MATH1251 Questions HELP (2 Viewers)

InteGrand

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Use double integration to find the area of the region bounded by y=x^3 and y=x^2

Just set up the integral please


(I wrote like "x = " and "y = " in the bounds for the integrals, but you don't need to write them, I just put them in because it may help with visualisation/understanding and keeping track of which variable is integrated when etc.)
 
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InteGrand

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Note that using these ideas, we could also set up and evaluate the integral in the other order (dx dy). It would be a good exercise to set it up that way too, to practise setting these integrals up. Oftentimes for multiple integrals, setting it up in some order makes the integral easy to evaluate, whereas in another order it may be extremely difficult or impossible to.
 
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leehuan

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I guess what I never understood (probably cause the tutorial problems are out of order) is that why are we integrating the unit constant function and not say 2. Why is it specifically the correct integrand?
 

seanieg89

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In a sense that double integral (technically that is an iterated integral, but Tonelli's/Fubini's tells us this is the same thing as the corresponding double integral) is how we define area, at least until one encounters measure theory.

Notice that this definition of area coincides with the usual definition of the area of a rectangle.

Indeed this is all that matters, because Riemann integration of function in R^n amounts to partitioning the domain into tiny boxes and forming a Riemann sum over this partition. In this case, with integrand 1, the contribution from each of the sub-boxes is simply the n-dimensional volume of that box.
 

Paradoxica

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Any ideas? I've tried comparison but to no avail. Certain methods also run into the issue of failing if and when all the an are less than 1.
 

leehuan

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To the guys in this course reminder that there's a review session next Thursday 3-6
 
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leehuan

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Ok I have a bit of confusion here. X is a matrix here, but they also call it an "invertible eigenvector".

My brain's getting lost in the words. Can an eigenvector be a matrix?
 

InteGrand

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Ok I have a bit of confusion here. X is a matrix here, but they also call it an "invertible eigenvector".

My brain's getting lost in the words. Can an eigenvector be a matrix?
An eigenvector is going to be some element from the domain vector space of the linear map (which are generally called "vectors", even if they may be matrices, polynomials, functions, whatever). So in this case, an eigenvector is a matrix, since the domain vector space is a space of matrices (so the vectors here are matrices. Remember the term "vector" in the general setting of vector spaces doesn't only refer to arrows or n-tuples.).

 
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InteGrand

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Ok I have a bit of confusion here. X is a matrix here, but they also call it an "invertible eigenvector".

My brain's getting lost in the words. Can an eigenvector be a matrix?
Here is a solution in spoilers:

 
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leehuan

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Trying to switch the order of integration. Please tell me that their answer is wrong.



Cause I think there's a split at y=1
 

Paradoxica

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Trying to switch the order of integration. Please tell me that their answer is wrong.



Cause I think there's a split at y=1
Indeed. The corresponding regions of integration don't represent identical bounds in the x-y plane, so the equality you have given above is false.
 

leehuan

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The double integral is equal to the iterated integral which is also equal to the product of the integrals...
Yeah, I think it was something I had seen before but why is it the product of the integrals again
 

leehuan

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<-2, 0, 1>, <1, 1, 0> and <-1, 1, -2> are eigenvectors of



with eigenvalues -1, -1 and 5 respectively.

 

seanieg89

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<-2, 0, 1>, <1, 1, 0> and <-1, 1, -2> are eigenvectors of



with eigenvalues -1, -1 and 5 respectively.

I won't do the calculation for you but here is the idea:

The -1-eigenspace and the 5-eigenspace are orthogonal. (This can be seen directly by computing dot products or as a consequence of A being symmetric.)

So being given these three linearly independent eigenvectors v1,v2,v3, we need to find an orthonormal basis of eigenvectors u1,u2,u3.

Normalise the third one, (so u3=v3/|v3|).

Normalise the first one to obtain u1.

To obtain u2, it is probably quickest to just take the cross product u2=u1xu3, because this will certainly satisfy the desired properties. (Note that althought u3 is unique up to sign, our choice of u1 and u2 are far from unique, any orthonormal basis of the 2-dimensional -1-eigenspace will suffice.)

Then the matrix Q with columns (u1|u2|u3) satisfies AQ=Q*diag(-1,-1,5) as required and is orthogonal by construction.

(Try to make sure you have a clear picture of the geometry of the situation here, as that is more important than the actual mechanical manipulations of diagonalisation.)
 
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leehuan

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This obviously converges but the ratio test is inconclusive, so how would you justify it

 

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