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

    0
    ReputationRep:
    I know how to do it but I keep getting the wrong answer. Could someone please post what you get . y=2/x^2 +x. The points A and B have coordinates (0.5) and (2) , these are x-coordinates. Find the area between line AB and the curve. The answer is 27/8 and I got 10.25 ...
    Offline

    16
    ReputationRep:
    Show us your working and we can then see where you've made your mistake.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by SophieL1996)
    I know how to do it but I keep getting the wrong answer. Could someone please post what you get . y=2/x^2 +x. The points A and B have coordinates (0.5) and (2) , these are x-coordinates. Find the area between line AB and the curve. The answer is 27/8 and I got 10.25 ...
    What are the y coordinates of A and B? It is impossible to answer without them!
    • Thread Starter
    Offline

    0
    ReputationRep:
    Ok so I found the area by: area of trapezium - integration of curve between limits of 2 and 0.5 area of trapezium = 0.5 x 1.5 (height) x (a+b) . a= 8.5 and b= 2.5. I found this by using the y= equation and sub in x coordinates. this gave an area of 8.25 I intergrated the curve equation to -2x^-1+2x which is -2/x + 2x I subbed in 2 and 0.5 which gave 3-5= -2 8.25- (-2)= 10.25
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mr M)
    What are the y coordinates of A and B? It is impossible to answer without them!
    sub x values into equation.... so I got 8.5 and 2.5
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by SophieL1996)
    sub x values into equation.... so I got 8.5 and 2.5
    What makes you think they are points of intersection?
    Offline

    11
    ReputationRep:
    (Original post by SophieL1996)
    Ok so I found the area by: area of trapezium - integration of curve between limits of 2 and 0.5 area of trapezium = 0.5 x 1.5 (height) x (a+b) . a= 8.5 and b= 2.5. I found this by using the y= equation and sub in x coordinates. this gave an area of 8.25 I intergrated the curve equation to -2x^-1+2x which is -2/x + 2x I subbed in 2 and 0.5 which gave 3-5= -2 8.25- (-2)= 10.25
    (Original post by SophieL1996)
    y=2/x^2 +x.
    The integral of x is not 2x . . .
    • Thread Starter
    Offline

    0
    ReputationRep:
    it shows you that on the sketch.... and how else are you going to work it out? As they give you little information.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by joostan)
    The integral of x is not 2x . . .
    Ah yes thank you I seemed to have written it down wrong when working out as I wrote +2 which intergrated to 2x
    Offline

    2
    ReputationRep:
    Ive just tried it and got close (not got correct answer tho)
    =2x^-2+x
    =S -(2x^-1)/-1 + (x^2)/2
    when X=2 S=1
    and when X=0.5 S= -27/8
    Where to go from there if it is right
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    The answer of 27/8 is correct.
    Offline

    0
    ReputationRep:
    (Original post by SophieL1996)
    sub x values into equation.... so I got 8.5 and 2.5
    That is only true if the question states that A and B lie on the curve
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mr M)
    The answer of 27/8 is correct.
    No I intergrated what should be x to 2x as I wrote it wrong....
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by kvohra)
    That is only true if the question states that A and B lie on the curve
    Yes, the sketch shows that...
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by SophieL1996)
    it shows you that on the sketch.... and how else are you going to work it out? As they give you little information.
    When posting questions you need to give us all the information. It is frustrating to be expected to guess.
    Offline

    2
    ReputationRep:
    Boom I have it for ya!
    =2x^-2+x
    =S -(2x^-1)/-1 + (x^2)/2
    when X=2 S=1
    and when X=0.5 S= -27/8
    1-(-)27/8= 35/8
    Area of trapezium= 8.25
    8.25-35/8= 27/8
    Ill take a fiver for that
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mr M)
    When posting questions you need to give us all the information. It is frustrating to be expected to guess.
    I thought it was obvious when I said find the area between AB and the curve, sorry I should have made it clearer.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by calm down)
    Boom I have it for ya!
    =2x^-2+x
    =S -(2x^-1)/-1 + (x^2)/2
    when X=2 S=1
    and when X=0.5 S= -27/8
    1-(-)27/8= 35/8
    Area of trapezium= 8.25
    8.25-35/8= 27/8
    Ill take a fiver for that
    Haha thanks and yes you deserve it!
    Offline

    11
    ReputationRep:
    (Original post by calm down)
    Boom I have it for ya!
    =2x^-2+x
    =S -(2x^-1)/-1 + (x^2)/2
    when X=2 S=1
    and when X=0.5 S= -27/8
    1-(-)27/8= 35/8
    Area of trapezium= 8.25
    8.25-35/8= 27/8
    Ill take a fiver for that
    8.25 - 35/8 =/= 27/8 . . .
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    For future reference you can do it all in one step:

    \displaystyle \int_{0.5}^2 ((10.5-4x) - (2x^{-2}+x)) \, dx
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
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

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.