Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    How do you find area under a curve. I know you have to integrate, but i forgot what you do

    The equation I have is: y=x^2

    it says:

    1. Find the area from x=0 to x=1
    2. Find the area from y=0 to y=1


    What do I need to do?
    Offline

    13
    Integrate.
    Offline

    15
    ReputationRep:
    ok intergrate it, then do [1 substituded for x in intergrated equation]-[0]

    we've just started covering this as were doing s1 and m1 first but i'm guessing for the y question you have to intergrate to get x=y something?

    someone plz explain rep available
    Offline

    14
    ReputationRep:
    Calculate the integral with the limits of the integral being 0 to 1 in the first case.

    In the second case, you also integrate using 0 to 1 as limits but you need to integrate the function that gives you x from y so if y=f(x), you're looking for the function g such that x=g(y) and you integrate that between 0 and 1
    Offline

    12
    ReputationRep:
    Integrate it, and the solve it for 1 and 0, and subtract.
    Offline

    15
    ReputationRep:
    (Original post by SamTheMan)
    Calculate the integral with the limits of the integral being 0 to 1 in the first case.

    In the second case, you also integrate using 0 to 1 as limits but you need to integrate the function that gives you x from y so if y=f(x), you're looking for the function g such that x=g(y) and you integrate that between 0 and 1

    so the second one you'd intergate x=rooty right?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Ekpyrotic)
    Integrate.
    I know i said that, but i forgot what you do

    for 1. I get:

    \displaystyle\int^1_0 x^2 \, dx

    = \frac{x^3}{3} from o to 1

    = \frac{1}{3}
    • Thread Starter
    Offline

    2
    ReputationRep:
    What do you do for the y?
    Offline

    13
    Idiotic post.
    Offline

    18
    ReputationRep:
    You're trying to find

    \int^1_0 \sqrt{y} \ \mathrm{d}y for the next part.
    Offline

    15
    ReputationRep:
    (Original post by Ekpyrotic)
    Integrate for x then flip.

    do you intergrate for x then flip or flip and theb intergrate.

    or are they the same thing?
    Offline

    13
    (Original post by Bateman)
    do you intergrate for x then flip or flip and theb intergrate.

    or are they the same thing?
    Correct idea is to ignore me, I was diff'ing.
    • Thread Starter
    Offline

    2
    ReputationRep:
    \displaystyle\int^1_0 y^\frac{1}{2}, dy

    = \frac{2}{3} \times y^\frac{3}{2} from o to 1

    = \frac{2}{3}

    ??
    Offline

    13
    The above is n. right.
    Offline

    14
    ReputationRep:
    (Original post by Bateman)
    so the second one you'd intergate x=rooty right?
    I concur. I hadn't actually seen that y=x^2 so indeed x=sqrt(y) and that's what you integrate between 0 and 1. I believe that gives you 2/3. (there's a chance I'm wrong about the result)
    Offline

    18
    ReputationRep:
    (Original post by G O D I V A)
    \displaystyle\int^1_0 y^\frac{1}{2}, dy

    = \frac{2}{3} \times y^\frac{3}{2} from o to 1

    = \frac{2}{3}

    ??
    Now sketch the curve and see why that's not the whole answer.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Swayum)
    Now sketch the curve and see why that's not the whole answer.

    huh?
    Offline

    18
    ReputationRep:
    Did you sketch y = x^2? What area do you think your integral worked out? Hint: x = -rooty is perfectly valid also.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Swayum)
    Did you sketch y = x^2? What area do you think your integral worked out? Hint: x = -rooty is perfectly valid also.
    I did sketch the graph, but i still dont get that
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (Original post by G O D I V A)
    I did sketch the graph, but i still dont get that
    As you've stated it, the question appears to want the area 'inside' the parabola from y = 0 to y = 1 (imagine pouring water into it up to the level of y = 1). But all you've done is integrate root y between 0 and 1, which will give you the area between the curve x = root y (the right-hand half of the parabola) and the y-axis. Shade this bit in on your diagram. What else do you need to find in order to work out the required area?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 9, 2008
The home of Results and Clearing

2,973

people online now

1,567,000

students helped last year

University open days

  1. Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 22 Aug '18
  2. University of Buckingham
    Postgraduate Open Evening Postgraduate
    Thu, 23 Aug '18
  3. University of Glasgow
    All Subjects Undergraduate
    Tue, 28 Aug '18
Poll
How are you feeling about GCSE results day?
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.