Results 1 to 13 of 13

Math Help - Find the definite integral using substitution

  1. #1
    Banned
    Joined
    Aug 2009
    Posts
    24

    Find the definite integral using substitution

    Suppose I have this function where a = 0 and b = .5 not sure on the latex tags for that

    I(x) = \int \sqrt{t^4+1}

    and I let

    u = t^4+1

    and

    du = 4t^3dx

    this is the part where I get confused...

    Do I

    \frac{1}{4}du = t^3dx

    or solve for dx completely

    If I do it at the step above and continue on to

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

    and plug in the corresponding values my answers do not come out correctly
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2009
    Posts
    13
    Quote Originally Posted by The Power View Post
    Suppose I have this function where a = 0 and b = .5 not sure on the latex tags for that

    I(x) = \int \sqrt{t^4+1}

    and I let

    u = t^4+1

    and

    du = 4t^3dx

    this is the part where I get confused...

    Do I

    \frac{1}{4}du = t^3dx

    or solve for dx completely

    If I do it at the step above and continue on to

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

    and plug in the corresponding values my answers do not come out correctly
    When you substitute u = t^4+1
    you need to get dt completely in terms of du, so you do

    du = 4t^3dt
    Further,
    t = (u-1)^{1/4}
    so
    t^3 = (u-1)^{3/4}

    and dt = \frac{du}{4(u-1)^{3/4}}

    and your integral now looks like

    \int u^\frac{1}{2} * \frac{du}{4(u-1)^{3/4}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by The Power View Post
    Suppose I have this function where a = 0 and b = .5 not sure on the latex tags for that

    I(x) = \int \sqrt{t^4+1}

    and I let

    u = t^4+1

    and

    du = 4t^3dx

    this is the part where I get confused...

    Do I

    \frac{1}{4}du = t^3dx

    or solve for dx completely

    If I do it at the step above and continue on to

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

    and plug in the corresponding values my answers do not come out correctly
    I doubt an answer exists in a closed form using elementary functions. Where has this integral come from?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Aug 2009
    Posts
    24
    From a text book we were told to make a table for the values of x = 0, 5, 1, 1.5

    if the integral is from b = x and a = 0

    And to the previous fella you lost me at t = (u-1)^1/4 where did that come from
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,560
    Thanks
    1425
    Quote Originally Posted by The Power View Post
    From a text book we were told to make a table for the values of x = 0, 5, 1, 1.5

    if the integral is from b = x and a = 0

    And to the previous fella you lost me at t = (u-1)^1/4 where did that come from
    If u = t^4 + 1

    u - 1 = t^4

    (u - 1)^{\frac{1}{4}} = t.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Aug 2009
    Posts
    24
    What other ways can I get the anti-derivative of this function to make a table of values
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by The Power View Post
    From a text book we were told to make a table for the values of x = 0, 5, 1, 1.5

    if the integral is from b = x and a = 0

    And to the previous fella you lost me at t = (u-1)^1/4 where did that come from
    Please type the question exactly as it appears in your textbook. Because it looks like you've omitted important imformation.


    Memo to all members: It helps if you type all of the problem rather than just the bits of it you think are important. (The fact that important information is either overlooked or considered unimportant is often a major reason why people can't do the question they post). This saves time for everyone.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    May 2009
    Posts
    471
    You aren't going to find an elementary antiderivative for that function... so unless you use estimation techniques or just put in into a calculator for the values of b=x that you want, I don't think you're going to get an answer


    You could use taylor series to get as good an estimation as you'd like
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Aug 2009
    Posts
    24
    I later stated the whole problem find I(x) which is the integral when x is = 0, .5 , 1, 1.5 and make a table of values it is in the section pertaining to the Second FTC
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by The Power View Post
    I later stated the whole problem find I(x) which is the integral when x is = 0, .5 , 1, 1.5 and make a table of values it is in the section pertaining to the Second FTC
    As has been said - twice - the integral cannot be evaluated in closed form using elementary functions. So either important information from the problem statement is missing or you're expected to use technology to get approximate values for the integral (in which case the question will have said so). The fact that you might be expected to apply the Second FTC only lends weight to the fact that something is missing from the question as posted.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Aug 2009
    Posts
    24
    Here Ill post the question in it's entirety instead of bits and pieces in this thread

    For x = 0, .5, 1, 1.5, 2 make a table of values for

    I(x)=\int\sqrt{t^4+1}

    and x is the upper limit and the lower limit is 0

    Thanks for your patience Mr.F
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by The Power and edited by Mr F (including an important omission in red) View Post
    Here Ill post the question in it's entirety instead of bits and pieces in this thread

    For x = 0, .5, 1, 1.5, 2 make a table of values for

    I(x)=\int_{0}^{x} \sqrt{t^4+1} \, {\color{red}dt}

    Thanks for your patience Mr.F
    Perhaps the following link will put in context the many attempts that have been made to tell you that the integral cannot be done in closed form using elementary functions: Wolfram|Alpha

    You will have to use technology to evaluate I(x) for x = 0.5, 1, 1.5, 2. Was that made clear in the question?

    Except for when x = 0 in which case obviously I(0) = 0.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Sep 2009
    Posts
    13
    Lets see...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Definite Integral By Substitution
    Posted in the Calculus Forum
    Replies: 8
    Last Post: June 5th 2011, 04:02 AM
  2. Replies: 6
    Last Post: May 29th 2011, 07:50 AM
  3. Replies: 13
    Last Post: November 21st 2010, 11:09 AM
  4. Replies: 2
    Last Post: November 20th 2010, 08:16 PM
  5. Definite integral substitution???
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 18th 2008, 08:33 PM

Search Tags


/mathhelpforum @mathhelpforum