Results 1 to 9 of 9

Math Help - quick question involving definite integral

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    229

    quick question involving definite integral



    I'm uncertain about something. I thought this was a simple problem where you simply bring up the denominator and apply the general power rule, but that doesn't seem to be right... Can anyone point me in the right direction?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Archduke01 View Post


    I'm uncertain about something. I thought this was a simple problem where you simply bring up the denominator and apply the general power rule, but that doesn't seem to be right... Can anyone point me in the right direction?
    Substitute u = 4x + 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    229
    \frac {1}{4} \int_6^{20} \frac {4}{\sqrt{4x+1}}

    I'm sorry for asking, but how do I proceed from there? Or atleast tell me why it isn't correct if it isn't.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Archduke01 View Post
    \frac {1}{4} \int_6^{20} \frac {4}{\sqrt{4x+1}}

    I'm sorry for asking, but how do I proceed from there? Or atleast tell me why it isn't correct if it isn't.
    Have you been taught the substitution technique.

    Otherwise you will just have to use the following result which I'm sure is somewhere in your class notes or textbook:

    \int (ax + b)^n \, dx = \frac{1}{a(n+1)} (ax + b)^{n + 1} provided n \neq -1 (and I have ignored the arbitrary constant since it's not needed for a definite integral).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie Homeomorphism's Avatar
    Joined
    Mar 2010
    Posts
    6
    For the first problem: Let u = \sqrt{4x+1}, then \dfrac{du}{dx} = \dfrac{2}{\sqrt{(4x+1)}} \Rightarrow dx = \dfrac{\sqrt{4x+1}}{2}\;{du} so that \int_{x=6}^{x=20}\dfrac{dx}{\sqrt{4x+1}} = \int_{u=5}^{u=9}\left(\dfrac{1}{u}\right)\left(\df  rac{u}{2}\right)\;{du}   and finish off.

    For the second problem: Let again u = \sqrt{4x+1}, then \dfrac{du}{dx} = \dfrac{2}{\sqrt{(4x+1)}} \Rightarrow dx = \dfrac{\sqrt{4x+1}}{2}\;{du} so that \dfrac{1}{4}\int_{x=6}^{x=20}\dfrac{4}{\sqrt{4x+1}  }\;{dx} = \dfrac{1}{4}\int_{u=5}^{u=9}\left(\dfrac{4}{u}\rig  ht)\left(\dfrac{u}{2}\right)\;{du} and finish off.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2009
    Posts
    229
    Where did you get the 2 from?
    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 Archduke01 View Post
    Where did you get the 2 from?
    Did you read my previous reply or did you decide it was easier to focus on a complete solution that is probably beyond your current level of comprehension?

    Do you understand that, in the context of my previous reply, n = -1/2?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Oct 2009
    Posts
    229
    Thank you, but I want to try the substitution method since I've never heard of your second one.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Archduke01 View Post
    Thank you, but I want to try the substitution method since I've never heard of your second one.
    Well, I'll ask again. Have you been taught the substitution method? If so, where are you having trouble making the substitution u = 4x + 1 (which brings us all the way back to post #2)?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 30th 2011, 08:46 PM
  2. Definite Integral Involving Absolute Value
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 14th 2010, 05:20 PM
  3. Replies: 3
    Last Post: September 21st 2010, 04:49 PM
  4. [SOLVED] Definite integral involving parts
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 10th 2010, 08:38 PM
  5. Quick help please!! definite integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 11th 2008, 11:43 AM

Search Tags


/mathhelpforum @mathhelpforum