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course Mth 277
4/13 2
Question: `q001. Evaluate the double integral Int( Int((x^2* e^xy) dy), 0, x) dx, 0 , 1)YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Int((x^2* e^xy) dy) = x^2*(e^(yx)/x) = xe^(xy), evaluated at limits of integral
xe^(x^2) - x
Int(xe^(x^2) - x dx) = Int(xe^(x^2) dx) - Int(x dx) = ((1/2)e^(x^2) - (x^2/2), at limit
[((1/2)e^(1^2) - (1^2/2)] - [((1/2)e^(0^2) - (0^2/2)]
=e/2 - (1/2) - (1/2) = (1/2)e - 1
= approx. .36
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Given Solution: First note, and be sure you understand, that an antiderivative of e^(x y) with respect to y is 1/x e^(x y). Then it should be clear that an antiderivative of x^2 e^(x y) with respect to y is x^2 * 1/x e^(x y) = x e^(x y).
The change in this antiderivative from y = 0 to y = x is x e^(x^2) - x.
We now integrate this expression from x = 0 to x = 1. An antiderivative of this expression is 1/2 e^(x^2) - x^2 / 2. Between x = 0 and x = 1 this changes by (1/2 e - 1/2) - (1/2) = 1/2 e - 1.
[Our antiderivative of x e^(x^2) is obtained by the substitution u = x^2 do that du = 2 x dx and x dx = du / 2, giving us int( 1/2 e^u du ) = 1/2 e^u. ]
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Self-critique (if necessary):
I it’s weird I was going to ask you if it was possible to when we have Int(xe^(x^2) - x dx), if we could use u sub for x^2 as stated then would we replace xe^(x^2) - x, with (1/2)e^(u) - (1/2). I guess that is not the right approach and because of some mathematical property it works but you should always separate it up and solve to make sure you have correct solution , right?
@& That substitution works only for x e^(x^2). It doesn't apply to integrating x (which is straightforward to integrate with respect to x).
In any case you always need to check to be sure the derivative of your antiderivative is equal to your integrand.*@
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Question: `q002. Integrate 4x over D with respect to A using a double integral where D is the region bounded by y = 4 - x^2, y = 3x, and x = 0.
y = 4 - x^2 is a parabola opening downward with vertex at (0, 4). It intersects the x axis when y = 0, so the x coordinates of the intersection are the solutions to the equation 0 = 4 - x^2. We easily conclude that the parabola intersects the x axis at (-2, 0) and at (2, 0). At this point we are confined to the part of the parabola above the x axis, which is the part lying between x coordinates -2 and 2.
y = 3 x intersects y = 4 - x^2 when 3 x = 4 - x^2, which occurs when x^2 + 3 x - 4 = 0. The equation is easily solved by factoring or by the quadratic formula, giving us x = -4 and x = 1. The x = -4 point lies outside our x interval [-2, 2] but the x = 1 point lies within it. At the x = 1 point our y value is 3, as seen by substituting x = 1 into either the quadratic or the linear function.
You should be able to plot the bounding functions, one of which is linear and the other quadratic, without the aid of a graphing calculator. The graph looks like this:
The region bounded by our functions can be described by x cross sections or y cross sections.
In terms of x cross sections:
-2 <= x <= 0, 0 <= y <= 4 - x^2 AND 0 <= x <= 1, 3x <= x <= 4-x^2.
In terms of y cross sections, we first note that if y = 4 - x^2 then x = +- sqrt(4 - y), and if y = 3 x then x = 1/3 y. Then our cross sections are easy to describe:
0 <= y <= 3, -sqrt(4 - y) <= x <= 1/3 y AND 3 <= y <= 4, -sqrt(4 - y) <= x <= sqrt( 4 - y).
We integrate 4x over the given region. Our result could be
int ( int ( 4 x dy), 0 , 4-x^2) dx, -2, 0 ) + int ( int ( 4 x dy), 3x, 4-x^2) dx, 0, 1 )
int ( int ( 4 x dx, -sqrt(4-y), y/3) , 0, 3 ) +int ( int ( 4 x dx, -sqrt(4-y),sqrt(4-y)) dy, 3, 4 )
In symbols:
Either way, the integral of 4 x over the region is -13. Integration is straightforward.
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Your solution:
I believe the solution is given with the question, which is good because I had some trouble seeing what’s going on but even with the solution I having a hard time seeing what’s going on.
Having trouble seeing what area I’m taking the integral of. Seems the easy part is taking int just finding the boundaries is causing me some trouble.
O.k. well its been a couple of hrs and I’ve thinking about it and I missed something that should have been VERY obvious. If we are trying to calculate and Integral of any type we are trying to find I specific area under a given curve. So we would need to find a partition of the graph that is isolated from the entire rest of the graph. that would be the are you were describing in this problem. That is defienitly a huge down fall to online courses not being able to understand and struggling to see things that should be very clear.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Question: `q003. Compute the following integral with the given order of integration and with the order changed: Int[ Int(4x^3 dy, 0, sqrt(x)) dx, 0, 4].
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Your solution:
Int[ Int(4x^3 dy, 0, sqrt(x)) dx, 0, 4]
For this Integral we type I we have a fn described with a top curve = `sqrt(x)and bottom curve at y=0 which runs along x axis and has ends at x= 0 and x=2 .
So our first integral is
Int( Int(4x^3 dy, 0, sqrt(x)) dx, 0, 4)
Int(4x^3 dy) = 4yx^3, evaluated at limits on integral
4*`sqrt(x)*x^3 - 0= 4x^(7/2)
Int(4x^(3/4)dx, 0, 4)
Int(4x^(7/2)dx) = (8/9)x^(9/2), evaluated at limits on integral
(8/9)*4^(9/2)- 0 = 4096/9 = approx. 455.1111
With the order changed we have integral of type II and fn has curve x = y^2 and and x = 0 is a restricting boundry and y is y=0 and y=2
Int( Int(4x^3 dx, 0,y^2) dy, 0, 2)
Int(4x^3 dx) = x^4, evaluated at limits on integral
y^6 - 0= y^6
@& x^4 evaluated at x = y^2 would be x^8, not x^6.*@
Int(y^6 dy, 0, 2)
Int(4x^(3/4)dx) = y^(7)/7, evaluated at limits on integral
(2^(7)/7)- 0 = approx. 18.2857
confidence rating #$&*:
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Given Solution:
The region is 0 <= x <= 4, 0 <= y <= sqrt(x).
This region was used in one of the q_a_ problems for section 12.1.
The bounding curve is y = sqrt(x), which can also be expressed as x = y^2, y > 0.
x values from 0 to 4 result in y values from sqrt(0) = 0 to sqrt(4) = 2.
y cross-sections run from the curve to the line x = 4, so our region can be described by
0 <= y <= 2, y^2 <= x <= 4.
Our integral is therefore
int( int( 4 x^3, x, y^2, 4), y, 0, 2) = 4096 / 9.
The two integrals would be written in standard form as
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Self-critique (if necessary):
I am having a hard time finding my boundaries on my integrals. Even when I’m looking at the pic of my graph I am having a hard time seeing what limits need to be to my area I’m calculating the integral wrt is causing me problems. Also the original fn we are using to take the integral of is simply what we take integral and no matter what as long as we need our boundaries straight we don’t have to think about what’s going on with this fn apart from finding the right antiderivative of it?
@& Right.*@
Everything was done right except I messed up the boundaries when changing the order and taking the integral wrt x first.
@& x runs from y^2 to 4, while y runs from 0 to 2.*@
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Question: `q004. Give two different ways to set up the integral of the area of the region D where D is the region in the first quadrant of the xy-plane bounded by y = 4/x^2 and y = 5 - x^2. Evaluate one of these integrals.
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Your solution:
With y = 4/x^2 and y = 5-x^2, in the 1st Quad intersects are (1,4) and (2,1) so we could use the double integral
????There would be no init fn to take integral right, because our area to calculate is not given wrt a fn??
@& When integrating to find the area, you are just integrating the function 1.*@
Int(Int( dy, 4/x^2 , 5 - x^2 ) dx, 1,2)
Int( dy ) = y, evaluating at limits of integral
5 - x^2 - 4/x^2
Int(5 - x^2 - 4/x^2 dx, 1,2)
Int(5 - x^2 - 4/x^2 dx) = 5x - x^3/3 + 4/x, evaluating at limits of integral
[10 - (8/3) + 2] - [5 - (1/3) + 4] = 2/3
Second integral is found by boundries wrt x, so x = `sqrt(5 - y)and x = `sqrt(4/y) because our curves are defined in x so we have type II wrt y is held constant and y runs on the interval from 1 to 4 on the curve. So we have the integral
Int(Int( dx, `sqrt(5 - y), `sqrt(4/y)) dy,1,4)
which would give us the same solution right, assuming I did the first double integral right
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Given Solution:
You should easily and without the aid of a graphing calculator be able to sketch reasonable graphs of these functions. You will see that they intersect in two points of the first quadrant.
To find the coordinates of these points:
The curves y = 4 / x^2 and y = 5 - x^2 intersect at the x coordinate such that 4 / x^2 = 5 - x^2.
Solving the equation:
4 / x^2 = 5 - x^2. Multiply both sides by x^2 to get
4 = 5 x^2 - x^4. Rearrange to get
x^4 - 5x^2 + 4 = 0. Let u = x^2 to get
u^2 - 5 u + 4 = 0. Factor to get
u = 4 or u = 1. Replace u with x^2 to get
x^2 = 4 or x = 1, which leads to solutions
x = 1 and x = 2.
The intersection points are (1, 4) and (2, 1)
Between x = 1 and x = 2 the parabola is the higher of the two curves.
The region could be described as
1 <= x <= 2
4/x^2 <= y <= 5 - x^2
with resulting area integral
int ( int ( dy, 4/x^2, 5-x^2) dx, 1, 2)
Alternatively we first note that y = 4 / x^2 and y = 5 - x^2 are equivalent in the first quadrant to the respective curves x = sqrt(y) / 2 and x = sqrt( 5 - y). y values between our two intersection points run from 1 to 4. So our region is
1 <= y <= 4
sqrt(y) / 2 <= x <= sqrt( 5 - y )
with resulting area integral
int ( int ( dx, 2/sqrt(y), sqrt( 5 - y) ), y, 1, 4)
Both integrals yield the same result, 2/3, which is therefore the area of the region.
The integrals in standard form, and the graph of the two function, are depicted below.
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Self-critique (if necessary):
Well on this problem I finally started to feel like I understand the process , after spending 4 hrs completing this Query and reviewing notes/book. I will probably go back and complete the qa on this section as well to get a little extra practice in.
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@& Good.
It might be worth working through the q_a_ and/or looking at the class notes, before reading the section and doing the query.*@
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Self-critique (if necessary):
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Self-critique rating:
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@& Good.
It might be worth working through the q_a_ and/or looking at the class notes, before reading the section and doing the query.*@
#*&!
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Self-critique (if necessary):
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Self-critique rating:
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@& Good.
It might be worth working through the q_a_ and/or looking at the class notes, before reading the section and doing the query.*@
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