Results 1 to 12 of 12

Math Help - u substitution problem

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    62

    u substitution problem

    integrate:

    [(1/3)x^3]*(1+x^4)^(1/2)

    u = 1 + x^4
    du = 4x^3dx


    but i have no idea how to finish it from here any help would be greatly appreciated...i took cal 1 about a year ago and i forgot how to properly use u substitution
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by maybnxtseasn View Post
    integrate:

    [(1/3)x^3]*(1+x^4)^(1/2)

    u = 1 + x^4
    du = 4x^3dx


    but i have no idea how to finish it from here any help would be greatly appreciated...i took cal 1 about a year ago and i forgot how to properly use u substitution
    \displaystyle \frac{1}{3} \int \frac{x^3}{\sqrt{1+x^4}} \, dx<br />

    u = 1 + x^4

    du = 4x^3 \, dx

    \displaystyle \frac{1}{12} \int \frac{4x^3}{\sqrt{1+x^4}} \, dx

    \displaystyle \frac{1}{12} \int \frac{1}{\sqrt{u}} \, du

    can you finish?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    62
    how did u get x^3 in the numerator? did u missread my problem?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by maybnxtseasn View Post
    integrate:

    [(1/3)x^3]*(1+x^4)^(1/2)

    u = 1 + x^4
    du = 4x^3dx


    but i have no idea how to finish it from here any help would be greatly appreciated...i took cal 1 about a year ago and i forgot how to properly use u substitution
    \displaystyle\frac{1}{3}\int{x^3\sqrt{1+x^4}}\;dx

    \displaystyle\ u=1+x^4\Rightarrow\frac{du}{dx}=4x^3\Rightarrow\fr  ac{du}{4}=x^3dx

    \displaystyle\int{x^3\sqrt{1+x^4}\;dx=\int{\sqrt{1  +x^4}\;x^3dx

    The integral becomes

    \displaystyle\frac{1}{12}\int{u^{\frac{1}{2}}du
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by maybnxtseasn View Post
    how did u get x^3 in the numerator? did u missread my problem?
    I did ... should be \displaystyle \frac{1}{3} \int x^3 \sqrt{1+x^4} \, dx

    same sub ...

    \displaystyle \frac{1}{12} \int 4x^3 \sqrt{1+x^4} \, dx

    \displaystyle \frac{1}{12} \int \sqrt{u} \, du

    thought I saw a negative on the 1/2 exponent.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    62
    ive worked my problem down to the last step both of u have...now do i just integrate the simple equation with u? and then plug in for u and du? to get my final awnser?

    so for example
    if i integrate your last equation i get
    (2/3)u^(3/2) du <---- do i just plug in (1+x^4) for u and im done?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Yes, as your final solution needs to be in terms of x.
    Also add on the constant of integration.
    The u-substitution allows a "convenient" solution.

    For the calculation of definate integrals (not necessary here), you may choose to return to x,
    or evaluate the definate integral using u with a change of limits to u.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2010
    Posts
    62
    (1/18)(1+x^4)^(3/2) du <----- do i just change du to dx or am i missing something for the final answer because that seems to easy?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by maybnxtseasn View Post
    (1/18)(1+x^4)^(3/2) du <----- do i just change du to dx or am i missing something for the final answer because that seems to easy?
    no "du" in the antiderivative.

    the antiderivative is ...

    \displaystyle \frac{1}{18}(1+x^4)^{\frac{3}{2}} + C

    take the derivative to confirm.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    \displaystyle \begin{aligned}  \frac{1}{3}\int{x^3\left(1+x^4}\right)^{\frac{1}{2  }}\;{dx} & = \frac{1}{4(3)}\int{\left(1+x^4}\right)^{\frac{1}{2  }}(1+x^4)'\;{dx} \\& = \frac{2}{3(4)(3)}\int{\left[\left(1+x^4}\right)^{\frac{3}{2}}\right]'\;{dx} \\& = \frac{1}{18}\left(1+x^4}\right)^{\frac{3}{2}}+k.\e  nd{aligned}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    @ maybenxtseasn: The Coffee Machine's point relates (I think) to your earlier question about u-substitution and the chain rule... to which the answer was yes, the first one is only for navigating the second. So if the first is bogging down you might want to try a different navigational aid (if only to see how the u-sub should be working). Here's my version of what TCM is pointing out...



    To find F, integrate with respect to the dashed balloon (which is like where you integrate with respect to u)...



    So you have...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    The general drift is...



    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by maybnxtseasn View Post
    (1/18)(1+x^4)^(3/2) du <----- do i just change du to dx or am i missing something for the final answer because that seems to easy?
    The purpose of the u-substitution was to make it easy!

    \displaystyle\frac{d}{dx}\left(x^4+C\right)=4x^3\R  ightarrow\ \int{d\left(x^4+C\right)}=\int{4x^3}\;dx\Rightarro  w\int{4x^3}dx=x^4+C

    \displaystyle\frac{d}{du}\left(u^4+C\right)=4u^3\R  ightarrow\int{4u^3}du=u^4+C


    \displaystyle\int{\sqrt{1+x^4}\;x^3dx=\int{\sqrt{u  }}\;du

    when u=1+x^4
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. please help with a substitution problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 2nd 2010, 12:42 PM
  2. substitution problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 19th 2009, 01:05 PM
  3. Another U-Substitution Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 29th 2009, 10:22 PM
  4. U Substitution problem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 4th 2008, 05:26 PM
  5. Need help with a u-substitution problem
    Posted in the Calculus Forum
    Replies: 12
    Last Post: September 23rd 2008, 05:09 PM

Search Tags


/mathhelpforum @mathhelpforum