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    (Original post by [email protected])
    What year is that?
    Jan 2012
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    (Original post by tingirl)
    How do you know when to use integration by substitution? Is there a rule or do you just use it when you don't think any other method would work?

    e.g. how would you know to use substitution for this: (or is there another method that works for this Q?) Attachment 425789
    not sure how to do all the funky symbols but i know the method.

    So u = 3x-2. New boundaries are 4 and 1 as they were 2 and 1, but putting them into u will change them.

    du/dx = 3 .... so rearranging that gives you
    du * 1/3 = dx

    so that leaves, (u)^-0.5 dx
    Integrating that gives us:
    2u^0.5. Now times the dx which is 1/3.
    So it's
    ((2u^0.5)/3). Now enter in the boundaries which are 4 and 1.
    To get: 4/3 - 2/3 which leaves us with 2/3.

    if you need more help explaining just say which bit you're stuck on.
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    (Original post by iAmanze)
    http://www.mei.org.uk/files/papers/2011_Jan_c3.pdf

    9(ii) Don't quite understand how to get g(x).
    You don't need g(x) for 9(ii).
    g(x) is there for 9(iii)
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    (Original post by messupp)
    You don't need g(x) for 9(ii).
    g(x) is there for 9(iii)
    I meant (iii) sorry :P
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    (Original post by iAmanze)
    I meant (iii) sorry :P
    Ok here's question 9iii...

    Substitute 0 into f(x) and it gets you 1.
    as 1/cos(0)^2 = 1
    So f(x) crosses at (0,1)

    now substitute n/4 into f(x) and it gets you 2.
    as 1/cos(n/4)^2 = 2.
    And now you divide it by 2 as it's g(x) = 0.5f(x+1/4π).
    So that proves that g(x) is also (0,1)


    Therefore they both cross at (0,1)
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    if you had a graph that has shifted or enlarged how would you go about finding the area given that youve worked out the area in the previous question
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    (Original post by Supergirlxxxxxx)
    Oh that makes so much sense thankyou!! Also I don't quite understand when it asked me to intergrate that curve between certain limits, why does that represent the area above the curve? I think when you integrate your finding the area below the curve?



    Posted from TSR Mobile
    Yes that's right, you're finding the area UNDER the curve when you integrate. if you equate the equations, then bring it all to 1 side, then integrate, you're then finding the area enclosed between the line and the curve (This is not expected normally), normally examiners make it easy for you by drawing shapes that you should know the area of.
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    (Original post by docdoc02)
    if you had a graph that has shifted or enlarged how would you go about finding the area given that youve worked out the area in the previous question
    If it has been stretched in both directions, multiply the scale factors of the stretches together, and then multiply that by the area before.

    For translations it depends on the specific translation, but usually you have to work out the area of a rectangle and add/subtract this to the area before.
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    (Original post by lizard54142)
    If it has been stretched in both directions, multiply the scale factors of the stretches together, and then multiply that by the area before.

    For translations it depends on the specific translation, but usually you have to work out the area of a rectangle and add/subtract this to the area before.
    thank you!
    how about this question last part which deals with translation
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    (Original post by docdoc02)
    thank you!
    how about this question last part which deals with translation
    This is the rectangle thing I was talking about! The curve has been shifted to the right by \pi/4, and up by 1. So what is the rectangle you have to add on?
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    People don't do C3 without a graphical calculator do they? How would you do C4?
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    (Original post by lizard54142)
    This is the rectangle thing I was talking about! The curve has been shifted to the right by \pi/4, and up by 1. So what is the rectangle you have to add on?
    oh so would the triangle be 1x 0.25pi and you need to add that on to the previous answer?
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    (Original post by docdoc02)
    oh so would the triangle be 1x 0.25pi and you need to add that on to the previous answer?
    Yes but it's a rectangle
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    (Original post by lizard54142)
    Yes but it's a rectangle
    haha omgsh
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    (Original post by docdoc02)
    haha omgsh
    :laugh: :banghead:
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    (Original post by AlecRobertson)
    People don't do C3 without a graphical calculator do they? How would you do C4?
    I don't use one for either
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    i think C3 will be slightly harder as last year C2 was veryyyy easy (i found C1 a bit harder than C2)
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    look at them asking to differentiate arctan
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    So I kinda screwed up the coursework, meaning i'm going to have to get 70-71/72 to get 90%. Anyone know whether the 90% in C3 and C4 is needed for both exams? Or is taken as an average?

    Also, can I redo my coursework without redoing my exams? If they go well that is.
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    anyone got a list of what each transformation means

    i.e. f(ax) is a strethc in the y direction by a?????
 
 
 
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