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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    How does the cube root of 2root2 = root 2 could someone show me how
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    (Original post by Mihael_Keehl)
    How does the cube root of 2root2 = root 2 could someone show me how
    \displaystyle \sqrt[3]{2\sqrt{2}}= \sqrt[3]{\sqrt{2^{2}\cdot 2}}=\sqrt[3]{\sqrt{2^{3}}}=\sqrt[3]{2^{\frac{3}{2}}}=\left ( 2^{\frac{3}{2}} \right )^{\frac{1}{3}} = 2^{\frac{3}{2}\times \frac{1}{3}} = 2^{\frac{1}{2}} = \sqrt{2}
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    (Original post by Azzer11)
    How do you integrate xe^x^2? It was in the c4 heinemann exercise book.
    By inspection you have f'x.e^fx so (1/2)(e^x^2) + c


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    (Original post by aymanzayedmannan)
    \displaystyle \sqrt[3]{2\sqrt{2}}= \sqrt[3]{\sqrt{2^{2}\cdot 2}}=\sqrt[3]{\sqrt{2^{3}}}=\sqrt[3]{2^{\frac{3}{2}}}=\left ( 2^{\frac{3}{2}} \right )^{\frac{1}{3}} = 2^{\frac{3}{2}\times \frac{1}{3}} = 2^{\frac{1}{2}} = \sqrt{2}
    Thank you, can I just ask how you know that 2root2 is the same as root 2^3, Like I get it is the root of 8 but do you know this by preactice, or is this a fact we are supposed to know etc. thanks
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    (Original post by Mihael_Keehl)
    Thank you, can I just ask how you know that 2root2 is the same as root 2^3, Like I get it is the root of 8 but do you know this by preactice, or is this a fact we are supposed to know etc. thanks
    It's just simple indices. You can write 2 as sqrt(4) and that makes things a lot easier.
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    (Original post by aymanzayedmannan)
    It's just simple indices. You can write 2 as sqrt(4) and that makes things a lot easier.
    I see lol, thanks.
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    (Original post by XxKingSniprxX)
    I would have thought you would use parts.
    (Original post by Azzer11)
    Oh yeah that'd probably work, I didn't think to use that because it was in one of the earlier exercises before parts.
    Parts will lead you to a dead end. This is by the form \displaystyle \int {f}'\left ( x \right )f\left ( x \right ) \mathrm{d}x, as BBeyond said.

    You know that \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x^{2}} \right ) = 2xe^{x^{2}}, so how does that help you find your integral?
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    (Original post by BBeyond)
    By inspection you have f'x.e^fx so (1/2)(e^x^2) + c


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    (Original post by BBeyond)
    By inspection you have f'x.e^fx so (1/2)(e^x^2) + c


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    (Original post by aymanzayedmannan)
    Parts will lead you to a dead end. This is by the form \displaystyle \int {f}'\left ( x \right )f\left ( x \right ) \mathrm{d}x, as BBeyond said.

    You know that \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x^{2}} \right ) = 2xe^{x^{2}}, so how does that help you find your integral?
    Oh yeah I've got it, thank you!
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    Has everyone finished c4?
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    (Original post by zaini101)
    Has everyone finished c4?
    No we still have some integration to do, so that means I can't do any past papers over Easter

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    (Original post by Mihael_Keehl)
    Thank you, can I just ask how you know that 2root2 is the same as root 2^3, Like I get it is the root of 8 but do you know this by preactice, or is this a fact we are supposed to know etc. thanks
    Just use the fact that for all non-negative a, b we have \sqrt{ab} = \sqrt{a}\sqrt{b}, so \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4}\sqrt{2} = 2\sqrt{2}.

    i.e: if you're given a number in surd form, you want to break it down into two factors, one of which is a square number. So, for example: \sqrt{80} = \sqrt{4 \times 20} = \sqrt{4}\sqrt{20} = 2\sqrt{20} = 2\sqrt{5 \times 4} = 2\sqrt{4}\sqrt{5} = 2\times 2 \times \sqrt{5} = 4\sqrt{5}.
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    (Original post by Zacken)
    Just use the fact that for all non-negative a, b we have \sqrt{ab} = \sqrt{a}\sqrt{b}, so \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4}\sqrt{2} = 2\sqrt{2}.

    i.e: if you're given a number in surd form, you want to break it down into two factors, one of which is a square number. So, for example: \sqrt{80} = \sqrt{4 \times 20} = \sqrt{4}\sqrt{20} = 2\sqrt{20} = 2\sqrt{5 \times 4} = 2\sqrt{4}\sqrt{5} = 2\times 2 \times \sqrt{5} = 4\sqrt{5}.
    thnx for this.
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    (Original post by zaini101)
    Has everyone finished c4?
    We have but I missed so much of it that I'm practically gonna have to self-teach myself everything over Easter..
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    (Original post by Azzer11)
    Oh yeah I've got it, thank you!
    Or by using the u = x^2 substitution

    1/2 du = x dx
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    Can anyone explain how to do part (b) of this question?
    I tried it but couldn't get the right answer, and can't seem to follow the mark scheme
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    (Original post by Louiseelg0rt)
    Can anyone explain how to do part (b) of this question?
    I tried it but couldn't get the right answer, and can't seem to follow the mark scheme
    There's probably a general derivative, I dunno if it's in formula booklet or not, but you can find the derivative of such expressions by expressing it as e^log(something)
    edit: Actually probs more natural to take log base 2 of each side then turn into ln and derive but whatever, just use logs in some way and it'll probably work out..
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    (Original post by Louiseelg0rt)
    Can anyone explain how to do part (b) of this question?
    I tried it but couldn't get the right answer, and can't seem to follow the mark scheme
    Let u=x^2, and use the chain rule, using the earlier result.
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    (Original post by Louiseelg0rt)
    Can anyone explain how to do part (b) of this question?
    I tried it but couldn't get the right answer, and can't seem to follow the mark scheme
    Use a substitution and part i.



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    physicsmaths EricPiphany 13 1 20 8 42


    Thanks for the help! I tried it and got it
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    (Original post by zaini101)
    Has everyone finished c4?
    Don't worry if you haven't and others have, there's still plenty of time until the exam! Loads of cramming to do
    Best of luck!!!
 
 
 
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