Assignment 8

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**  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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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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For an integral of a typical function over a finite interval this condition could work.

However the interval isn't finite

** 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. **

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The integral of x^-p is undefined at 0, but its integral over this interval converges. Being undefined at an endpoint does not guarantee divergence.

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.

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The region x^2 + y^2 = 10 is a circle of radius sqrt(10).

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.

 

y values in the first quadrant run from y = 0 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 the substitution

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 cos^2(theta)

thus sqrt(10 – y^2) becomes sqrt(10) cos(theta)

y = 0 corresponds to sin(theta) = 0 so that theta = 0

y = sqrt(10) becomes sqrt(10) sin(theta) = sqrt(10) or sin(theta) = 1, so that 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 Ό the area of the circle x^2 + y^2 = 10.

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 the substitution

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 cos^2(theta)

thus sqrt(10 – y^2) becomes sqrt(10) cos(theta)

x = 0 corresponds to sin(theta) = 0 so that theta = 0

x = sqrt(10) becomes sqrt(10) sin(theta) = sqrt(10) or sin(theta) = 1, so that 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 Ό the area of the circle x^2 + y^2 = 10.

 

DER

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This is a good decimal approximation but it's not exact.

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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 / 2.

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.

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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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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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You've certainly got some of the basic ideas. You don't show enough detail to tell whether you completely understand the process. Remember your solutions have to be completely documented.

I've inserted extensive notes. Let me know if there's anything you don't understand.