Year 13 Maths Help Thread

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    (Original post by Palette)
    Consider I=\int^b_a f(x) dx, b>a. How can one intuitively explain why \int^a_b f(x) dx=-I?
    I just imagine measuring the area from b to a, which would make the difference between the two negative as you're moving backwards (so to speak), and the area would be multiplied by a negative, hence the negative I.

    Feels weird to explain it like this...

    I think you'd have to refer back to what integration actually does by considering infinitesimal changes on the rectangles which approximate the area, and the width each rectangle is a negative infinitesimal while height is positive. Hope that makes sense.
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    (Original post by Palette)
    Consider I=\int^b_a f(x) dx, b>a. How can one intuitively explain why \int^a_b f(x) dx=-I?
    We have, by FToC:

    \displaystyle 

\begin{equation*}I = \int_a^b f(x) \, \mathrm{d}x= F(x)\bigg|_{a}^{b} = F(b) - F(a) \end{equation*}

    whilst

    \displaystyle 

\begin{equation*}\int_b^a f(x) \, \mathrm{d}x = F(x)\bigg|_b^a =  F(a) - F(b) = -(F(b) - F(a)) = -I \end{equation*}

    See, definite integration is only defined for b \geq a so at some, we need to define that flipping the limits gives us the negative of the integral. One justification for this is that we'd like the integral identity:

    \displaystyle 

\begin{equation*} \int_a^b f(x) \mathrm{d}x + \int_b^c f(x) \mathrm{d}x = \int_a^c f(x) \mathrmm{d}x \end{equation*}

    to hold for arbitrary a,b,c so taking a=c gives us

    \displaystyle

\begin{equation*}\int_a^b f(x) \mathrm{d}x + \int_b^a f(x) \mathrm{d}x = \int_a^a f(x) \mathrm{d}x = 0 \end{equation*}
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    (Original post by Palette)
    Consider I=\int^b_a f(x) dx, b>a. How can one intuitively explain why \int^a_b f(x) dx=-I?
    Well F(b)>F(a) so F(b)-F(a) is positive and F(a)-F(b) is negative
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    (Original post by metrize)
    Well F(b)>F(a) so F(b)-F(a) is positive and F(a)-F(b) is negative
    Huh? Says who?

    Take f(x) = \sin x then F(x) = -\cos x so \int_{\pi}^{2\pi} f(x) = F(2\pi) - F(\pi) = -1 - 1 = -2  < 0. So how is F(b) - F(a) positive?
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    (Original post by Zacken)
    Huh? Says who?

    Take f(x) = \sin x then F(x) = -\cos x so \int_{\pi}^{2\pi} f(x) = F(2\pi) - F(\pi) = -1 - 1 = -2  < 0. So how is F(b) - F(a) positive?
    Well I should have said "if" F(b)>F(a), my bad and then I should have said the vice versa for when F(a)>F(b)


    Thanks I didn't consider that
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    Name:  Image.jpeg
Views: 62
Size:  201.3 KB stuck on first part.
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    (Original post by Coolsul98)
    stuck on first part.
    A bit awkward to explain.

    We know the k part of the vector because the info tells us the top of the roof is 5m above the base.

    You can work out the i part of the vector by considering the lengths of 14m and 6m.

    You can work out j part by considering the triangle OCD and the length along the j direction.
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    A BMO 2 question some years ago asked:

    Find positive integer solutions to \sqrt{a}+\sqrt{b}=\sqrt{2009}.

    I happen to be good at detecting stuff to do with square numbers (writing down lists of square numbers was a hobby of mine when I was younger) so I took a shortcut by writing \sqrt{2009}=\sqrt{2025-16}=\sqrt{45^2-4^2}=\sqrt{(45+4)(45-4)}=\sqrt{(49)(41)}=7\sqrt{41}.

    As 41 is prime we know that \sqrt{41} can't be simplified further so we can deduce that both \sqrt{a} and \sqrt{b} is in the form m\sqrt{41}, n\sqrt{41} respectively, m,n \in \mathbb{N}. Hence m+n=7 which simplifies the problem a lot.

    But I don't know if this is 'cheating' (I,e. missing what the examiner was testing) or not.
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    (Original post by Palette)
    A BMO 2 question some years ago asked:

    Find positive integer solutions to \sqrt{a}+\sqrt{b}=\sqrt{2009}.

    I happen to be good at detecting stuff to do with square numbers (writing down lists of square numbers was a hobby of mine when I was younger) so I took a shortcut by writing \sqrt{2009}=\sqrt{2025-16}=\sqrt{45^2-4^2}=\sqrt{(45+4)(45-4)}=\sqrt{(49)(41)}=7\sqrt{41}.

    As 41 is prime we know that \sqrt{41} can't be simplified further so we can deduce that both \sqrt{a} and \sqrt{b} is in the form m\sqrt{41}, n\sqrt{41} respectively, m,n \in \mathbb{N}. Hence m+n=7 which simplifies the problem a lot.

    But I don't know if this is 'cheating' (I,e. missing what the examiner was testing) or not.
    I wouldn't say so. I think you're expected to know your squares and make use of them wherever appropriate in a BMO so that's a valid method.
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    (Original post by Palette)
    But I don't know if this is 'cheating' (I,e. missing what the examiner was testing) or not.
    It isn't. The only other thing you could do is the proof that a and b have to be of the form root(41)*k, but I suspect that isn't even necessary.
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    (Original post by ValerieKR)
    It isn't. The only other thing you could do is the proof that a and b have to be of the form root(41)*k, but I suspect that isn't even necessary.
    what's the alternative way to do that question?
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    (Original post by metrize)
    what's the alternative way to do that question?
    What I said was in he's hand-wavily said that a and b have to have a factor of 41^(2n+1), which is a true thing to say but some of the marks may require that you prove that (although I suspect they won't).

    (As in the proof that ksqrt(a) + msqrt(b) can't equal lsqrt(c) with none of a, b, c being equal.

    I can't think of an alternative way to do it off the top of my head.
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    (Original post by ValerieKR)
    What part of it are you weak at? Making sense of breaking down the recurrence system? If so 15, 17 and 35 of advanced problems in mathematics are good

    http://www.mathshelper.co.uk/110501_...athematics.pdf
    A shame that he took down his website as it was a useful bank of STEP and AEA papers.
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    (Original post by Palette)
    A shame that he took down his website as it was a useful bank of STEP and AEA papers.
    Hi I see you posting quite a bit of these types of questions, are you going to study Maths at uni?
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    (Original post by metrize)
    Hi I see you posting quite a bit of these types of questions, are you going to study Maths at uni?
    I am undergoing mathematical training via solving various questions from different papers so that I can make the transition to maths at university as smooth as possible. Some of the questions are merely a product of my own curiosity.
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    (Original post by Palette)
    I am undergoing mathematical training via solving various questions from different papers so that I can make the transition to maths at university as smooth as possible. Some of the questions are merely a product of my own curiosity.
    Ah cool, what unis you thinking of applying to?
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    (Original post by Palette)
    Liverpool John Moores, Glyndwr, Southampton Solent, London South Bank and Swansea Metropolitan.

    Ok, it's actually Oxford (probably won't get in), Imperial (uncertain), UCL (hopefully yes), Bristol (hopefully yes) and Manchester (ditto).

    P.S. I think Glyndwr has a circumflex accent on top of the w.
    I see, good luck!
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    (Original post by metrize)
    I see, good luck!
    Where are you applying to?
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    (Original post by Palette)
    Where are you applying to?
    Cambridge Bristol Bath Imperial Loughborough for engineering
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    (Original post by metrize)
    Cambridge Bristol Bath Imperial Loughborough for engineering
    Engineering sounds very cut-throat- I believe it's the most competitive course at Cambridge.

    Is it mechanical engineering you're interested in?
 
 
 
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