Results 1 to 3 of 3

Math Help - Evaluating an integral interpreted as area...

  1. #1
    Newbie ILovePizza's Avatar
    Joined
    Apr 2013
    From
    East L.A
    Posts
    7

    Question Evaluating an integral interpreted as area...

    Finishing the last chapter of Calculus 1, so go easy on me you guys :|

    I know this is like the easiest and most obvious thing. I just can't figure it out though; I need help.

    I don't know why the answer is 6π.

    Evaluate ∫ from -2 to 2 of (x + 3)(4 - x^2)1/2 dx by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.


    Why? And how is this done?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2013
    From
    Green Bay
    Posts
    68
    Thanks
    16

    Re: Evaluating an integral interpreted as area...

     \int_{-2}^2 (x+3)(4 - x^2)^{\frac{1}{2}} dx
    \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} + 3*(4 - x^2)^{\frac{1}{2}} dx
    \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx + \int_{-2}^2 3*(4 - x^2)^{\frac{1}{2}} dx
    \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx + 3 * \int_{-2}^2 (4 - x^2)^{\frac{1}{2}} dx

    Now, note that \int_{-2}^2 (4 - x^2)^{\frac{1}{2}} is the semi unit circle of radius 2. Thus, the area is  \frac{1}{2} \pi r^2 = \frac{1}{2} \pi 2^2 = 2 \pi

    Note, use direct subsitution  u = 4 - x^2 to evaluate \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx

    Then  u(-2) = u(2) = 0 and  du = -2x dx

    Then  \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx = \int_{0}^0 \frac{-1}{2} u du which equals zero.

    Therefore,

    \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx + \int_{-2}^2 3*(4 - x^2)^{\frac{1}{2}} dx = 0 + 3 * 2 \pi = 6 \pi
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie ILovePizza's Avatar
    Joined
    Apr 2013
    From
    East L.A
    Posts
    7

    Thumbs up Re: Evaluating an integral interpreted as area...

    Omg

    I have an exam on this stuff soon. This helped a lot!

    Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: February 10th 2013, 10:21 PM
  2. Evaluating area under the curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 13th 2010, 05:30 AM
  3. Help evaluating an integral
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 17th 2009, 12:47 PM
  4. Replies: 3
    Last Post: February 6th 2009, 02:58 PM
  5. Replies: 6
    Last Post: May 18th 2008, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum