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    Suppose you had the following integral. The [ "(" represents the integral notation ]

    (dy = (3x dx

    I know that this just means to take the sum of a bunch of rectangles with dimensions dx by 3x(y). But what sense does it make, that when these rectangles are summed up, that they will equal 3*(x^2) )/ 2. ?

    Also, suppose you have [ the ( means the integral sign ]

    ( 6*y dy

    Why is it that when you have a dy then everything should be expressed in terms of y
    and when you have dx then everything should be expressed in terms of x?
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    Have you seen the definition of integration in terms of the supremum and infimum of the sets of different partitions of a given graph? If not, don't worry, http://en.wikipedia.org/wiki/Integral read that and post back if you don't understand still.

    With regards to the second part, the y is simply a dummy variable, it doesn't matter what you call it. It's the process that's important. You could say  \int_a^b 6x dx = \int_a^b 6y dy .
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    When you have something like \int6ydy then you're not asking about an area, you're just asking 'what function can be differentiated with respect to y to get 6y?'. It so happens that there are several functions that do this, for example 3y^2, 3y^2+11, 3y^2-8/3 etc. So for your answer, you give a general solution which is 3y^2+c.

    When you do integration with limits then you're finding an area. If you work out \int_a^bf(x)dx then the Fundamental Theorem of Calculus tells us that the answer is F(b)-F(a) where F(x) is a function that can be differentiated with respect to x to get f(x).
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    this question made me re-visit the basics of integration as the summation of increasingly thin rectangles. But I am getting stuck with it.

    given
    f(x)=3x
    The area of each rectangle of width \delta x between y=0 and the function is 3x\delta x, but I can't figure out how to get these to sum to \frac{3}{2}x^2 (which is the indefinite integral of 3x.

    As \delta x \rightarrow 0 why doesn't each rectangle reduce to a zero area? I realise that in the limit this is presumably counterbalanced by the fact that there is an infinite number of them, but how is this represented mathematically?
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    (Original post by Plato's Trousers)
    this question made me re-visit the basics of integration as the summation of increasingly thin rectangles. But I am getting stuck with it.

    given
    f(x)=3x
    The area of each rectangle of width \delta x between y=0 and the function is 3x\delta x, but I can't figure out how to get these to sum to \frac{3}{2}x^2 (which is the indefinite integral of 3x.

    As \delta x \rightarrow 0 why doesn't each rectangle reduce to a zero area? I realise that in the limit this is presumably counterbalanced by the fact that there is an infinite number of them, but how is this represented mathematically?
    You might want to work out \int_0^b3xdx. Imagine you're chopping up the interval [0,b] into n equally sized segments, so you have n rectangles with their bottom-right corners at b/n, 2b/n, 3b/n, ..., nb/n=b. Each rectangle has width b/n and heights 3b/n, 3*2b/n, etc.

    So to calculate the integral you would work out \sum_{r=1}^n 3(br/n) (b/n) and let n tend to infinity. So here you have each rectangle tending to zero area counterbalanced by the number of rectangles tending to infinity, which doesn't tell you much about what the overall limit is. However with some algebra you can rewrite this summation in a way that makes it more obvious what the limit is:

    \sum_{r=1}^n 3(br/n) (b/n) =3\frac{b^2}{n^2} \sum_{r=1}^nr = 3\frac{b^2}{n^2} \frac{1}{2}n(n+1) = \frac{3}{2}b^2\times\frac{n+1}{n  } = \frac{3}{2}b^2\times(1+\frac{1}{  n})\rightarrow \frac{3}{2}b^2 as n tends to infinity.
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    (Original post by ttoby)
    You might want to work out \int_0^b3xdx. Imagine you're chopping up the interval [0,b] into n equally sized segments, so you have n rectangles with their bottom-right corners at b/n, 2b/n, 3b/n, ..., nb/n=b. Each rectangle has width b/n and heights 3b/n, 3*2b/n, etc.

    So to calculate the integral you would work out \sum_{r=1}^n 3(br/n) (b/n) and let n tend to infinity. So here you have each rectangle tending to zero area counterbalanced by the number of rectangles tending to infinity, which doesn't tell you much about what the overall limit is. However with some algebra you can rewrite this summation in a way that makes it more obvious what the limit is:

    \sum_{r=1}^n 3(br/n) (b/n) =3\frac{b^2}{n^2} \sum_{r=1}^nr = 3\frac{b^2}{n^2} \frac{1}{2}n(n+1) = \frac{3}{2}b^2\times\frac{n+1}{n  } = \frac{3}{2}b^2\times(1+\frac{1}{  n})\rightarrow \frac{3}{2}b^2 as n tends to infinity.
    That's exactly what I was looking for. Thanks!
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    (Original post by ttoby)
    You might want to work out \int_0^b3xdx. Imagine you're chopping up the interval [0,b] into n equally sized segments, so you have n rectangles with their bottom-right corners at b/n, 2b/n, 3b/n, ..., nb/n=b. Each rectangle has width b/n and heights 3b/n, 3*2b/n, etc.

    So to calculate the integral you would work out \sum_{r=1}^n 3(br/n) (b/n) and let n tend to infinity. So here you have each rectangle tending to zero area counterbalanced by the number of rectangles tending to infinity, which doesn't tell you much about what the overall limit is. However with some algebra you can rewrite this summation in a way that makes it more obvious what the limit is:

    \sum_{r=1}^n 3(br/n) (b/n) =3\frac{b^2}{n^2} \sum_{r=1}^nr = 3\frac{b^2}{n^2} \frac{1}{2}n(n+1) = \frac{3}{2}b^2\times\frac{n+1}{n  } = \frac{3}{2}b^2\times(1+\frac{1}{  n})\rightarrow \frac{3}{2}b^2 as n tends to infinity.
    ( This all concerns the last line of the explanation )

    I understand everything till the part where the sigma sign disappears and the
    (1/2*n)*(n+1) comes in. How did that come to be?

    Also, when you multiplied (3br/n) by(b/n) , why did (3b^2/n^2) appear before the sigma sign and the r after the sigma sign?

    Sorry for the questions
    Either way, thank you for the detailed explanation
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    (Original post by chowderspin)
    ( This all concerns the last line of the explanation )

    I understand everything till the part where the sigma sign disappears and the
    (1/2*n)*(n+1) comes in. How did that come to be?

    Also, when you multiplied (3br/n) by(b/n) , why did (3b^2/n^2) appear before the sigma sign and the r after the sigma sign?

    Sorry for the questions
    Either way, thank you for the detailed explanation
    \sum_{r=1}^nr=\frac{1}{2}n(n+1) because of a standard summation identity for an arithmetic series with first term 1, common difference 1 and n terms. You can prove it by induction if you like or derive it from the formulas for an arithmetic series. I'll answer your other question in a few minutes.
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    (Original post by Plato's Trousers)
    That's exactly what I was looking for. Thanks!
    You could also generalise for any x^n but you would have to use a slightly different approach....
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    Ok I'm back (I just had to catch a bus)

    Yeah, the problem with my method is that it doesn't work very well for higher powers because the summation identities can get quite hard to work out. But it shows you the general idea.

    As for your second question, it's a bit like 'taking out a factor':

    \sum_{r=1}^n 3(br/n) (b/n) =  \sum_{r=1}^n3\frac{b^2}{n^2}r = 3\frac{b^2}{n^2} + 3\frac{b^2}{n^2}2 + 3\frac{b^2}{n^2}3 + 3\frac{b^2}{n^2}4 + \cdots +3\frac{b^2}{n^2}n = 3\frac{b^2}{n^2}(1+2+3+4+ \cdots +n) =3\frac{b^2}{n^2} \sum_{r=1}^nr

    The reason why you can't take out the r in a similar way is because r has a special meaning. The limits on the sigma tell you that it's being summed from 1 to n so if you took r out of the sigma then it would lose that meaning.
 
 
 
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