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

    17
    ReputationRep:
    (Original post by kiiten)
    Then i subbed in 3 and 1 separately then took them away to get 25? :dontknow:

    So 14 x 3 - 1/3 × 3^2 + 9 × 3^-1 for the first term (3)
    for 3 it should be  \displaystyle 14(3) - \frac{3^3}{3} + 9 \times 3^{-1}
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by DylanJ42)
    for 3 it should be  \displaystyle 14(3) - \frac{3^3}{3} + 9 \times 3^{-1}
    Thanks, careless mistake :sigh:

    Posted from TSR Mobile
    Offline

    17
    ReputationRep:
    (Original post by kiiten)
    Thanks, careless mistake :sigh:

    Posted from TSR Mobile
    dw we all make them
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by DylanJ42)
    dw we all make them
    Wait, that was a typo. I still get 36 then 11 when i sub in 1? (Answer: 25) :s

    Posted from TSR Mobile
    Offline

    4
    ReputationRep:
    (Original post by kiiten)
    Wait, that was a typo. I still get 36 then 11 when i sub in 1? (Answer: 25) :s

    Posted from TSR Mobile
    What seems to be the problem? Sorry I wasn't part of the conversation so I don't know what's going on!

    Also: it would be good if you could post your working out, that way we may be able to pinpoint where you're making mistakes!
    Furthermore, I sometimes find it helpful that when a question is getting out of hand and messy, start it again on a new piece of paper

    This is where I think we've gotten up to so far:

    So you need to find \displaystyle \int^{3}_{1} 14-x^{2}-9x^{-2} \ \mathrm{d}x

    Your integration is correct, we now have \displaystyle \int_1^3 14-x^{2}-9x^{-2} \ \mathrm{d}x = \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right]_1^3

    First you substitute in the limits (3 and 1), then you subtract what you get for the lower limit from what you get with the upper limit, so you have

    =\ \left[14(3)-\dfrac{3^{3}}{3}+\dfrac{9}{3}\ \right] - \left[14-\dfrac{1}{3}+9\right]

    Which is the area enclosed between x= 1 and x= 3 :

    When x=3\ \ \rightarrow\ \ \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right] = 36

    This is correct, however:

    When x=1\ \ \rightarrow\ \ \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right] \not= 11

    As \left[ 14 + 9 - \dfrac{1}{3} \right] = \dfrac{68}{3}
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by edothero)
    What seems to be the problem? Sorry I wasn't part of the conversation so I don't know what's going on!

    Also: it would be good if you could post your working out, that way we may be able to pinpoint where you're making mistakes!
    Furthermore, I sometimes find it helpful that when a question is getting out of hand and messy, start it again on a new piece of paper

    This is where I think we've gotten up to so far:

    So you need to find \displaystyle \int^{3}_{1} 14-x^{2}-9x^{-2} \ \mathrm{d}x

    Your integration is correct, we now have \displaystyle \int_1^3 14-x^{2}-9x^{-2} \ \mathrm{d}x = \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right]_1^3

    First you substitute in the limits (3 and 1), then you subtract what you get for the lower limit from what you get with the upper limit, so you have

    =\ \left[14(3)-\dfrac{3^{3}}{3}+\dfrac{9}{3}\ \right] - \left[14-\dfrac{1}{3}+9\right]

    Which is the area enclosed between x= 1 and x= 3 :

    When x=3\ \ \rightarrow\ \ \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right] = 36

    This is correct, however:

    When x=1\ \ \rightarrow\ \ \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right] \not= 11

    As \left[ 14 + 9 - \dfrac{1}{3} \right] = \dfrac{68}{3}
    Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
    Offline

    4
    ReputationRep:
    (Original post by kiiten)
    Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
    Sorry I'm really busy right now. Look at the graph. Area under the curve from x= 1 to 3 is a rectangle + curve.
    Zacken can help you here. Sorry again.
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
    As edothehero said, the area you've found using integration is the area under the curve. But you want the shaded area so you want to subtract the area of that rectangle.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    As edothehero said, the area you've found using integration is the area under the curve. But you want the shaded area so you want to subtract the area of that rectangle.
    As i said before you have to minus 8 (4x2) from 40/3 to get 5 1/3. But, where does the 2 come from?

    Posted from TSR Mobile
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    As i said before you have to minus 8 (4x2) from 40/3 to get 5 1/3. But, where does the 2 come from?

    Posted from TSR Mobile
    What's the area of the rectangle? Width * height, no...?
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    What's the area of the rectangle? Width * height, no...?
    Height is 4 and i dont know about the width - maybe 3? The question says that the shaded area has to equal 5 1/3 so i did 40/3 - 5 1/3 to get 8. Thats why i thought the width might be 2?
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    Height is 4 and i dont know about the width - maybe 3? The question says that the shaded area has to equal 5 1/3 so i did 40/3 - 5 1/3 to get 8. Thats why i thought the width might be 2?
    You're integrating from 1 to 3, the rectangle has width 3-1 = 2.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    You're integrating from 1 to 3, the rectangle has width 3-1 = 2.
    Oh, i must have got confused from the diagram because the curve goes inwards at the top so i thought it would be less. But, i see what you mean. Thanks
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    Oh, i must have got confused from the diagram because the curve goes inwards at the top so i thought it would be less. But, i see what you mean. Thanks
    The curve is, the rectangle isn't.
 
 
 
  • 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
    Would you like to hibernate through the winter months?
    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

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