin in my head and I cant figure out what im doing. it seems that integral and differential are the same though.

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This is just because you are trying to really sort this out and understand this. It's natural to be somewhat confused at this point. Most students who aren't are just trying to do it rather than to understand it.

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so

2 * 1=2, 5 * 2= 10, 17 * 2 = 34, = 46

5 * 2 = 10, 17 * 2 = 34, 37 * 2 = 74, = 118

difference 72

???so what is the difference between this and the integral, the integral is the amount of change, and we know the period of time, 0-6, so with an assumption of 0 at 0, the value would be the integral, 78, and doesnt the left and right sums give us a ""range"" to pick from, so what is the point of doing those, if the integral gives us the value.

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The left and right sums are used to prove that the change in the antiderivative function is in fact the value of the integral of the given function.

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The given function is y = x^2 + 1.

An antiderivative is x^3 / 3 + x. Any antiderivative is called an integral of the given function. (Note that x^3 / 3 + x + 14 pi is also an antiderivative of the given function, since its derivative is als x^2 + 1; this will be the case for any function of the form x^3 / 3 + x + c, where c is a constant number.).

The general antiderivative of x^2 + 1 is x^3 / 3 + x + c, consistent with the preceding.

The definite integral of x^2 + 1 between x = 0 and x = 6 is the change in the value of an antiderivative function between those values of x. In this case the change in the value of the antiderivative x^3 / 3 + x is 78. You should verify that the change in the value of the other antiderivative function mentioned above is also 78. You should also verify that if c is a constant number, the change in the general antiderivative function x^3 / 3 + x + c is also 78.

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confidence rating #$&*:1

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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. **

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Self-critique (if necessary):

had a little trouble with this, didnt know exactly what my data tells me, and what I can do with it.

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Self-critique Rating:OK

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Question: `q query problem 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?

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Your solution:

Problem 31

(a)Using Figure 5.38 find integral between -3 and 0 f(x)dx

(b)

well given the way the function is, i cant set up a formula to find it, but since the integral is the area, i can cut it into pieces i know how to find the area for

from -3 to 0 is a trapezoid. 1/2(1 + 3) * 1 so 2. and its below the x-axis so -2

from 0 to 3 the next is another trapezoid, identical but flipped over the x-axis, so 2.

the next part is a little different, the entire shape is parabola like, but i only need to its vertex, which is almost a line, I cant tell, but it looks like its vertex lays on -1, and it goes over one unit, so thats a line cutting a square in half, so .5, but its curved ever so slightly, so it would be a little more than .5, dont know by how much so lets just go to .6. it is below the x-axis so its -.6

so add those all together and you get -.6. and since the integral is the area of shapes, adding those together should give you the integral over the entire segment.

confidence rating #$&*:2

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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. **

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Self-critique (if necessary):

I understand this bit a lot better than the previous question.

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Self-critique Rating:OK

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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?

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Your solution:

???just for clarity, its technially giving me the rate of change, which id have to find the antiderivative for, which is just the same in this case, correct?

by average value, does it mean rate of change? like per second?

this is e^10 - e^0 = 22026.46579

if it means by per second, or the linear path. we did it from 0-10 so 10-0= 10

2202.646579/unit

confidence rating #$&*:2

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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. **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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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?

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Your solution:

I dont have that problem in my book, or at least can't locate it. but just looking at it, and assuming that from point to point, each line is close to linear, or just curved to the value. that whatever the t value is at their point at (2/6,0) is the farthert away, because all the values are negative or neutral to that point, and after that it goes positive or neutral.

confidence rating #$&*: 3

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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. **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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

the first problem caused me great trouble, but i understood everything else relatively fine, so i dont think it is as much the concept i dont get, as just definitions, and precise meaning. the importance."

Self-critique (if necessary):

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Self-critique rating:

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

the first problem caused me great trouble, but i understood everything else relatively fine, so i dont think it is as much the concept i dont get, as just definitions, and precise meaning. the importance."

Self-critique (if necessary):

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Self-critique rating:

#*&!

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You're doing well, and asking very good questions. You'll have an excellent understanding once you get it sorted out, as you will do.

Check my notes.

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