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    (Original post by physicsmaths)
    Lol.
    Hahahahaha


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    From your experience, which is harder: BMO 1 geometry or STEP I geometry?
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    (Original post by Palette)
    From your experience, which is harder: BMO 1 geometry or STEP I geometry?
    BMO1
    STEP doesn't have a huge amount of geometry tbh
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    Can someone check if this is right so far...

    NOTE: sorry the pic is the wrong way around it wouldnt let me rotate it. Ill take it properly next time

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    (Original post by kiiten)
    Can someone check if this is right so far...


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    Yes it is
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    (Original post by MathMoFarah)
    Yes it is
    then

    1/2 (2) - ln2 - 1/2
    1 - ln2 - 1/2
    -ln2 + 1/2

    ?

    NOTE: im asked to prove the area is ln2 - 1/2
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    (Original post by kiiten)
    then

    1/2 (2) - ln2 - 1/2
    1 - ln2 - 1/2

    ?

    NOTE: im asked to prove the area is ln2 - 1/2
    Well your working looks right, and what you've got is 1/2-ln2
    Are you sure the question in the book isn't the integral of 2-e^2x or something?
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    (Original post by MathMoFarah)
    Well your working looks right, and what you've got is 1/2-ln2
    Are you sure the question in the book isn't the integral of 2-e^2x or something?
    Sometimes the book gives the wrong answer. So is mine right?

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    (Original post by kiiten)
    Sometimes the book gives the wrong answer. So is mine right?

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    It's that it's the area of the shaded region - the region is all below the x axis so comes out as it's negative
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    (Original post by kiiten)
    Sometimes the book gives the wrong answer. So is mine right?

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    The area is below the x-axis therefore you will get the negative area. Mult. by -1 to switch signs and make it positive.
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    (Original post by RDKGames)
    The area is below the x-axis therefore you will get the negative area. Mult. by -1 to switch signs and make it positive.
    Do you mean ive worked out the negative area or the area is supposed to be negative
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    (Original post by kiiten)
    Do you mean ive worked out the negative area or the area is supposed to be negative
    You've evaluated the integral, but as there is a net area below the x axis, the integral is negative, but the actual area itself cannot be negative.
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    (Original post by ValerieKR)
    x
    I have to admit that the questions you've given me for the recurrence relations are slightly beyond my ability at the moment.

    However, I did manage to do the question on the Reuleaux heptagon which is also in the booklet (along with a number of other questions from various past STEP papers).
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    (Original post by kiiten)
    Do you mean ive worked out the negative area or the area is supposed to be negative
    Negative area doesn't make sense. From C1 you should know that if you evaluate any function f(x) between x=a and x=b, and get a negative value, then f is negative for a<x<b. Either that, or there is more negative area than positive area in this interval

    For example, if you were to find the area of \sin(x) between 0 and 2\pi then you would say \displaystyle \int_0^{2\pi} \sin(x) .dx

    You would find it to be 0, where obviously this is not the case. This is because the two areas cancel each other out in the interval 0 \leq x \leq 2\pi .
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    (Original post by RDKGames)
    Negative area doesn't make sense. From C1 you should know that if you evaluate any function f(x) between x=a and x=b, and get a negative value, then f is negative for a<x<b. Either that, or there is more negative area than positive area in this interval

    For example, if you were to find the area of \sin(x) between 0 and 2\pi then you would say \displaystyle \int_0^{2\pi} \sin(x) .dx

    You would find it to be 0, where obviously this is not the case. This is because the two areas cancel each other out in the interval 0 \leq x \leq 2\pi .
    So its always positive unless the area underneath the x axis is greater than the area above the x-axis?

    If its negative (like the question i posted) you multiply by -1 to make it positive?
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    (Original post by Palette)
    From your experience, which is harder: BMO 1 geometry or STEP I geometry?
    You cant compare the geometry as its different areas (euclid vs coordinate) but in terms of problem solving difficulty BMO1 for sure but even then BMO1 geometry isnt hard anyway.
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    (Original post by kiiten)
    So its always positive unless the area underneath the x axis is greater than the area above the x-axis?
    Correct.

    (Original post by kiiten)
    If its negative (like the question i posted) you multiply by -1 to make it positive?
    Not always. Only if your function is purely negative between your two limits, which is the exact case with your question.

    Otherwise, you can have situations where the function begins positive at one end of the interval, and is negative at the other end. The integral evaluated between these two boundaries would give you the net area, not the overall one, so you'd have some positive area plus some negative area which is not giving you the total area. Of course, this is something you'd investigate further before proceeding with on different integrals concerning area.
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    Have i gone wrong in simplifying this?
    The answer is supposed to be in the form a + bln2

    I got it as 28 + ln 16???

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    (Original post by kiiten)
    Have i gone wrong in simplifying this?
    The answer is supposed to be in the form a + bln2

    I got it as 28 + ln 16???

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    Ques attached (c)

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    (Original post by kiiten)
    Have i gone wrong in simplifying this?
    The answer is supposed to be in the form a + bln2

    I got it as 28 + ln 16???

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    Correct working but you need to turn 4 and 8 into bases of 2 with exponents. From there on in you should be fine.

    The ln(16) is incorrect.
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    (Original post by RDKGames)
    Correct working but you need to turn 4 and 8 into bases of 2 with exponents. From there on in you should be fine.

    The ln(16) is incorrect.
    does that make the ln part

    ln (2^12 / 2^18)

    ln (2^ -6) ?

    I think i misunderstood what you said - i still get ln16 ( ln(2^6 / 4) )
 
 
 
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