Results 1 to 12 of 12

Math Help - TRICKY u substitution - are my answers correct

  1. #1
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161

    Cool TRICKY u substitution - are my answers correct

    QUESTION ONE  \int z^3\sqrt{z^2 + 1}dz

    u = z^2 + 1

     \frac{du}{dz} = 2z

    \frac{z^2}{2}\int u^\frac{1}{2}\ du

    \frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}

     \frac{z^2}{3}\sqrt{(z^2+1)^3} + c

    QUESTION TWO

     \int y\sqrt{y+1}\ dy

     u = y + 1

     \frac{du}{dy} = 1

     y\int u^\frac{1}{2}\ du

     \frac{2y}{3}u^\frac{3}{2} + c

    \frac{2y}{3}\sqrt{(y+1)^3} + c


    QUESTION THREE
     \int \frac{x}{(x-1)^3}\ dx

     u = x -1

    \frac{du}{dx}=1

     x\int u^3\ du

      x\frac{u^4}{4} + c

     \frac{x(x-1)^4}{4}+c

    QUESTION FOUR  \int \frac{x}{\sqrt{2x-1}}\ dx

     \frac{du}{dx} = 2

     \frac{x}{2}\int u^\frac{-1}{2}\ du

    \frac{x}{2}.2u^\frac{1}{2}  + c

     x\sqrt{2x -1} + c

    If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2009
    Posts
    397
    Hi differentiate
    Quote Originally Posted by differentiate View Post
    QUESTION ONE [tex]

    \frac{z^2}{2}\int u^\frac{1}{2}\ du

    \frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}
    This step is wrong. You have to state z^2 in term of u before integrating it.

    QUESTION TWO

     y\int u^\frac{1}{2}\ du

     \frac{2y}{3}u^\frac{3}{2} + c
    This step is wrong. You have to state y in term of u before integrating it.

    QUESTION THREE
     x\int u^3\ du

      x\frac{u^4}{4} + c
    This step is wrong. You have to state x in term of u before integrating it.

    QUESTION FOUR  \int \frac{x}{\sqrt{2x-1}}\ dx

     \frac{du}{dx} = 2

     \frac{x}{2}\int u^\frac{-1}{2}\ du

    \frac{x}{2}.2u^\frac{1}{2}  + c

     x\sqrt{2x -1} + c

    If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

    Thank you.
    What is your u ? ( I suppose it's 2x-1). And you have the same mistake as previous questions.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    ohh thanks songoku! I get it now
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    alright, I got question 2 and 3 correct now. The problem is question 1 and question 2.

    Question One  \int z^3\sqrt{z^2 + 1}\ dz

     u = z^2 + 1

     z^3 = (u-1)^\frac{3}{2}

     \int u^\frac{1}{2}(u-1)^\frac{3}{2}\ du

    I'm stuck now. What should I do next?

    QUESTION FOUR  \int \frac{x}{\sqrt{2x-1}}\ dx

     u = 2x - 1

     \frac{u + 1 }{2} = x

    \int \frac{\frac{u+1}{2}}{u^\frac{1}{2}}\ du

     \int \frac{u+1}{2u^\frac{1}{2}}

    \int \frac{u^\frac{1}{2}}{2} + \frac{1}{2}u^\frac{-1}{2}\ du

    \frac{1}{3}u^\frac{3}{2} + u^\frac{1}{2} + c

    \frac{1}{3}\sqrt{(2x-1)^3} + \sqrt{2x-1} + c
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2009
    Posts
    397
    Quote Originally Posted by differentiate View Post
    QUESTION ONE  \int z^3\sqrt{z^2 + 1}dz

    u = z^2 + 1

     \frac{du}{dz} = 2z

    \frac{z^2}{2}\int u^\frac{1}{2}\ du
    It's good until this step. Now substitute z^2. Do the same for question 2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    hmm.. I still couldn't get question 1
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    Is this the way you do it?

    \int z^3(z^2+1)\ dz

     z^3 = (u-1)^\frac{3}{2}

    \int (u-1)^\frac{3}{2}u^\frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jul 2009
    Posts
    397
    Hi differentiate
    Quote Originally Posted by differentiate View Post
    QUESTION ONE  \int z^3\sqrt{z^2 + 1}dz

    u = z^2 + 1

     \frac{du}{dz} = 2z

    \frac{z^2}{2}\int u^\frac{1}{2}\ du
    I'll continue your work.


    \frac{z^2}{2}\int u^\frac{1}{2}\ du

    =\frac{u-1}{2}\int u^\frac{1}{2}\ du

    =\frac{1}{2}\int (u-1)u^\frac{1}{2}\ du

    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    Thanks skeeter. so continuing on... could someone check this for me?

    = \frac{1}{2}\int u^\frac{3}{2} - u^\frac{1}{2}\ du

     \frac{1}{2}\left [ \frac{2}{5}u^\frac{5}{2} - \frac{2}{3}u^\frac{3}{2} \right ]

    \frac{1}{5}u^\frac{5}{2} - \frac{1}{3}u^\frac{3}{2}

    \frac{1}{5}(z^2+1)^\frac{5}{2} - \frac{1}{3}(z^2+1)^\frac{3}{2} + c

    Hence, evaluate \int_{0}^{\sqrt{2}} z^3\sqrt{z^2 +1}

     [\frac{1}{5}(z^2+1)^\frac{5}{2} - \frac{1}{3}(z^2+1)^\frac{3}{2} ]_0^\sqrt{2}

     = \frac{1}{5}(2+1)^\frac{5}{2} - \frac{1}{3}(2+1)^\frac{3}{2} - \frac{1}{5} + \frac{1}{3}

    = \frac{\sqrt{3^5}}{5} - \frac{\sqrt{3^3}}{3} + \frac{2}{15}

     = \frac{9\sqrt{3}}{5} - \sqrt{3} + \frac{2}{15}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391
    Quote Originally Posted by differentiate View Post
    QUESTION ONE  \int z^3\sqrt{z^2 + 1}dz

    u = z^2 + 1

     \frac{du}{dz} = 2z

    \frac{z^2}{2}\int u^\frac{1}{2}\ du

    \frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}

     \frac{z^2}{3}\sqrt{(z^2+1)^3} + c

    QUESTION TWO

     \int y\sqrt{y+1}\ dy

     u = y + 1

     \frac{du}{dy} = 1

     y\int u^\frac{1}{2}\ du

     \frac{2y}{3}u^\frac{3}{2} + c

    \frac{2y}{3}\sqrt{(y+1)^3} + c


    QUESTION THREE  \int \frac{x}{(x-1)^3}\ dx

     u = x -1

    \frac{du}{dx}=1

     x\int u^3\ du

     x\frac{u^4}{4} + c

     \frac{x(x-1)^4}{4}+c

    QUESTION FOUR  \int \frac{x}{\sqrt{2x-1}}\ dx

     \frac{du}{dx} = 2

     \frac{x}{2}\int u^\frac{-1}{2}\ du

    \frac{x}{2}.2u^\frac{1}{2} + c

     x\sqrt{2x -1} + c

    If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

    Thank you.
    Q.2

     \int{ y\sqrt{y+1}\, dy}

    Let u = y + 1 so that \frac{du}{dy} = 1.

    Also note that y = u - 1.


    So the integral becomes

    \int{(u - 1)\sqrt{u}\,\frac{du}{dy}\,dy}

     = \int{(u - 1)u^{\frac{1}{2}}\,du}

     = \int{u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du}

     = \frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}} + C

     = \frac{2}{5}(y + 1)^{\frac{5}{2}} - \frac{2}{3}(y + 1)^{\frac{3}{2}} + C

     = \frac{2}{5}\sqrt{(y + 1)^5} - \frac{2}{3}\sqrt{(y + 1)^3} + C.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391
    Quote Originally Posted by differentiate View Post
    QUESTION ONE  \int z^3\sqrt{z^2 + 1}dz

    u = z^2 + 1

     \frac{du}{dz} = 2z

    \frac{z^2}{2}\int u^\frac{1}{2}\ du

    \frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}

     \frac{z^2}{3}\sqrt{(z^2+1)^3} + c

    QUESTION TWO

     \int y\sqrt{y+1}\ dy

     u = y + 1

     \frac{du}{dy} = 1

     y\int u^\frac{1}{2}\ du

     \frac{2y}{3}u^\frac{3}{2} + c

    \frac{2y}{3}\sqrt{(y+1)^3} + c


    QUESTION THREE  \int \frac{x}{(x-1)^3}\ dx

     u = x -1

    \frac{du}{dx}=1

     x\int u^3\ du

     x\frac{u^4}{4} + c

     \frac{x(x-1)^4}{4}+c

    QUESTION FOUR  \int \frac{x}{\sqrt{2x-1}}\ dx

     \frac{du}{dx} = 2

     \frac{x}{2}\int u^\frac{-1}{2}\ du

    \frac{x}{2}.2u^\frac{1}{2} + c

     x\sqrt{2x -1} + c

    If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

    Thank you.
    Q.3

    \int{\frac{x}{(x - 1)^3}\,dx}.

    Let u = x - 1 so that \frac{du}{dx} = 1.

    Also note that x = u + 1.


    Therefore, the integral becomes

    \int{\frac{u + 1}{u^3}\,\frac{du}{dx}\,dx}

     = \int{(u + 1)u^{-3}\,du}

     = \int{u^{-2} + u^{-3}\,du}

     = -u^{-1} - \frac{1}{2}u^{-2} + C

     = -\frac{1}{x - 1} - \frac{1}{2(x - 1)^2} + C.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,492
    Thanks
    1391
    Quote Originally Posted by differentiate View Post
    QUESTION ONE  \int z^3\sqrt{z^2 + 1}dz

    u = z^2 + 1

     \frac{du}{dz} = 2z

    \frac{z^2}{2}\int u^\frac{1}{2}\ du

    \frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}

     \frac{z^2}{3}\sqrt{(z^2+1)^3} + c

    QUESTION TWO

     \int y\sqrt{y+1}\ dy

     u = y + 1

     \frac{du}{dy} = 1

     y\int u^\frac{1}{2}\ du

     \frac{2y}{3}u^\frac{3}{2} + c

    \frac{2y}{3}\sqrt{(y+1)^3} + c


    QUESTION THREE  \int \frac{x}{(x-1)^3}\ dx

     u = x -1

    \frac{du}{dx}=1

     x\int u^3\ du

     x\frac{u^4}{4} + c

     \frac{x(x-1)^4}{4}+c

    QUESTION FOUR  \int \frac{x}{\sqrt{2x-1}}\ dx

     \frac{du}{dx} = 2

     \frac{x}{2}\int u^\frac{-1}{2}\ du

    \frac{x}{2}.2u^\frac{1}{2} + c

     x\sqrt{2x -1} + c

    If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

    Thank you.
    Q.4

    \int{\frac{x}{\sqrt{2x - 1}}\,dx}.

    Let u = 2x - 1 so that \frac{du}{dx} = 2.

    Also note that x = \frac{1}{2}(u + 1).


    So the integral becomes

    \frac{1}{2}\int{\frac{2x}{\sqrt{2x - 1}}\,dx}

     = \frac{1}{2}\int{\frac{u + 1}{2\sqrt{u}}\,\frac{du}{dx}\,dx}

     = \frac{1}{4}\int{(u + 1)u^{-\frac{1}{2}}\,du}

     = \frac{1}{4}\int{u^{\frac{1}{2}} + u^{-\frac{1}{2}}\,du}

     = \frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}} + 2u^{\frac{1}{2}} + c\right]

     = \frac{1}{6}(2x - 1)^{\frac{3}{2}} + \frac{1}{2}(2x - 1)^{\frac{1}{2}} + C

     = \frac{1}{6}\sqrt{(2x - 1)^3} + \frac{1}{2}\sqrt{2x - 1} + C.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: September 27th 2009, 11:49 AM
  2. Are my answers correct
    Posted in the Geometry Forum
    Replies: 1
    Last Post: September 26th 2009, 05:58 PM
  3. Are my answers correct?
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: August 30th 2009, 02:01 PM
  4. Are my answers correct?
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 27th 2009, 02:46 PM
  5. Which two answers are correct?
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 11th 2007, 06:33 AM

Search Tags


/mathhelpforum @mathhelpforum