course Phy 201
assignment #008
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Physics II
07-01-2008
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10:18:06
query explain the convergence or divergence of series (no summary needed)
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A series converges if as the limit b -> inf. int[f(x)dx b, a] is a finite number.
Otherwise it diverges when b grows without bound.
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10:23:00
explain how we know that the integral from 1 to infinity of 1 / x^p converges if p > 1, diverges for p < 1, diverges for p = 1
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I do not truly understand the converge/diverge thing. I was wondering if, other than practice, there was an easier way to describe or learn this part of chapter seven.
I suggest that you work through the series of questions I sent you. By the time this is posted you might already have done so.
** 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]. **
** 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. **
The numbers don't tell you much. The key is the antiderivative. If p < 1 then the antiderivative is a positive power of x and hence, since positive powers approach infinity as x approaches infinity, the integral diverges. If p > 1 then the antiderivative is a negative power of x and, since negative power approach 0 as x approaches infinity, the integral converges.
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10:23:06
explain how we know that the integral from 0 to 1 of 1 / x^p diverges if p > 1, converges for p < 1, diverges for p = 1
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** If p > 1 then the antiderivative is a negative-power function, which approaches infinity as x -> 0. So the limit as b -> 0 of the integral from b to 1 of 1 / x^p would be infinite.
If p < 1 the antiderivative is a positive-power function which approaches zero as x -> 0. So the limit as b -> 0 of the integral from b to 1 of 1 / x^p would be divergent.
These integrals are the basis for many comparison tests. **
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10:23:09
explain how we know that the integral from 0 to infinity of e^(-a x) converges for a > 0.
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** You don't want to try compare the behavior of e^(-a x) with that of a y = 1 / x^p function. Both of these functions provide a basis for comparison with other functions, but comparing these functions themselves is difficult.
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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10:38:45
query problem 7.8.18 integral of 1 / (`theta^2+1) from 1 to infinity
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I rewrote the function to read int[1 / (sqrt(theta^2) + sqrt(1)) = int[ 1 / (theta + 1) = int[ (1 / theta) + 1]
My guess is that it diverges, but I am not sure if my work proves it.
The integral of 1 / theta is the integral of 1 / theta^1, and by the p test it diverges. However your otherwise good attempt doesn't quite work because of a common algebra error. sqrt(a+b) is not the same thing as sqrt(a) + sqrt(b). For example sqrt(16 + 9) = sqrt(25), whereas sqrt(16)+sqrt(9) = 4 + 3 = 7.
** 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 1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we will prove 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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10:38:50
does the integral converge or diverge, and why?
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10:39:58
If you have not already stated it, with what convergent or divergent integral did you compare the given integral?
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I compared it with int(1 / x^p) from 1 -> inf.
converges for p > 1
diverges for p <= 1
For which specific value of p did you make the comparison, and what is the specific connection between your integral the the p test?
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10:48:57
query problem 7.8.19 (3d edition #20) convergence of integral from 0 to 1 of 1 / `sqrt(`theta^3 + `theta)
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The book had the integral from 2 -> inf, but I'll answer both.
I rewrote the integral to read int[ 1 / (theta^(3/2) + 1)]
Again right idea, but exponentiation does not distribute over addition. sqrt(theta^3 + theta) is not sqrt(theta^3) + sqrt(theta).
I then rewrote it again to read from 2 -> inf int(1 / theta^(3/2) + int(1)
From 2 -> inf the integral converges because (3/2) > 1
From 0 -> 1 the integral diverges because (3/2) > 1
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.
So the integral converges by the p test.
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10:49:02
does the integral converge or diverge, and why?
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10:51:41
If you have not already stated it, with what convergent or divergent integral did you compare the given integral?
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I compared it with from 1 to inf int(1 / x^p) dx converges for p > 1 and diverges for p <= 1
and I compared it with from 0 to 1 int(1 / x^p) dx converges for p < 1 and diverges for p >= 1
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11:52:30
Query problem 8.1.5. Riemann sum and integral inside x^2 + y^2 = 10 within 1st quadrant, using horizontal strip of width `dy.
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I first solved for the Area using A = pi * r * r, where sqrt(10) = r to find A = pi * 10, but since it is the first quadrant, A = ('pi * 10) / 4
Area of the strip = w * 'dy
I used the Example #3 on page 370 to help with this problem.
I drew a triangle underneath the strip using the bottom of the strip as the top of the triangle. So the triangle has a height of h, a length of w, and a hypoteneuse of sqrt(10). Using the Pythagorean theorem I solved for w.
h^2 + w^2 = sqrt(10) ^ 2
w = sqrt(10) - sqrt(y^2)
Area of the 1st Quadrant approx. R-sum (w * 'dy) = R-sum(sqrt(10) - sqrt(y^2) * 'dy)
sqrt(10-y^2) is not the same thing as sqrt(10) - sqrt(y^2).
Taking the limit as 'dy -> 0 gives the integral: int[sqrt(10) - sqrt(y^2)]
I used the chart in the back of the book, Part VI, number 30, to find int[sqrt(10 - y^2)] = (1/2)y(sqrt(10 - y^2) + 10int(1 / (sqrt(10 - y^2)))
int(1 / (sqrt(10 - y^2))) = int(1/sqrt(10)) - int(1/sqrt(y^2)
= 2sqrt(10) - ln(y)
so
(1/2) * y * sqrt(10) - y + 10 (2 * sqrt(10) - ln(y))
My interval was from 0 -> sqrt(10) so I used L(sqrt(10)) - L(0) to find the answer, but it did not work out
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 Ό 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 Ό the area of the circle x^2 + y^2 = 10.
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11:52:34
Give the Riemann sum and the definite integral it approaches.
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11:52:37
Give the exact value of your integral.
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12:02:16
Query problem 8.1.12. Half disk radius 7 m thickness 10 m, `dy slice parallel to rectangular base.
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I could not figure out a way to find the Riemann sums of these slices because it is like a cylinder on its side, instead of sitting on its base. It just confused me. I did although figure a volume for the cylinder to be 'pi * 7^2 * 10 = 1539.3804
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12:02:20
Give the Riemann sum and the definite integral it approaches.
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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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12:02:24
Give the exact value of your integral.
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12:19:42
query problem 8.2.11 arc length x^(3/2) from 0 to 2
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Arc length = from a to b int[sqrt(1 + (3/2)x^(1/2)^2)]
= int[sqrt(1) + sqrt((3/2)x)]
=int(1) + int((3/2) x^ (1/2))
= x + (3/2)(3/2) * x * (2/3)
= x + (3/2)x
Using the fundamental theorem F(2) - F(0) = 5
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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12:19:45
what is the arc length?
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12:19:49
What integral do you evaluate obtain the arc length?
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12:29:32
What is the approximate arc length of a section corresponding to an increment `dx near coordinate x on the x axis?
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I am not really sure what this is asking me to do. I know to fidn the approx. arc length, but I am not sure how.
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12:32:58
What is the slope of the graph near the graph point with x coordinate x?
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I always thought the slope of a graph at a point on the graph was the derivative of the graph at that point.
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12:37:58
How is this slope related to the approximate arc length of the section?
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If the approx. arc length is the R-sum [sqrt (1 + (f'(x))^2)], and the slope of a graph at any point is f'(x) then they have some relation.
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13:01:56
query problem 8.2.31 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
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I am not real sure how to get the volume since I do not know hhow tall the object is. I did however draw out a good graph of the base, with the correct bounds.
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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13:02:01
what is the volume of the region?
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13:02:04
What integral did you evaluate to get the volume?
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13:10:02
What is the cross-sectional area of a slice perpendicular to the x axis at coordinate x?
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I am not really exactly positive on what a cross section is. From my understanding it is a slice of the solid that we are trying to find the volume of.
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13:10:02
What is the cross-sectional area of a slice perpendicular to the x axis at coordinate x?
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13:10:39
What is the approximate volume of a thin slice of width `dx at coordinate x?
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13:10:43
How the you obtain the integral from the expression for the volume of the thin slice?
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13:18:36
Query Add comments on any surprises or insights you experienced as a result of this assignment.
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Alot of this assignment was difficult and surprising. I felt like I was doing some of the text problems correctly, but I never got the same answer as in the back of the book. I will practice more, but I am looking forward to your corrections. Usually your currections help me more than the book.
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In several places you made the error of treating exponentiation as if it distributes over addition.
It is not the case that (a + b)^c = a^c + b^c. There are some combinations of distinct, specific values a, b and c for which this happens to be true, but it's almost never so.
If c = .5 this tells us that sqrt(a + b) is not equal to sqrt(a) + sqrt(b), an error you made more than once.
This error is the primary reason you didn't get the correct answers on some problems.
Your approach to the problems was generally not bad; you mainly got bogged down in the details.