query 8

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course mth 174

5/13 11:55 pm

174assignment # 8

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Question:

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Question: explain the convergence or divergence of a p series; that is, explain why the p series converges for p > 1 and diverges for p <= 1.

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Your solution:

A p series is of the form 1/(n^p) we know that if p is greater than 1, then the series converges, or has a finite value. For values of p less than or equal to 1, we know that the series diverges.

For example if we had the integral from 1 to infinity of 1/x^2 our p is equal to 2 which is greater than 1 so the series converges or approaches a number

So the antiderivative is 1/x and when we evaluate at infinity it is 0 and at 1 the value is 1 so 0-(-1) = -1 so this proves convergeing (approaching a specific number)

For values of p less than or equal to 1 we look at example of integral from 1 to infinity of 1/squareroot of x where p would then be equal to ½ and since ½ is less than one the series should diverge. So we take antiderivative which would be 2x^(1/2) now when we evaluate this function at infinity we see that the numbers get larger and larger and approach infinity instead of a set number where when evaluated at 1 the value is 2 so this would show divergent.

confidence rating #$&*: 3

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Given Solution:

The key is the antiderivative. For p > 1 the antiderivative is still a negative power and approaches 0 as x -> infinity. For p < 1 it's a positive power and approaches infinity--hence diverges. For p = 1 the antiderivative is ln x, which also approches infinity. **

** An integral converges if the limit of its Riemann sums can be found and is finite. Roughly this corresponds to the area under the curve (positive for area above and negative for area below the axis) being finite and well-defined (consider for example the area under the graph of the sine function, from 0 to infinity, which could be regarded as finite but it keeps fluctuating as the sine goes positive then negative, so it isn't well-defined). **

** There is very little difference in the rapidity with which 1/x^1.01 approaches zero and the rapidity of the approach of 1 / x^.99. You would never be able to tell by just looking at the partial sums which converges and which diverges. In fact just looking at partial sums you would expect both to converge, because they diverge so very slowly.

If you graph 1 / x^.99, 1 / x and 1 / x^1.01 you won't notice any significant difference between them. The first is a little higher and the last a little lower past x = 1. You will never tell by looking at the graphs that any of these functions either converge or diverge.

However if you find the antiderivatives you can tell pretty easily. The antiderivatives are 100 x^.01, ln(x) and -100 x^-.01.

On the interval from, say, x = 1 to infinity, the first antiderivative 100 x^.01 will be nice and finite at x = 1, but as x -> infinity is .01 power will also go to infinity. This is because for any N, no matter how great, we can find a solution to 100 x^.01 = N: the solution is x = (N / 100) ^ 100. For this value of x and for every greater value of x, 100 x^.01 will exceed N. The antiderivative function grows without bound as x -> infinity. Thus the integral of 1 / x^.99 diverges.

On this same interval, the second antiderivative ln(x) is 0 when x = 0, but as x -> infinity its value also approaches infinity. This is because for any N, no matter how large, ln(x) > N when x > e^N. This shows that there is no upper bound on the natural log, and since the natural log is the antiderivative of 1 / x this integral diverges.

On the same interval the third antiderivative -100 x^-.01 behaves very much differently. At x = 1 this function is finite but negative, taking the value -100. As x -> infinity the negative power makes x-.01 = 1 / x^.01 approach 0, since as we solve before the denominator x^.01 approaches infinity. Those if we integrate this function from x = 1 to larger and larger values of x, say from x = 1 to x = b with b -> infinity, we will get F(b) - F(1) = -100 / b^.01 - (-100 / 1) = -100 / b^.01 + 100. The first term approaches 0 and b approaches infinity and in the limit we have only the value 100. This integral therefore converges to 100.

We could look at the same functions evaluated on the finite interval from 0 to 1. None of the functions is defined at x = 0 so the lower limit of our integral has to be regarded as a number which approaches 0. The first function gives us no problem, since its antiderivative 100 x^.01 remains finite over the entire interval [0, 1], and we find that the integral will be 100. The other two functions, however, have antiderivatives ln(x) and -100 x^-.01 which are not only undefined at x = 0 but which both approach -infinity as x -> 0. This causes their integrals to diverge.

These three functions are the p functions y = 1 / x^p, for p = .99, 1 and 1.01. This example demonstrates how p = 1 is the 'watershed' for convergence of these series either on an infinite interval [ a, infinity) or a finite interval [0, a]. **

More generally:

** If p > 1 then the antiderivative is a negative-power function, which approaches 0 as x -> infinity. So the limit as b -> infinity of the integral from 1 to b of 1 / x^p would be finite.

If p < 1 the antiderivative is a positive-power function which approaches infinity as x -> infinity. So the limit as b -> infinity of the integral from 1 to b of 1 / x^p would be divergent.

These integrals are the basis for many comparison tests. **

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Self-critique (if necessary):

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Self-critique Rating:3

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Question:

`q explain how we know that the integral from 0 to infinity of e^(-a x) converges for a > 0.

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Your solution:

We know that the integral from 0 to infinity of e^(-ax) converges for values of a greater than zero because we can tell this behavior by evaluating the antiderivative. For example

The integral from 0 to infinity of e^(-5x) would have an antiderivative of -1/5 e ^(-5x) so that when you evaluate at infinity the value is 0 and when you evaluate at 1 the value is -1/5 so that 0 -(-1/5) is 1/5 which shows that it converges because it approaches 1/5.

The same would be true when evaluating the integral from 0 to infinity of e^(-100x) where the antiderivative is -1/100 e^(-100x) again when evaluating at infinity the value is 0 and at 0 the value is -(1/100) so it is approaching the finite number 1/100 which is convergant.

We know that any values of a greater than 0 in the integral from 0 to infinity of e^(-a x) will always be approaching 0 because larger and larger negative power of e in the antiderivative approach 0, thus evaluating the antiderivative at 1 will cause the function to approach a finite number.

confidence rating #$&*: 3

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Given Solution:

The convergence of the integral of e^-(ax) from 0 to infinity is plausible because this is a decreasing function. However 1 / x^.99 is also decreasing on the same interval but it doesn't converge.

The reason e^(- a x) converges is that it antiderivative is -1/a e^(-a x), and as x -> infinity this expression approaches zero.

Thus the integral of this function from 0 to b is -1/a e^(-a b) - (-1/a e^0) = -1/a e^(-ab) + 1.

If we allow the upper limit b to approach infinity, e^-(a b) will approach 0 and we'll get the finite result 1. **

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Self-critique (if necessary):

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Self-critique Rating:3

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Question:

`q Section 7.8 Problem 3 7.8.18

convergence of integral of 1 / sqrt (`theta^2+1) from 1 to infinity

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Your solution:

Beginning with the integral of 1 / sqrt (`theta^2+1) from 1 to infinity we need to be able to compare this integral to a similar integral about which it is easier to show convergence/divergence. We know that the integral of 1/squareroot of theta^2 would be easier to evaluate and the plus 1 shouldn’t make too much of a difference. So we rewrite at the integral from 1 to infinity of 1/theta^ (2/2) or 1/theta^1 since the p value is less than or equal to 1, we know this would diverge so now we have to decide if our comparison integral is greater than or equal to our original integral to see if we can prove divergence of original.

So 1/squareroot (theta^2 +1) and 1/squareroot (theta ^2)

Square both sides and then get a common denominator and we see that for the numerators we have

Theta^2 for the original and (theta^2)+1 for the comparison since the original is less than the divergent comparison integral we can’t conclude anything, we need the original to be more than the divergent comparison so we half the comparison integral

1/(2theta)^1 so this would still be divergent since p=1

So now we compare again

1/(squareroot theta^2 +1) and 1/2theta

By squaring both sides and getting a common denominator we see that we have the original as 4theta ^2 and the comparison as theta^2 +1 so it is easy to see that the original is now greater than the divergent comparison and we can thus prove that the original is also divergent.

confidence rating #$&*: 3

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Given Solution:

** If it wasn't for the 1 in the denominator this would be 1 / `sqrt(`theta^2) or just 1 / `theta, which is just 1 / `theta^p for p = 1. As we know this series would diverge (antiderivative of 1 / `theta in ln(`theta), which appropaches infinity as `theta -> infinity).

As `theta gets large the + 1 shouldn't matter much, and we expect that 1 /`sqrt (`theta^2 + 1) diverges. If this expression is shown to be greater than something that diverges, then it must diverge.

However, 1 / `sqrt ( `theta^2 + 1) < 1 / `theta, and being less than a divergent function that diverges does not prove divergence.

We can adjust our comparison slightly:

Since 1 / `theta gives a divergent integral, half of this quantity will yield half the integral and will still diverge. i.e., 1 / (2 `theta) diverges.

• So if we can show that 1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we will have proved the divergence of 1 / `sqrt(`theta^2 + 1).

We prove this. Starting with

1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we square both sides to get

1 / (4 `theta^2) < 1 / (`theta^2+1). Multiplying by common denominator we get

`theta^2 + 1 < 4 `theta^2. Solving for `theta^2 we get

1 < 3 `theta^2

`sqrt(3) / 3 < `theta.

This shows that for all values of `theta > `sqrt(3) / 3, or about `theta > .7, the function 1 / `sqrt(`theta^2+1) exceeds the value of the divergent function 1 / (2 `theta). Our function 1 / `sqrt(`theta^2 + 1) is therefore divergent. **

STUDENT ERROR:

This integral converges because 1/sq rt(theta^2) approaches 0 rapidly.

INSTRUCTOR COMMENT:

** It will indeed converge but your argument essentially says that it converges because it converges. Way too vague. You have to use a comparison test of some kind. **

You have the right idea, but being less than a converging comparison function would prove convergence; being greater than a diverging function would prove divergence.

However being greater than a converging function, or being less than a diverging function, does not prove anything at all about the convergence of the original function.

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Self-critique (if necessary):

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Self-critique Rating:3

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Question:

**** query problem 7.8.20 convergence of integral from 0 to 1 of 1 / `sqrt(`theta^3 + `theta) **** does the integral converge or diverge, and why?

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Your solution:

We begin with integral from 0 to 1 of 1 / `sqrt(`theta^3 + `theta) so we will need to do a comparison test of a simpler integral so I will choose 1/(squareroot theta^3) which is 1/theta^(3/2) since 3/2 is greater than 1, the comparison integral would diverge.

Now we need to compare the original and the comparison in hopes that the original will be greater than the divergent comparison

So we have

1 / `sqrt(`theta^3 + `theta) and 1/(squareroot theta^3)

By squaring and getting a common denominator we see the numerators are theta is less than theta cubed plus theta

So this doesn’t prove anything.

I can try to see if I can somehow make the second integral less than the first but nothing I am trying is working so I assume I need to try a different comparison so now I will compare the original to 1/squareroot of theta which since p is less than 1 this comparison integral will converge. So if the original is less than the comparison integral, we can prove convergence

By squaring both sides and getting a common denominator we are left with numerators theta is less than theta cubed + theta so we can prove the original integral in convergent.

confidence rating #$&*: 2

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Given Solution:

On the interval 1 to infinity the integral would converge:

1 / `sqrt(theta^3 + `theta) < 1 / `sqrt(`theta^3) = 1 / `theta^(3/2).

This is an instance of 1 / x^p for x =`theta and p = 3/2.

The integral of 1 / theta^(3/2) converges by the p test.

The given integrand is less than 1 / theta^(3/2), so the original integral also converges, by the comparison test.

On the interval from 0 to 1 the result might be surprising, in the context of the previous result:

We might expect that the integral diverges, since the integral of 1 / theta^(3/2) diverges on this interval.

All we would need to do is show that our original integrand is greater than 1 / theta^(3/2)

However this isn't the case, our integrand is less than 1 / theta^(3/2) on the interval 0 < theta < 1.

We could patch this up by showing that our original integrand is greater than some fixed multiple of 1 / theta^(3/2) (e.g., maybe 1 / sqrt (theta^3 + theta) is greater than .000000001 * 1 / theta^(3/2)) on this interval), but it just isn't so. Whatever constant multiple we choose, there is some neighborhood of x = 0 where the original function is less than our comparison function (you are invited to prove this).

It turns out that as we approach zero, it is theta^3 that becomes insignificant, not theta. So the integrand must be compared 1 / theta^(1/2), or at least with a constant multiple of this expression.

The integral 1 / theta^(1/2) on 0 <= theta <= 1 is convergent (its antiderivative is 2 theta^(1/2), which can be evaluated at 0 and 1; the value of the integral is just 2).

Conveniently, 1 / (theta^3 + theta) is less than 1 / theta^(1/2), so the convergence of the latter ensures the convergence of the former..

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Self-critique (if necessary):

### I think my answer is correct, however I am not sure how I am supposed to choose which part of the original is important to use in the comparison and which one isn’t. Since I couldn’t get the original to be more than the divergent comparison when using the 1/squareroot of theta cubed for the comparison I chose to use the convergent 1/squareroot of theta and then the original was less than the convergent so I could prove the convergence of the original. Can you provide any tips on how to choose the significant portion of the denominator when setting up comparison integral?????

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Self-critique Rating:3

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One way to think about the present example:

The possibility of divergence arises because the integrand approaches infinity as theta approaches zero.

Near theta = 0, theta^3 becomes insignificant compared to theta.

So it's the behavior of theta, rather than that of theta^3, that will be more influential as theta approaches zero.

Now, in general:

If the given integrand is less than that of some convergent integral then the given integral converges.

If the given integrand is greater than that of some divergent integral then the given integral diverges.

So you would want to either show that the given integrand is less than that of some known convergent integral (or a constant multiple thereof) or that it is greater than that of a known divergent integral (or constant multple therefore).

It can't do both. So whichever part of the given expression works, that's the one you'll use. If one part works, the other won't.

Applying this to the present example:

1/sqrt(theta^3) = theta^(-3/2), which integrated between 0 and 1 would be divergent (its antiderivative would be -2/3 theta^(-1/2), which approaches infinity as theta approaches zer).

1 / sqrt(theta) = theta^(-1/2), which would be convergent (antiderivative is 2 sqrt(theta), which is 0 at theta = 0 and 2 at theta = 1).

Between 0 and 1, the given integrand 1 / sqrt( theta^3 + theta ) is less than the divergent 1 / sqrt( theta^3). So this doesn't tell us anything.

We can, however, say that between 0 and 1 the given integrand 1 / sqrt( theta^3 + theta ) is less than the convergent 1 / sqrt(theta). This proves that the integral converges.
*@

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Question:

query 8.1.5: x^2 + y^2 = 10 in 1st quadrant

strip `dh and y position y above center at origin ****

What is your expression for the Riemann sum?

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Your solution:

We are told that we are looking for a strip dh and y position above the center of the origin and are given x^2+y^2 =10 in the first quadrant.

We solve x^2+y^2 =10 for x getting

X^2=+-squareroot (10-y^2)

Since we are told the strip is in the first quadrant, we know the x value is positive so we have that the height of the strip is x=squareroot of (10-y^2), the width of the strip is dy

So the area of the strip is height * width or squareroot (10-y^2) * dy

So now we set up our integral from 0 to squareroot of 10 for (10-y^2) * dy (this is Reimann sum)

Here we use the tables and get

½(y*squareroot(10-y^2) + 10 * integral 1/(squareroot 10-y^2)

Now we use the tables again for the second part and get

1/2y 8 squareroot (10-y^2) + 5 arcsin y/squareroot 10

We evaluate from 0 to squareroot of 10

½(squareroot 10)(squareroot (10-squareroot 10^2) + 5 arcsin squareroot 10/squareroot 10 -[1/2*0*squareroot 10-0 + 5 arcsin 0/squareroot 10

7.85-0

7.85

confidence rating #$&*: 2

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Given Solution:

****** ******

FOR HORIZONTAL STRIPS

The solution for the x of the equation x^2 + y^2 = 10 is x = +- sqrt(10 - y^2). In the first quadrant we have x > = 0 so the first-quadrant solution is x = +sqrt(10 - y^2).

A vertical strip at position y extends from the y axis to the point on the curve at which x = sqrt(10 - y^2), so the ‘altitude’ of the strip is sqrt(10 - y^2). If the width of the strip is `dy, then the strip has area

`dA = sqrt(10 - y^2) `dy.

The curve extends along the y axis from y = 0 to the x = 0 point y^2= 10, or for first-quadrant y values, to y = sqrt(10). If the y axis from y = 0 to y = sqrt(10) is partitioned into subintervals the the total area of the subintervals is approximated by

A = sum(`dA) = sum ( sqrt(10 - y^2) `dy), and as interval width approaches zero we obtain the area

A = integral ( sqrt(10 - y^2) dy, y, 0, sqrt(10)).

The integral is performed by letting y = sqrt(10) sin(theta) so that dy = sqrt(10) cos(theta) * dTheta; 10 - y^2 will then equal 10 - 10 sin^2(theta) = 10 ( 1 - sin^2(theta)) = 10 sin^2(theta) and sqrt(10 - y^2) becomes sqrt(10) cos(theta); y = 0 becomes sin(theta) = 0 so that theta = 0; y = sqrt(10) becomes sqrt(10) sin(theta) = sqrt(10) or sin(theta) = 0, giving us theta = pi / 2.

The integral is therefore transformed to

Int(sqrt(10) cos(theta) * sqrt(10) cos(theta), theta, 0, pi/2) =

Int(10 cos^2(theta) dTheta, theta, 0, pi/2).

The integral of cos^2(theta) from 0 to pi/2 is pi/4, so that the integral of our function is 10 * pi/4 = 5/2 pi.

Note that this is ¼ the area of the circle x^2 + y^2 = 10.

FOR VERTICAL STRIPS

The solution for the y of the equation x^2 + y^2 = 10 is y = +- sqrt(10 - x^2). In the first quadrant we have y > = 0 so the first-quadrant solution is y = +sqrt(10 - x^2).

A vertical strip at position x extends from the x axis to the point on the curve at which y = sqrt(10 - x^2), so the ‘altitude’ of the strip is sqrt(10 - x^2). If the width of the strip is `dx, then the strip has area

`dA = sqrt(10 - x^2) `dx.

The curve extends along the x axis from x = 0 to the y = 0 point x^2= 10, or for first-quadrant x values, to x = sqrt(10). If the x axis from x = 0 to x = sqrt(10) is partitioned into subintervals the the total area of the subintervals is approximated by

A = sum(`dA) = sum ( sqrt(10 - x^2) `dx), and as interval width approaches zero we obtain the area

A = integral ( sqrt(10 - x^2) dx, x, 0, sqrt(10)).

The integral is performed by letting x = sqrt(10) sin(theta) so that dx = sqrt(10) cos(theta) * dTheta; 10 - x^2 will then equal 10 - 10 sin^2(theta) = 10 ( 1 - sin^2(theta)) = 10 sin^2(theta) and sqrt(10 - x^2) becomes sqrt(10) cos(theta); x = 0 becomes sin(theta) = 0 so that theta = 0; x = sqrt(10) becomes sqrt(10) sin(theta) = sqrt(10) or sin(theta) = 0, giving us theta = pi / 2.

The integral is therefore transformed to

Int(sqrt(10) cos(theta) * sqrt(10) cos(theta), theta, 0, pi/2) =

Int(10 cos^2(theta) dTheta, theta, 0, pi/2).

The integral of cos^2(theta) from 0 to pi/2 is pi/4, so that the integral of our function is 10 * pi/4 = 5/2 pi.

Note that this is ¼ the area of the circle x^2 + y^2 = 10.

DER

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Self-critique (if necessary):

###I believe my answer is ok although I used a 7.85 rather than 5/2 pi???

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Self-critique Rating:

@&
Rounded values are often acceptable, but exact values are always preferable.
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Question:

Query problem 8.1.12. Half disk radius 7 m thickness 10 m, `dy slice parallel to rectangular base.

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Your solution:

We are given that a Half disk radius 7 m thickness 10 m, `dy slice parallel to rectangular base. We take a slice delta h thick.

So we need to find the width of the slice A triangle is formed with hypotenuse 7, one side x^2 and one side yi^2 so we use phythagrean theorem getting x^2+yi^2 =7^2 solving for x we get x=squareroot(49-y^2) which is the width.

So now we go back to the volume formula of rectangle which is length * width* thickness or

10*squareroot (49-y^2) * dy this is our integral from 1 to n for the reimann sum

So lim as x approaches infinity integral from 0 to 7 of 10*sqaureroot(49-yi^2) dy

@&
In the limit you don't have y_i values anymore; you have just the continuous values of y. So your integrand would be 10 sqrt( 49 - y^2).
*@

10*integral from 0 to 7 of squareroot of (49-yi^2)

Use talbes

10[1/2y*sqaureroot(49-yi^2) + 49 times integral 1/squareroot (49-y^2)

Use tables again

5y*squareroot(49-yi^2) + 490 arcsin (yi/7)

Substituting in for 7 and 0 and subtracting we get 769.70

confidence rating #$&*: 2

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Given Solution:

STUDENT SOLUTION

here we used the pythagorean Theorem which is

y^2 + (7/2)^2 = 7^2

y = square root of 49 - (7/2)

Then we use the area of a semicircle which is w * delta h = 49 - (7/2) * delta h cm^2

then we get 49 arcsin 1 = 49 / 2 pi = 65.23 cm^2

INSTRUCTOR RESPONSE

You have to do the Riemann sum, get the integral then perform the integration.

A slice of the region parallel to the y axis is a rectangle. If the y coordinate is y = c_i then the slice extends to an x coordinate such that x^2 + (c_i)^2 = 7^2, or x = sqrt(49 - (c_i)^2).

So if the y axis, from y = -7 to y = 7, is partitioned into subintervals y_0 = -7, y1, y2, ..., y_i, ..., y_n = 7, with sample point c_i in subinterval number i, the corresponding 'slice' of the solid has thickness `dy, width sqrt(49 - (c_i)^2) and length 10 so its volume is 10 * sqrt(49 - (c_i)^2) * `dy and the Riemann sum is

sum(10 * sqrt(49 - (c_i)^2) * `dy, i from 1 to n).

The limit of this sum, as x approaches infinity, is then

integral ( 10 * sqrt(49 - y^2) dy, y from 0 to 7) =

10 integral (sqrt(49 - y^2) dy, y from 0 to 7).

Using the trigonometric substitution y = 7 sin(theta), analogous to the substitution used in the preceding solution, we find that the integral is 49 pi / 2 and the result is 10 times this integral so the volume is

10 * 49 pi / 2 = 490 pi / 2 = 245 pi.

A decimal approximation to this result is between 700 and 800. Compare this with the volume of a 'box' containing the figure:

The 'box' would be 14 units wide and 7 units high, and 10 units long, so would have volume 14 * 7 * 10 = 980.

The solid clearly fills well over half the box, so a volume between 700 and 800 appears very reasonable.

DER

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Self-critique (if necessary):

###I have a little bit of difficulty with these, please let me know if my answer is wrong and any suggestions for improvement???

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Self-critique Rating:3

@&
Your approach is good. Check my note, as you probably already have, for some details.
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Question:

query problem 8.2.14 (previously 8.2.11) arc length x^(3/2) from 0 to 2

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Your solution:

Ok we are given the function x^(3/2) ans asked to find the arclength from 0 to 2

We first graph the function with values (0,0) (1,1) and (2, 2.8)

After graphing the curve we draw a triangle coming from the bottom of the curve

We know y=x^(3/2) so delta y is f’x * delta x

So that would be 3/2 x^(1/2) delta x

So the length would be squareroot fo (1+f’x^2 *delta x)

So we have squareroot of 1+ (3/2x^.5)^2

Or squareroot of 1+9/4x

We evaluate this at the integral 0 to 2

U equals 9/4 x +1

Du = 9/4x

4/9du=dx

So 4/9 * integral from 0 to 2 of u^1/2

4/9*2/3*u^(3/2)

We then evaluate at 2 and subtract the evaluation at 0

We get 3.82-.296 which is 3.524

confidence rating #$&*: 3

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Given Solution:

On an interval of length `dx, containing x coordinate c_i, the ‘slope triangle’ at the top of the approximating trapezoid has slope approximately equal to f ‘ (c_i). The hypotenuse of this triangle corresponds to the arc length.

A triangle with ‘run’ `dx and slope m has ‘rise’ equal to m * `dx. So its hypotenuse is sqrt((`dx)^2 + (m `dx)^2) = sqrt( (1 + m^2) * (`dx)^2) = sqrt( 1 + m^2) * `dx. For small `dx, the hypotenuse is very close to the curve so its length is very near the arc length of the curve on the given interval.

Since the slope here is f ‘(c_i), we substitute f ‘ (c_i) for m and find that the contribution to arc length is

`dL_i = sqrt(1 + f ‘ ^2 (c_i) ) * `dx

So that the Riemann sum is

Sum(`dL_i) = sum ( sqrt(1 + f ‘ ^2 (c_i) ) * `dx ),

where the sum runs from i = 1 to i = n, with n = (b - a) / `dx = (2 - 0) / `dx. In other words, n is the number of subintervals into which the interval of integration is broken.

This sum approaches the integral of sqrt(1 + (f ' (x)) ^2, over the interval of integration.

In general, then, the arc length is

arc length = integral(sqrt(1 + (f ' (x)) ^2) dx, x from a to b).

In this case

f(x) = x^(3/2) so f ' (x) = 3/2 x^(1/2), and (f ' (x))^2 = (3/2 x^(1/2))^2 = 9/4 x.

Thus sqrt( 1 + (f ‘ (x))^2) = sqrt(1 + 9/4 x) and we find the integral of this function from x = 0 to x = 2.

The integral is found by letting u = 1 + 9/4 x, so that u ‘ = 9/4 and dx = 4/9 du, so that our integral becomes

Integral ( sqrt( 1 + 9/4 x) dx, x from 0 to 2)

= Integral ( sqrt(u) * 4/9 du, x from 0 to 2)

= 4/9 Integral ( sqrt(u) du, x from 0 to 2)

Our antiderivative is 4/9 * 2/3 u^(3/2), which is the same as 8/27 (1 + 9/4 x) ^(3/2). Between x = 0 and x = 2, the change in this antiderivative is

8/27 ( 1 + 9/4 * 2) ^(3/2) - 8/27 ( 1 + 9/4 * 0) ^(3/2)

= 8/27 ( ( 11/2 )^(3/2) - 1)

= 3.526, approximately.

Thus the arc length is

integral(sqrt(1 + (f ' (x)) ^2) dx, x from a to b

= Integral ( sqrt( 1 + 9/4 x) dx, x from 0 to 2)

= 3.526.

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Self-critique (if necessary):

### I believe I am ok on this on I got the correct answer????

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Self-critique Rating:3

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Question:

query problem 8.2.41 (31 4th edition, 21 3d edition) volume of region bounded by y = e^x, x axis, lines x=0 and x=1, cross-sections perpendicular to the x axis are squares

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We begin with graphing y=e^x and having it bounded by the x axis, x =0 and x=1. So we can see the integral will go from 0 to 1.

The height of this region is e^x and it is a square so the area can be found by (e^x)^2 or e^2x

We would divide this region into n intervals with a sample point, si being in the i interval. The thickness of this region is dx.

So the area of the sample point is e^2x but now we can substitute in si for x and get e^2si

The volume is area times thickness so we have e^2si * dx

Now we can integrate from 0 to 1 e^2si

u= si

du= 2 dx

1/2du = dx

So we have 1/2 integral from 0 to 1 of e^2si

When we evaluate at 1 we get 3.69

When we evaluate at 0 we get ½

So 3.69-.5 = 3.19

confidence rating #$&*: 3

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Given Solution:

x runs from 0 to 1.

At a given value of x the 'altitude' of the y vs. x graph is e^x, and a slice of the solid figure is a square with sides e^x so its area is (e^x)^2 = e^(2x).

If we partition the interval from x = 0 to x = 1 into subintervals defined by x_0 = 0, x_1, x_2, ..., x_i, ..., x_n, with interval number i containing sample point c_i, then the slice corresponding to interval number i has the following characteristics:

the thickness of the 'slice' is `dx

the cross-sectional area of the 'slice' at the sample point x = c_i is e^(2x) = e^(2 * c_i)

so the volume of the 'slice' is e^(2 * c_i) * `dx.

The Riemann sum is therefore

sum(e^(2 * c_i * `dx) and its limit is

integral(e^(2 x) dx, x from 0 to 1).

Our antiderivative is easily found to be 1/2 e^(2 x) and the change in our antiderivative is

1/2 e^(2 * 1) - 1/2 e^(2 * 0) = 1/2 e^2 - 1/2 = 1/2 (e^2 - 1).

The approximate value of this result about 3.19.

A 'box' containing this region would have dimensions 1 by e by e, with volume e^2 = 7.3 or so. The figure fills perhaps half this box, so our results are reasonably consistent.

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Self-critique (if necessary):

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Self-critique Rating:

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Question:

**** 8.1.28 (3d edition 8.1.29) volume of dam base 1400 m long, 160 m wide, 150 m high, narrows to 10 m wide at top. **** What is the volume of the dam?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We begin knowing that we have a dam base of 1400m, width 160m, height 150 m, top width 10 m

We know that when y is 0, the height is 160

And when y is 150, the height is 10 so we will need to try to figure out the function we can use to make both of these statements true.

We find that f(x) = 160-y

So to find the volume of the dam we would take a strip with height yi and width dyi

The area of this strip would be (160-y)*dyi

So the volume would be 1400(160 - yi)dyi

So the integral from 0 to 150 of 1400*(160-y)

We move the 1400 to the front and integrate getting

1400(160y - .5y^2) we evaluate at 150 and get 33,588,750 and at 0 and get 0

33,588,750 - 0= 33,588,750

confidence rating #$&*: 2

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Given Solution:

** The width at height y is a linear function equal to 160 when y = 0 and to 10 when y = 150. This linear function is 160 - y.

'So a strip at altitude yi having width `dyi, through the dam from front to back, has area (160 - yi) `dyi', and the volume corresponding to this strip is 1400 (160 - yi) `dyi.

This leads to the integral of 1400 (160 - y) with respect to y, for y = 0 to y = 150.

Antiderivative is 1400 (160 y - y^2/2); evaluating at limits we get 1400 (160 * 150 - 150^2 / 2) - 0 = 1400 ( 24000 - 22500/2) = 1400 (24000 - 11250) = 1400 ( 22750) = 30 million, approx. **

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Self-critique (if necessary):

###I do ok with these and sometimes get stuck part of the way through, but when I look at the answers, the steps are clear to me. I guess I just need to keep practicing???

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Self-critique Rating:3

@&
Practice is the answer. You learn mathematics by solving problems.

You appear to be doing very well with this.
*@

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Question:

problem 7.8.24 convergence of integral from 1 to infinity of

(2x^2+1)/(4x^4+4x^2-2) and

[ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We begin with the first integral from 1 to infinity of (2x^2 +1) /(4x^4 + 4x^2 -2) we need to find a more simple integral to compare this to. We could try to compare 2x^2/4x^4 which would simplfy to 1/2x^2 this would converge since p is greater than one but we have to prove that the original is less than the comparison in order to prove the convergence of the original.

So 2x^2 +1/(4x^4 + 4x^2 -2) is less than? 1/(2x^2)

We get a common denominator and the numerators become 4x^4 + 2x^2 and 4x^4 +4x^2 -2

We can try to solve for x to determine what values of x would make the statement true

So we get 0 is less than 2x^2 -2

2x^2 is greater than 2

X^2 is greater than 1

X is greater than +- 1

@&
You should consider reviewing quadratic inequalities.

x^2 > 1 has solution

x < -1 OR x > 1.

To show this rigorously, note that

sqrt( x^2 ) = | x |, which is equal to x if x is positive, and to -x if x is negative.

Since x^2 is always positive, x^2 > 1 is equivalent to

sqrt(x^2) > 1

If x is positive, sqrt(x^2) = x so that

x > 1.

If x is negative, sqrt(x^2) = - x so that

-x > 1, from which we conclude that

x < -1.

*@

@&
Now, the integral is from x = 1 to infinity, so we can disregard values of x which are < -1.

You have therefore shown that

(2 x^1 + 1) / (4 x^4 + 4 x^2 - 2) < 1 / (2 x^2) for x > 1.

Since the integral of 1/ (2 x^2) converges on the given interval, it follows that the integral of (2x^2+1)/(4x^4+4x^2-2) also converges on this interval.
*@

so would this prove convergence???? I’m a little confused on how to proceed???

confidence rating #$&*: 2

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Given Solution:

** The first thing we see is that (2x^2+1)/(4x^4+4x^2-2) acts for large x like the ratio of the leading terms in the numerator and denominator: 2 x^2 / ( 4 x^4) = 1 / (2 x^2), which is quite convergent so we expect that the integral will converge. All we have to do is spell out the details to be sure.

There's a little sticking point here, because of the -2 in the denominator and +1 in the numerator we can't say that the given expression is always less than 1 / (2 x^2). So let's solve to see where (2x^2 + 1) / ( 4 x^4 + 4 x^2 - 2 ) < 1 / ( 2 x^2).

This expression is equivalent to 2 x^2 ( 2 x^2 + 1) < 4 x^4 + 4 x^2 - 2, or to 4 x^4 + 2 x^2 < 4 x^4 + 4 x^2 - 2, which is in turn equivalent to 0 < 2 x^2 - 2, equiv to 0 < x^2 - 1 which occurs for x > 1 or for x < -1.

So for x > 1 we have (2x^2+1)/(4x^4+4x^2-2) < 1 / (2 x^2).

Noting that 1 / ( 2 x^2) < 1 / x^2, and that in [1, infinity) x^2 is a convergent p series, we see that (2x^2+1)/(4x^4+4x^2-2) is convergent on [1, infinity).

Now [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) acts a whole lot like ( 1 / (2x^2) )^(1/4) = 1 / 2^(1/4) * 1 / x^(1/2) . We are thus looking at comparison with a p series with p = .5, which diverges on [1, infinity). We therefore suspect that our function is divergent.

The problem is that [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) < ( 1 / (2x^2) )^(1/4) = 1 / 2^(1/4) * 1 / x^(1/2), and being < something that diverges doesn't imply divergence. However the function is so close to the divergent function that we know in our hearts that it doesn't matter. Our hearts don't prove a thing, but they can sometimes lead us in the right direction.

We need to show that [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) is greater than something that diverges. Fortunately this is easy to do. All we need is something just a little smaller than 1 / 2^(1/4) * 1 / x^(1/2). If we change the 1 / 2^(1/4) to 1/2 we get 1 / (2 x^(1/2)), which is still divergent but a bit smaller than the original divergent expression. Now we hypothesize that

[ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) > 1 /(2 x^(1/2) ).

Solving this inequality for x we first take the 4th powe of both sides to obtain

[ (2x^2+1)/(4x^4+4x^2-2) ] > 1 /(16 x^2 ) which we rearrange to get

(2x^2+1) * (16 x^2) > (4x^4+4x^2-2) which we expand to get

32 x^4 + 16 x^2 > 4 x^4 + 4x^2 - 2 or

28 x^2 + 12 x + 2 > 0.

If we evaluate the discriminant of the quadratic function on the left-hand side we find that it is negative so that it can never be zero. Since for x = 0 the quadratic is equal to 2, it must always be positive. Thus the inequality is satisfied for all x.

It follows that [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) > 1 /(2 x^(1/2) ) for all x and, since the right-hand side is a multiple of a divergent p-series, the original series diverges. **

17:28:47

Note: The questions below are 'followup questions'. You need not insert answers to these questions, but you may do so. If you do, please precede your answer with **** and add #$&* in the line immediately after your insertion. You will be able to recognize similar 'followup questions' on many questions in subsequent assignments.

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does the first integral converge? If so what is an upper bound for the integral?

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*****the first integral is 2x^2 +1/(4x^4 + 4x^2 -2) to find out if it converges we tried to compare it to a simpler interval above but we could also plug in infinity and 1

For infinity I will test a large value of x like 100 so I would get (2(100^2) +1)/4(100^4)+4(100^2)-2=.00000499

Then I would try an even larger value of x like 1000 (2(1000^2) +1)/4(1000^4)+4(1000^2)-2=.0000000078

It appears that at infinity this is approaching 0 and when I put in 1 I get 3/6 or ½ which would mean this is converging and the upper bound is ½???

#$&*

17:28:47

17:29:01

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Does the second integral converge? If so what is an upper bound for the

integral?

****** (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4)

To find out if this converges or diverges on integral 1 to infinity we substitute in large values of x for infinity to see what we get

For infinity I will test a large value of x like 100 so I would get (2(100^2) +1)/4(100^4)+4(100^2)-2=.00000499^(1/4)= .047

Then I would try an even larger value of x like 1000 (2(1000^2) +1)/4(1000^4)+4(1000^2)-2=.0000000078^(1/4)= .009

It appears that this intergral would be divergent because as x goes to infinity it doesn’t appear that the amounts are approaching any given number. So there would be no upper bound?????

#$&*

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17:29:01

17:29:07

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Explain how you obtained your results.

****I obtained my results by plugging in large numbers to signify approaching infinity for x and if the numbers appeared to be getting closer and closer to zero or a certain number we could assume it was convergent whereas if the numbers approaching infinity gave no movement towards a significant number and showed no pattern of getting closer to zero then I would assume they were divergent????

#$&*

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17:29:07

I graphed x^2/(x^4 + x^2) which approaches 0

** 1/x approaches zero too but its integral doesn't converge. Approaching zero doesn't show anything. And you can't tell about convergence from a graph on your calculator. **

17:30:09

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Is it true that (2x^2+1)/(4x^4+4x^2-2) < 1 / (2 x^2) for 1 < x?

****we want to know if 2x^2 +1/(4x^4 + 4x^2 -2) is less than? 1/(2x^2)

We get a common denominator and the numerators become 4x^4 + 2x^2 and 4x^4 +4x^2 -2

If we want to check the accuracy of the statement for values of x greater than 1 we could try

10, 100, and 50 to see if the statement is true by comparing the numerators

4x^4 + 2x^2 and 4x^4 +4x^2 -2

4(10^4) + 2(10^2) 4(10^4) + 4(10^2) -2= 40200 is less than 40398

4(100^4) + 2(100^2) 4(100^4) + 4(100^2) -2=400,020,000 is less than 400,039,998

4(50^4) + 2(50^2) 4(50^4) + 4(50^2) -2=25,005,000 is less than 25,009,998

Yes it appears to be true that the first is less than the second which would prove the convergence

#$&*

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17:30:09

17:32:06

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The given inequality is true. How the you use this inequality to place an

upper bound on the first integral?

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***we know that is the comparison integral is convergent and if the original integral is less than the convergent integral which is true from the work in the previous question, then the original integral would also be convergent. Since this is true the upper bound would be the value of the function evaluated at 1 which is the upper portion of the integral since when the function is evaluated at zero the result is 0 so zero - (value at 1) would be the upper bound???

#$&*