Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Is this integrable? [(-y^2)-(2y)+3]^(1/2)

    How would i go about doing this?
    Offline

    15
    \sqrt{-y^2 - 2y + 3}.

    Complete the square on -y^2 - 2y + 3, and then try making a suitable hyperbolic/trigonometric substitution. I've not tried it, but I think the following substitution will work:

    Spoiler:
    Show
    y + 1 = 2 \sin u

    (after obtaining -y^2 - 2y + 3 = 4 - (y + 1)^2)
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hey thanks for the reply, i used the subsitution y+1=2sinu (as u said) and i got and answer of
    (1/2)(y+1)[(4-(y+1)^2)] + sin^-1 [(y+1)/2]

    do you reckon it looks right?

    sorry if ur having trouble with reading my answer (im not good with that math typing software thing)
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by dopemann)
    Hey thanks for the reply, i used the subsitution y+1=2sinu (as u said) and i got and answer of
    (1/2)(y+1)[(4-(y+1)^2)] + sin^-1 [(y+1)/2]

    do you reckon it looks right?

    sorry if ur having trouble with reading my answer (im not good with that math typing software thing)
    I think it should be 2 sin^-1 etc.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Another question, if i was to put the limits of integration between y=1 and y=0 , would i get my answer in degrees? because sin^-1 [(y+1)/2] would give degrees if i put in y=1 or y=0?
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    No, you should get your answer in radians (which are dimensionless).
    • Thread Starter
    Offline

    0
    ReputationRep:
    Yeh sorry mr m, i forgot to write that
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    Ok, no problem, as the general suggested, when working with differentiation and integration, you should ALWAYS work in radian measure.
    • Thread Starter
    Offline

    0
    ReputationRep:
    are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by dopemann)
    are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
    The value of that integral with those limits is 1.23 to 3 sf.

    Imagine the area represented by that integral. The value of the function when y = 0 is \sqrt3 and its value when y = 1 is . If you calculate the area of a rectangle of height \sqrt3 and width 1 you will only obtain \sqrt3 as the answer and the area of the integral must clearly be less than this.
    • Thread Starter
    Offline

    0
    ReputationRep:
    how did you get that? that looks like the right answer but i dont understand how to get that...
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (Original post by dopemann)
    are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
    Is your calculator in degree mode?
    • Thread Starter
    Offline

    0
    ReputationRep:
    o i understand now... i must use radian mode when plugging in y=1 and y=0 because after all these are in radians (obviously - silly me). so yeh i get the answer 1.23.

    Thankyou everyone, appreciate the help
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    You are welcome.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 15, 2008
The home of Results and Clearing

2,816

people online now

1,567,000

students helped last year

University open days

  1. Sheffield Hallam University
    City Campus Undergraduate
    Tue, 21 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 22 Aug '18
  3. University of Buckingham
    Postgraduate Open Evening Postgraduate
    Thu, 23 Aug '18
Poll
A-level students - how do you feel about your results?
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.