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    (Original post by LegendX)
    See attachment
    ahhh I was getting stuck at differentiating xlna, but lna is just a constant which is why x -> 1 or rather, x disappears

    thanks very much
    (is that biology? lol)
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    (Original post by PythianLegume)
    Integration finds the total area, and area below the x-axis is considered 'negative area' therefore, if a graph has sections above and below the x-axis, you have to split it into integrals and take the modulus of the negative integral.
    Thank you

    I see how that would work for something like a sine graph (where it's easier to split the parts of the curve) but what about a graph where limits for both areas (above and below) are the same? How would we split it then?
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    (Original post by Fiyinad)
    Yh i get dat bit buh wah if cot3x came up in the exam? would it be 1/3ln[sin(3x)]?
    Yeh.
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    (Original post by MedMed12)
    its all about the MBE, OBEs at the moment
    Surely in a maths exam it's all about the ODE's?

    How's everyone feeling for this tomorrow?
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    (Original post by Supes180)
    If you swap limits by substitution and you end up with limits that have a different size order, do you still go from high to low/top to bottom? For example \int^5_3 \ dx turns into \int^2_4 \ du Is that right? Or should it be \int^4_2 \ du
    Keep them the first way.
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    (Original post by Frankster)
    y = a^x

    dy/dx = a^x lna

    y = a^x

    ln y = xlna

    1/y . dy/dx = ln a

    dy/dx = y ln a

    (From y = a^x )

    Therefore: dy/dx = a^x ln a
    can someone explain to me why the x goes in the bolded line?
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    (Original post by Mallika)
    Thank you

    I see how that would work for something like a sine graph (where it's easier to split the parts of the curve) but what about a graph where limits for both areas (above and below) are the same? How would we split it then?
    Well, if the graph is symmetrical above and below the x-axis, you double the answer - all you have found is the area above the x-axis (assuming you didn't get an answer of zero).
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    (Original post by Kreayshawn)
    can someone explain to me why the x goes in the bolded line?
    It's been differentiated.
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    (Original post by arcturus7)
    Surely in a maths exam it's all about the ODE's?

    How's everyone feeling for this tomorrow?
    haha good one!!
    Good I like c4 just so tired
    wbu?
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    n an experiment a scientist considered the loss of mass of a collection of picked leaves. The
    mass M grams of a single leaf was measured at times t days after the leaf was picked.
    The scientist attempted to find a relationship between M and t. In a preliminary model she
    assumed that the rate of loss of mass was proportional to the mass M grams of the leaf.
    (a) Write down a differential equation for the rate of change of mass of the leaf, using this
    model.
    (2)
    (b) Show, by differentiation, that M = 10(0.98)^t
    satisfies this differential equation.

    I am really stuck on part b, can anyone help?

    I've got this far: dm/dt=-kv
    For the diffrentiation part: dM/dt: 10x0.98^(t)xln(0.98)
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    Someone PLEASE post their solution to question 7c from paper E!!! Emailed my teacher and they havent replied -_-
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    (Original post by Tikara)
    ahhh I was getting stuck at differentiating xlna, but lna is just a constant which is why x -> 1 or rather, x disappears

    thanks very much
    (is that biology? lol)
    Yeah lol it's biology, I'm running out of paper so using whatever I can find lol haha.
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    Is everyone ready for C4, i can't wait for it to be over :/ It's boring doing the same maths again and again and again and again.....:/
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    An integral a day keeps the doctor away...
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    (Original post by Econ1994)
    Someone PLEASE post their solution to question 7c from paper E!!! Emailed my teacher and they havent replied -_-
    post the question and ill try and help i dont know what paper E is tho
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    Name:  C4.png
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Size:  77.6 KBName:  C4.png
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Size:  77.6 KB

    Can someone please help me with part b?
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    (Original post by PythianLegume)
    Well, if the graph is symmetrical above and below the x-axis, you double the answer - all you have found is the area above the x-axis (assuming you didn't get an answer of zero).
    Just one more question

    What if it wasn't symmetrical? Or would we never such a graph in C4? (There's a question like this in the mixed exercise for integration which is why I'm asking)
    Thank you again
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    I haven't done C3 for a year so can anybody think of anybody i should read through thats in c3 that could potentially come up to tomorrow, or something to just remind myself about.
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    (Original post by LegendX)
    Look at the attachment.
    For the last part, when we got our H value, n u subbed it back in, did u still use 120 as your dv over dt
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    Name:  ImageUploadedByStudent Room1371498992.588221.jpg
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    How do you integrate this?


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