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    \displaystyle\int_1^2\ -4x^5} = \left[ \frac{ -4x^6}{6}\right]_1^2

    just wondering what the next step would be

    i guess it would be -x^6 on top of the next fraction, but i dunno about on the bottom?
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    What...?

    Do you want to evaluate the integral because you are effectively there...
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    If you have \displaystyle \left[ \text{stuff} \right]^a_b then you do \text{stuff(a) - stuff(b)}.

    So it's \dfrac{-4}{6} \times 2^6 - (\dfrac{-4}{6} \times 1^6).
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    (Original post by Jimny)
    What...?

    Do you want to evaluate the integral because you are effectively there...
    complete the following integrations, it says

    i'm a maths noob :p: :o:
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    (Original post by robinson999)
    complete the following integrations, it says

    i'm a maths noob :p: :o:
    Ah. Well, that general rule is:

    \displaystyle \int x^n = \frac{x^{n+1}}{n+1} (for n not equal to 1).

    In your case n = 5, so insert this into the formula, remembering the multiplier of 4 and then use my post above.

    Maths repulses me.
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    i think i get it, i don't think my lecture notes help at all
    can't see where the other numbers came from in the example they use

    \dfrac{-4}{6} \times 2^6 - (\dfrac{-4}{6} \times 1^6)

    would they just come out as -42?
    or 64
    i'll take -42 as the post below says :p:
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    yep
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    (Original post by Glutamic Acid)
    Ah. Well, that general rule is:

    \displaystyle \int x^n = \frac{x^{n+1}}{n+1} (for n not equal to 1).

    In your case n = 5, so insert this into the formula, remembering the multiplier of 4 and then use my post above.

    Maths repulses me.
    GA do you know how to derive it? I know it should involve Riemann sum notation.
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    (Original post by Simplicity)
    GA do you know how to derive it? I know it should involve Riemann sum notation.
    I can do it using inequalities (there was an old question on find x^5; it could be easily generalized for x^n), not sure about Riemann sum notation, though. [Actually, the aforementioned question is a definite integral between 0 and 1, but I think that's fixable.]
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    (Original post by Glutamic Acid)
    (for n not equal to 1)
    *stabs*

    n not equal to -1.
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    (Original post by generalebriety)
    *stabs*

    n not equal to -1.
    Oh, oops. :o:

    (+1)
 
 
 
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