Turn on thread page Beta
    • Thread Starter
    Offline

    13
    ReputationRep:
    I'm a bit stuck on this integral:

    ∫[(x e^(-x)]^2

    As 0 → 2

    Thanks for any help.
    Offline

    8
    ReputationRep:
    Ok, you want the integral of [(x e^(-x)]^2 with the limits of from x=0 to x=2.

    I=∫ [(x e^(-x)]²
    I=∫ x².e^(-2x)

    Integrating by parts,

    I = -½.x².e^(-2x) + ∫ (x.e^(-2x)
    I = -½.x².e^(-2x) + {-½.x.e^(-2x) + ½ ∫ (e^(-2x))}
    I = -½.x².e^(-2x) - ½.x.e^(-2x) -¼.e(-2x)

    puting in the limits of x=0 and x=2, we get

    I = {(-½.4.e^(-4) - ½.2.e^(-4) - ¼.e^(-4)) - (0 - 0 - ¼)}
    I = {(-2 -1 - ¼) e^(-4) + ¼}
    I = ¼ - 3¼e^(-4)
    ===========
    Offline

    18
    ReputationRep:
    He's using pi-notation, doesn't that indicate a product?
    • Thread Starter
    Offline

    13
    ReputationRep:
    Amazing, I understand fully now. Thanks very much.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by ZJuwelH)
    He's using pi-notation, doesn't that indicate a product?
    Well, more of a constant. It doesn't really matter.
    Offline

    18
    ReputationRep:
    (Original post by 2776)
    Well, more of a constant. It doesn't really matter.
    And the answer is a constant, I see.
    Offline

    1
    ReputationRep:
    Integration by parts
    x^2 e^-2x
    u= x^2 du/dx= 2x
    v= -1/2 e ^-2x dv/dx= e^-2x
    x^2(-o.5e^(-2x))-the integral of (-0.5e^(-2x))(2x) dx
    Integrate the bold expression by parts:
    u= 2x du/dx=2
    v= 1/4e^(-2x) dv/dx=(-o.5e^(-2x))
    [-0.5 x^2e^(-2x)-0.5xe^(-2x)+1/4e^(-2x)] between 0 and 2
    final answer= pi(-11/4e^-4+1/4)
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by zazy)
    Integration by parts
    x^2 e^-2x
    u= x^2 du/dx= 2x
    v= -1/2 e ^-2x dv/dx= e^-2x
    x^2(-o.5e^(-2x))-the integral of (-0.5e^(-2x))(2x) dx
    Integrate the bold expression by parts:
    u= 2x du/dx=2
    v= 1/4e^(-2x) dv/dx=(-o.5e^(-2x))
    [-0.5 x^2e^(-2x)-0.5xe^(-2x)+1/4e^(-2x)] between 0 and 2
    final answer= pi(-11/4e^-4+1/4)
    You've gone wrong numerically I think. Its (-13/4)pi...not (-11/4)pi...
    Offline

    1
    ReputationRep:
    (Original post by 2776)
    You've gone wrong numerically I think. Its (-13/4)pi...not (-11/4)pi...
    Aha, so what's new,clumsy me...
    • Thread Starter
    Offline

    13
    ReputationRep:
    I'm stuck on another integration here:

    ∫(x sin3x)

    As 0 → (1/3)pi

    Unless I made an error, it doesn't matter whether I make (sin3x) u or dv/dx it seems that I am trapped by ever widening integrals.
    Offline

    18
    ReputationRep:
    (Original post by 2776)
    I'm stuck on another integration here:

    ∫(x sin3x)

    As 0 → (1/3)pi

    Unless I made an error, it doesn't matter whether I make (sin3x) u or dv/dx it seems that I am trapped by ever widening integrals.
    INT xsin3x

    Let u=x, dv/dx = sin3x, then du/dx=1 and v=-cos3x/3 so
    INT xsin3x = -xcos3x/3 + INT cos3x/3
    = xsin3x/3 + sin3x/9

    I think.
    Offline

    0
    ReputationRep:
    (Original post by 2776)
    I'm stuck on another integration here:

    ∫(x sin3x)

    As 0 → (1/3)pi

    Unless I made an error, it doesn't matter whether I make (sin3x) u or dv/dx it seems that I am trapped by ever widening integrals.
    If you set u = x, and dv/dx = sin3x, it seems to go straight to something you integrate for me. :/
    Offline

    1
    ReputationRep:
    (Original post by ZJuwelH)
    INT xsin3x

    Let u=x, dv/dx = sin3x, then du/dx=1 and v=-cos3x/3 so
    INT xsin3x = -xcos3x/3 + INT cos3x/3
    = xsin3x/3 + sin3x/9

    I think.
    That's what i get...(too lazy to type)
    the only differenc ebeing that my final answer is;
    -1/3x(cos3x)+1/9sin3x
    Offline

    0
    ReputationRep:
    Just the final line, am I being silly, or did you copy down the sin term instead of the cos term?
    Offline

    2
    ReputationRep:
    (Original post by fishpaste)
    Just the final line, am I being silly, or did you copy down the sin term instead of the cos term?
    yeah, final line should read (i think):

    -(xcos3x)/3 + 1/9(sin3x) (between limits.....)
    • Thread Starter
    Offline

    13
    ReputationRep:
    Cheers people.

    Say if you wish to integrate the square of the above function i.e.:

    ∫(x.sin3x)^2

    How would you go about it?
    Offline

    18
    ReputationRep:
    (Original post by 2776)
    Cheers people.

    Say if you wish to integrate the square of the above function i.e.:

    ∫(x.sin3x)^2

    How would you go about it?
    I'd say:

    INT xsin3x = INT x²sin²3x, substitute sin²3x for 2sin3xcos3x and get one messy mother****er of an integration.
    Offline

    11
    ReputationRep:
    (Original post by ZJuwelH)
    I'd say:

    INT xsin3x = INT x²sin²3x, substitute sin²3x for 2sin3xcos3x and get one messy mother****er of an integration.
    I'd also rewrite it as x²sin²3x, but I'd replace sin²3x with ½-½cos6x instead. Then use integration by parts.
    Offline

    8
    ReputationRep:
    (Original post by ZJuwelH)
    I'd say:

    INT xsin3x = INT x²sin²3x, substitute sin²3x for 2sin3xcos3x and get one messy mother****er of an integration.
    Use the identity,

    sin²ø = ½(1-cos2ø)
    Offline

    18
    ReputationRep:
    (Original post by Fermat)
    Use the identity,

    sin²ø = ½(1-cos2ø)
    Of course! I was trying to think of it while doing the post but couldn't so I suggested the other substitution. But is mine OK or not?
 
 
 

University open days

  1. University of Cambridge
    Christ's College Undergraduate
    Wed, 26 Sep '18
  2. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  3. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
Poll
Which accompaniment is best?

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.