Turn on thread page Beta
    • Thread Starter
    Offline

    17
    ReputationRep:
    Name:  hiya.GIF
Views: 13
Size:  32.3 KB

    honest to god, i think i've tried every flaming possible method to answering this question and each time i've gotten it wrong.

    Please can i have some help on this question?

    Methods i've tried

    ->Seperate integration i tried integrating each part seperatly and then adding the 2 areas together. Didnae work.

    ->Rearranging to get x because they're y coordinates, i tried rearranging the equation so i could sub in the y coordinates. that didn't work either?

    ->Using x i tried integrating with (1/2,0) using x coordinates but that didn't work either.

    Idk I'm really panicing that i can't answer this, it's due tommorow and i have a test, help needed thank you!!


    btw the answer should be (1/12 + 1/2e )
    • Community Assistant
    Online

    20
    ReputationRep:
    Community Assistant
    (Original post by ZiggyStardust_)

    honest to god, i think i've tried every flaming possible method to answering this question and each time i've gotten it wrong.

    Please can i have some help on this question?

    Methods i've tried

    ->Seperate integration i tried integrating each part seperatly and then adding the 2 areas together. Didnae work.

    ->Rearranging to get x because they're y coordinates, i tried rearranging the equation so i could sub in the y coordinates. that didn't work either?

    ->Using x i tried integrating with (1/2,0) using x coordinates but that didn't work either.

    Idk I'm really panicing that i can't answer this, it's due tommorow and i have a test, help needed thank you!!


    btw the answer should be (1/12 + 1/2e )
    Not sure why your first method didn't work unless you added the incorrect areas together. The correct answer is given by \displaystyle \int_{0}^{1/2} (1-2x)^5 .dx + \int_{0}^{1/2} -(e^{2x-1}-1) .dx

    Why the negative? Because the latter integral is negative as it's below the y-axis, but we're interested in the positive quantity of it since area is defined to be positive. - so the negative sign turns the negative quantity positive, before we add the two.

    The above doesn't yield the question's answer so I disagree with it, but that's the method.

    EDIT: I think you meant it as \frac{1}{12}+\frac{1}{2e} and not \frac{1}{12}+\frac{1}{2}e? In which case, the two answers are equivalent, you just need to rearrange it in the wanted form. Also means you should probably use some brackets if you don't want people misinterpreting what you write...
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by RDKGames)
    Not sure why your first method didn't work unless you added the incorrect areas together. The correct answer is given by \displaystyle \int_{0}^{1/2} (1-2x)^5 .dx + \int_{0}^{1/2} -(e^{2x-1}-1) .dx

    Why the negative? Because the latter integral is negative as it's below the y-axis, but we're interested in the positive quantity of it since area is defined to be positive. - so the negative sign turns the negative quantity positive, before we add the two.

    The above doesn't yield the question's answer so I disagree with it, but that's the method.

    EDIT: I think you meant it as \frac{1}{12}+\frac{1}{2e} and not \frac{1}{12}+\frac{1}{2}e? In which case, the two answers are equivalent, you just need to rearrange it in the wanted form. Also means you should probably use some brackets if you don't want people misinterpreting what you write...
    LOL so I forgot to integrate the curves. I was subbing in the coordinates into the original equation (silly). the amount of dumb mistakes i make in maths is laughable....

    this method makes plenty of sense though, i did it slightly differently tho, i used the y coordinates to integrate separately. anyhow, thank you very very much for providing the method, i think i might actually understand now. I appreciate it

    Also yes, i meant the second one,, that's the answer. Alright I'll use brackets better next time. thank you!!! )
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 8, 2017
Poll
Do you think parents should charge rent?
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.