Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    11
    ReputationRep:
    Q
    https://docs.google.com/viewer?a=v&q...3wzMVVFonZ1aMg
    MS
    https://docs.google.com/viewer?a=v&q...aCGtRKx9DaqGGg

    In question 8 I chose the route of saying area of S= area below straight line - area below curve . Then I worked out both the area below curve and area below the straight line. I then I'd the subtraction. Isn't this right? This is the textbook method that I understand. In the MS they did something different which I have never seen before in my C2 book.
    Offline

    14
    Your method is fine.
    Offline

    0
    ReputationRep:
    The method is good. The only difference with this is that they have taken the equation of the curve away from the equation of the line and integrated this equation to save you the hassle of finding two different areas.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Coakley95)
    The method is good. The only difference with this is that they have taken the equation of the curve away from the equation of the line and integrated this equation to save you the hassle of finding two different areas.
    But I cannot visually understand how their method works. But I can understand my method above. I want to visually understand their method.
    Offline

    0
    ReputationRep:
    What you do is integrate equation of both the curve and the line and use the definite integral to get to different values for area. Once you have done this you take the value for the area under the curve and take it away from the area under the line. What they have done is minus the equation of the curve from the equation of the line to give you one equation to integrate. Once you integrate this equation you have already taken away the area under the curve essentially so that when you use the definite integral you get the final answer. I hope this helps.
    Offline

    14
    (Original post by krisshP)
    But I cannot visually understand how their method works. But I can understand my method above. I want to visually understand their method.
    You've done

    \displaystyle\int_{-5}^{2} (3x + 20) \text{d}x - \displaystyle\int_{-5}^{2} (x^2 + 6x + 10) \text{d}x

    They've done

    \displaystyle\int_{-5}^{2} (3x + 20) - (x^2 + 6x + 10) \text{d}x
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by tommm)
    You've done

    \displaystyle\int_{-5}^{2} (3x + 20) \text{d}x - \displaystyle\int_{-5}^{2} (x^2 + 6x + 10) \text{d}x

    They've done

    \displaystyle\int_{-5}^{2} (3x + 20) - (x^2 + 6x + 10) \text{d}x
    So in a general way is this right
    Integral of A between limits c and v - integral of B between limit c and v

    = integral of (A-B) between limits c and v
    ?
    Thanks
    Offline

    14
    (Original post by krisshP)
    So in a general way is this right
    Integral of A between limits c and v - integral of B between limit c and v

    = integral of (A-B) between limits c and v
    ?
    Thanks
    Yep
    Offline

    2
    ReputationRep:
    Find the area of the area beneath the straight line and then integrate the curve between the same parameters. Take away the area from beneath the curve from the area beneath the straight line

    Posted from TSR Mobile
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by tommm)
    ....
    Sorry to bother you, but I was wondering if you could help me.
    http://homework.king-ed.suffolk.sch....attach_id=3458
    http://www.google.com/url?sa=t&rct=j...45368065,d.d2k

    See the last question part b. They keep on changing their mark scheme method for finding the area and it's so frustrating. I used the same method in the previous paper on this question and I ended up with the correct answer . But the method for this question is different. Are examiners just looking for a correct answer with some logical working out?

    Thanks
    Offline

    1
    ReputationRep:
    No it dosent matter if your meathod was to find area under both curves then minus them.As-long as you show working it should be fine.
    Offline

    16
    ReputationRep:
    (Original post by krisshP)
    Are examiners just looking for a correct answer with some logical working out?
    yes
    Offline

    14
    (Original post by krisshP)
    Sorry to bother you, but I was wondering if you could help me.
    http://homework.king-ed.suffolk.sch....attach_id=3458
    http://www.google.com/url?sa=t&rct=j...45368065,d.d2k

    See the last question part b. They keep on changing their mark scheme method for finding the area and it's so frustrating. I used the same method in the previous paper on this question and I ended up with the correct answer . But the method for this question is different. Are examiners just looking for a correct answer with some logical working out?

    Thanks
    If you've managed to get the right answer with a method that makes sense, you needn't worry.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by TenOfThem)
    yes

    (Original post by tommm)
    If you've managed to get the right answer with a method that makes sense, you needn't worry.
    Thank you
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.