Results 1 to 2 of 2

Math Help - Integral of udu

  1. #1
    Member integral's Avatar
    Joined
    Dec 2009
    From
    Arkansas
    Posts
    200

    Integral of udu

    \textrm{given}\,\,\,\int 3(5+3x)dx
    \textrm{say that}\,\,\, u=(5+3x) \,\,\,\textrm{and}\,\,\, u'=\frac{d(3x)}{dx} \therefore
    \int 3(5+3x)^6dx=\int u^6u'dx

    \textrm{ This is were I get lost. It says that the answer to this would be}\,\,\,\frac{u^7}{7}+C

    \textrm{But u' is a scalar so:} \,\,\, \int u^6u'dx=u'\int u^6dx
    \textrm{and thus}\,\,\,\frac{d(3x)}{dx}\frac{u^7}{7}=3\frac{u^  7}{7}

    \textrm{So why does this book say}\,\, \frac{u^7}{7}\,\, \textrm{and not:} \,\, 3\frac{u^7}{7}

    \textrm{thanks in advance}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by integral View Post
    \textrm{given}\,\,\,\int 3(5+3x)dx
    \textrm{say that}\,\,\, u=(5+3x) \,\,\,\textrm{and}\,\,\, u'=\frac{d(3x)}{dx} \therefore
    \int 3(5+3x)^6dx=\int u^6u'dx

    \textrm{ This is were I get lost. It says that the answer to this would be}\,\,\,\frac{u^7}{7}+C

    \textrm{But u' is a scalar so:} \,\,\, \int u^6u'dx=u'\int u^6dx
    \textrm{and thus}\,\,\,\frac{d(3x)}{dx}\frac{u^7}{7}=3\frac{u^  7}{7}

    \textrm{So why does this book say}\,\, \frac{u^7}{7}\,\, \textrm{and not:} \,\, 3\frac{u^7}{7}

    \textrm{thanks in advance}
    Hi integral,

    u=5+3x

    \frac{du}{dx}=3

    du=3dx

    Hence

    \int{u^6}3dx=\int{u^6}du
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Volume integral to a surface integral ... divergence theorem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: October 7th 2010, 07:11 PM
  2. Replies: 2
    Last Post: August 31st 2010, 08:38 AM
  3. Replies: 1
    Last Post: June 2nd 2010, 03:25 AM
  4. Replies: 0
    Last Post: May 9th 2010, 02:52 PM
  5. Replies: 0
    Last Post: September 10th 2008, 08:53 PM

Search Tags


/mathhelpforum @mathhelpforum