Results 1 to 3 of 3

Math Help - Integration...having neither x-values

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    7

    Talking Integration...having neither x-values

    given that ∫15x^2 dx = 3430, and x=a and x=-a. Find the value of the constant a.

    I'm not quite sure how to approach solving for a.
    I(x) = 5x^3

    So would I set that equal to 3430?
    5x^3=5430
    x^3=686
    x= approx. 8.8?

    The answer is that a=7.

    7^3 is 343 ...

    3430/10 = 343 .. but how could I get to that (justify it anyways)

    Thanks so much, I really want to understand how to do this, I'm just confused on where to go from here.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by rubbersoul923 View Post
    given that ∫15x^2 dx = 3430, and x=a and x=-a. Find the value of the constant a.

    I'm not quite sure how to approach solving for a.
    I(x) = 5x^3

    So would I set that equal to 3430?
    5x^3=5430
    x^3=686
    x= approx. 8.8?

    The answer is that a=7.

    7^3 is 343 ...

    3430/10 = 343 .. but how could I get to that (justify it anyways)

    Thanks so much, I really want to understand how to do this, I'm just confused on where to go from here.

    Thanks!
    \int_{-a}^{a}15x^2dx=3430 \iff 5x^{3}\bigg|_{-a}^{a}=3430 \iff 5(a^3-(-a)^3)=3430

    This gives

    10a^3=3430 \iff a^3=343

    just solve from here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    Using symmetry solve for a where \int_0^a 15x^2~dx = \frac{3430}{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration by Parts with Discrete Values
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2010, 08:38 AM
  2. Integration by parts.. Find values of a and b.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 17th 2010, 06:03 AM
  3. Replies: 1
    Last Post: May 24th 2009, 05:16 AM
  4. volume: figuring out integration values.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 21st 2009, 08:57 AM
  5. Integration with absolute values?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 9th 2008, 02:45 AM

Search Tags


/mathhelpforum @mathhelpforum