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    Integrate the cube root of x (edited)

    I think it's the same as x^(3/2) and then you add one to the power making it 5/2 so it becomes x^(5/2)/5/2?

    The markscheme says its 3/4(x)^4^/^3, how?
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    (Original post by Magenta96)
    Integrate 3(root)x.

    I think it's the same as x^(3/2) and then you add one to the power making it 5/2 so it becomes x^(5/2)/5/2?

    The markscheme says its 3/4(x)^4^/^3, how?
    Are you integrating 3\sqrt{x} or, as the answer suggests ^3\sqrt{x}
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    i think you mean "integrate the cube root of x"...

    well this is x1/3 so add 1 to the power and divide by the new power... and add c
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    (Original post by the bear)
    i think you mean "integrate the cube root of x"...

    well this is x1/3 so add 1 to the power and divide by the new power... and add c
    oh right yeah it so I get up to this part:
    x^4^/^3 divided by 4/3, how do they manage to get 3/4 as the co-efficient?
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    Reverse power rule!

     \int x^n  dx = \frac {x^{n + 1}}{n + 1} + C

    They get 3/4 as the coefficient as it's been switched around from division to multiplication.
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    (Original post by Frankster)
    Are you talking about: 3x^1^/^2 ?

    Therefore, add one to the power: 3/2. And also divide 3 by 3/2 = 2

    so u get 2x^3^/^2 + c
    I've edited the q, it's actually asking to integrate the cube root of x
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    (Original post by Magenta96)
    oh right yeah it so I get up to this part:
    x^4^/^3 divided by 4/3, how do they manage to get 3/4 as the co-efficient?
    if you divide by a fraction you multiply by its reciprocal
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    (Original post by the bear)
    if you divide by a fraction you multiply by its reciprocal
    ohhh right, thanks I forgot all about that
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    Ok, so to integrate, you add one to the power and divide by the new power.







    Because



    Get it?
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    (Original post by Banny Dyrne)
    Ok, so to integrate, you add one to the power and divide by the new power.







    Because



    Get it?
    yes, thank you
 
 
 
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