Results 1 to 11 of 11

Math Help - Evaluate

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    25

    Evaluate

    Hi I am in the middle of this question and im not sure if a. im right or b. where to go from here.

    Q. Evaluate:

    \int \frac{s}{\sqrt{s^2+4^2}} between the limits of 3 and 0

    So far I have:

    \int \frac{s}{\sqrt{s^2+4^2}}

    let.P =\sqrt{s^2+4^2} then dp =  \frac{s}{\sqrt{s^2+16}} ds

    when S = 3 then P = 5

    and

    when S = 0 then P = 4

    Am I right so far, if so what is the next step?

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    You forgot the ds in your integral.
    I think the substitution u=s^2+4^2 is better.
    Do not forget to change the limits of the integration according to this substitution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    25
    Ok so, a few changes:


    \int \frac{s}{\sqrt{s^2+4^2}}ds

    u =\sqrt{s^2+4^2} then du =  \frac{s}{\sqrt{s^2+16}} ds

    when S = 3 then u = 5

    and

    when S = 0 then u = 4

    With the limits now between 5 and 4

    Ok, that looks a bit better but im still not sure how to take this further, its been ages since ive done anything like this and my mind is blank.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    No.
    I said u=s^2+4^2
    Without the square root.
    This does not mean your solution is wrong.
    but this substitution is easier.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    25
    Sorry you have lost me, what happens to the sqrt then?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by anothernewbie View Post
    Ok so, a few changes:


    \int \frac{s}{\sqrt{s^2+4^2}}ds

    u =\sqrt{s^2+4^2} then du = \frac{s}{\sqrt{s^2+16}} ds

    when S = 3 then u = 5

    and

    when S = 0 then u = 4

    With the limits now between 5 and 4

    Ok, that looks a bit better but im still not sure how to take this further, its been ages since ive done anything like this and my mind is blank.
    You changed the limits correctly.
    The next step is changing your integral to another integral in terms of u according to your substitution.
    To do that.
    First you should find ds in terms of u
    You need to solve your substitution for s then differentiate it with respect to s.
    u=\sqrt{s^2+4^2}
    s^2=u^2-4^2
    s=\sqrt{u^2-4^2}
    Differentiate with respect to s.
    ds=\frac{u}{\sqrt{u^2-4^2}}
    So, For the original integral, we found ds in terms of u.
    the denominator is u(Which is just our the substitution you did).
    we have s in the numerator.
    Actually, we found it above.
    s=\sqrt{u^2-4^2}
    You have everything in the original integral in terms of u.
    Can you replace them and try to solve the resulting integral ?


    -------

    Quote Originally Posted by anothernewbie View Post
    Sorry you have lost me, what happens to the sqrt then?


    Sir, You are using the substitution method to solve this integral, Right?

    I said your substitution u=\sqrt{u^2+4^2} will make it hard a little bit.

    So, I gave you another substitution to solve it, There is no relation between them.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2009
    Posts
    25
    Quote Originally Posted by General View Post

    Sir, You are using the substitution method to solve this integral, Right?

    I said your substitution u=\sqrt{u^2+4^2} will make it hard a little bit.

    So, I gave you another substitution to solve it, There is no relation between them.
    Yes, thanks I realised this I just wasn't sure how you would use your method. Thanks for the help I should be able to finish it now.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2009
    Posts
    25
    Right, sorry im still not 100%, does this mean that it works out to be:

    \frac {\sqrt{5^2-16}}{\sqrt{\sqrt{(5^2-16)^2}+16}} - \frac {\sqrt{4^2-16}}{\sqrt{\sqrt{(4^2-16)^2}+16}}

    Or am I totally wrong, again!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by anothernewbie View Post
    Right, sorry im still not 100%, does this mean that it works out to be:

    \frac {\sqrt{5^2-16}}{\sqrt{\sqrt{(5^2-16)^2}+16}} - \frac {\sqrt{4^2-16}}{\sqrt{\sqrt{(4^2-16)^2}+16}}

    Or am I totally wrong, again!
    Show your steps for evaluating the resulting integral which bacame from the substitution.
    And also, specify which substitution did you use.
    And doubling posts is not allowed in this forum.
    For the substitution u=s^2+4^2
    u(0)=16 and u(3)=25
    So the new Upper/Lower limits the integral are 16 and 25.
    And we have du=2sds
    i.e. \frac{1}{2}du=sds
    Integral will be :
    \frac{1}{2}\int_{16}^{25}u^{\frac{-1}{2}}du=\sqrt{25}-\sqrt{16} = 5 - 4 = 1.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Oct 2009
    Posts
    25
    Thanks again, but
    Quote Originally Posted by General View Post
    And doubling posts is not allowed in this forum.
    I don't see where I did that?????? All my posts are genuine
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by anothernewbie View Post
    Thanks again, but


    I don't see where I did that?????? All my posts are genuine
    You did before my last post.
    You wrote 2 posts.
    Anyway, Forget it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. evaluate
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 18th 2010, 12:16 PM
  2. evaluate
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 21st 2009, 03:26 AM
  3. Evaluate
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 26th 2009, 04:56 AM
  4. how do i evaluate...
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 4th 2008, 10:55 PM
  5. Please evaluate
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 14th 2007, 04:36 PM

Search Tags


/mathhelpforum @mathhelpforum