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    How do you integrate:

    \displaystyle \int \frac{x^3+3x^\frac{1}{2}}{x} dx

    How do you remove the denominator? I know how to integrate but I don't know what to do with the x in the denominator.

    Also how would you integrate:

    \displaystyle \int_2^a 6x^{-4} dx where a is a constant greater than 2.

    What does 'where a is a constant greater than 2.' mean exactly? How do I manipulate the a?

    Thanks, sorry my maths is awful.
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    For the first one, split the fraction and use the laws of indices.

    For the second one, integrate it like you would normally and then just put a in instead of a numerical value for the upper limit. "a is a constant greater than 2" just means that a \ge 2... but that has no effect on this integral -- all it means is that your integral's not screwed up*; notice that if a<2 then your upper limit is lower than your lower limit.

    *In fact having an upper limit lower than your lower limit is fine, but if this is C2 it's not worth worrying about, which is probably why they imposed the condition a>2.
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    (Original post by nuodai)
    For the first one, split the fraction and use the laws of indices.

    For the second one, integrate it like you would normally and then just put a in instead of a numerical value for the upper limit. "a is a constant greater than 2" just means that a \ge 2... but that has no effect on this integral -- all it means is that your integral's not screwed up*; notice that if a<2 then your upper limit is lower than your lower limit.

    *In fact having an upper limit lower than your lower limit is fine, but if this is C2 it's not worth worrying about, which is probably why they imposed the condition a>2.
    \displaystyle \int \frac{x^3+3x^\frac{1}{2}}{x} dx= \displaystyle \int \frac{x^3}{x}+ \frac{3x^\frac{1}{2}}{x}dx=\frac  {1}{3}x +

    I'm not so sure on the last term.
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    (Original post by I'm clever)
    \displaystyle \int \frac{x^3+3x^\frac{1}{2}}{x} dx= \displaystyle \int \frac{x^3}{x}+ \frac{3x^\frac{1}{2}}{x}dx=\frac  {1}{3}x +

    I'm not so sure on the last term.
    You haven't applied the laws of indices; when you integrate \dfrac{x^3}{x} you don't get \dfrac{x}{3}, that's for sure. Simplify \dfrac{x^3}{x} and \dfrac{3x^{\frac{1}{2}}}{x} first.
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    (Original post by nuodai)
    You haven't applied the laws of indices; when you integrate \dfrac{x^3}{x} you don't get \dfrac{x}{3}, that's for sure. Simplify \dfrac{x^3}{x} and \dfrac{3x^{\frac{1}{2}}}{x} first.
    \dfrac{x^3}{x}=x^2

    \dfrac{3x^{\frac{1}{2}}}{x}=3 \times \frac{\sqrt x}{x}
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    (Original post by nuodai)
    You haven't applied the laws of indices; when you integrate \dfrac{x^3}{x} you don't get \dfrac{x}{3}, that's for sure. Simplify \dfrac{x^3}{x} and \dfrac{3x^{\frac{1}{2}}}{x} first.
    \dfrac{x^3}{x}=x^2

    \dfrac{3x^{\frac{1}{2}}}{x}=3 \times \frac{\sqrt x}{x}=\frac{3}{\sqrt x}
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    (Original post by I'm clever)
    \dfrac{x^3}{x}=x^2
    That's right. So what do you get when you integrate that? [Hint: not x/3]

    (Original post by I'm clever)
    \dfrac{3x^{\frac{1}{2}}}{x}=3 \times \frac{\sqrt x}{x}
    You can simplify \dfrac{\sqrt{x}}{x}. Notice that x = \sqrt{x} \times \sqrt{x}. But really you could have just used the laws of indices; you should know from C1 that \dfrac{x^a}{x^b} = x^{a-b}.

    EDIT: Just seen your updated reply; you're right with the second bit, but you're better off writing the power of x as a number so that it's easier to integrate.
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    (Original post by nuodai)
    That's right. So what do you get when you integrate that? [Hint: not x/3]



    You can simplify \dfrac{\sqrt{x}}{x}. Notice that x = \sqrt{x} \times \sqrt{x}. But really you could have just used the laws of indices; you should know from C1 that \dfrac{x^a}{x^b} = x^{a-b}.
    I'm not doing A level maths. I'm only in year 10 but I find integration interesting

    EDIT: I have simplified it further.
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    (Original post by I'm clever)
    I'm not doing A level maths. I'm only in year 10 but I find integration interesting

    EDIT: I have simplified it further.
    Okay, see my edit :p: I think the laws of indices is a GCSE thing anyway, but if you're in Y10 you might not have covered it yet... I don't know what the syllabus is these days.
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    (Original post by nuodai)
    Okay, see my edit :p: I think the laws of indices is a GCSE thing anyway, but if you're in Y10 you might not have covered it yet... I don't know what the syllabus is these days.
    The answer to the second question is \frac{a^3}{4} -2

    I can't seem to get that one too.
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    (Original post by I'm clever)
    The answer to the second question is \frac{a^3}{4} -2

    I can't seem to get that one too.
    That's definitely not the right answer to the second bit, assuming you've not made a typo. You should be aiming for \dfrac{1}{4} - \dfrac{2}{a^3}.
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    (Original post by nuodai)
    That's definitely not the right answer to the second bit, assuming you've not made a typo. You should be aiming for \dfrac{1}{4} - \dfrac{2}{a^3}.
    Yea I understand it all clearly now.

    Thanks!
 
 
 
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