Results 1 to 5 of 5

Math Help - area of the region

  1. #1
    Member
    Joined
    Nov 2005
    Posts
    172

    area of the region

    Decide whether to integrate with respect to x or y. Then find the area of the region

     y = 7 \cos x, y = (6 sec(x))^2, x = -\pi / 4, x = \pi / 4

    i used the formula:  A = \int_a^b(f(x)-g(x)) dx

     A = \int_a^b ((6 sec(x))^2 - 7 cosx)dx

     a = -\pi / 4 , b = -\pi / 4

    not sure what to do next
    Follow Math Help Forum on Facebook and Google+

  2. #2
    fgn
    fgn is offline
    Newbie
    Joined
    Nov 2006
    Posts
    6
    Quote Originally Posted by viet View Post
     A = \int_a^b ((6 sec(x))^2 - 7 cosx)dx

     a = -\pi / 4 , b = -\pi / 4
    You almost had it.
    But  \int_a^bf(x)\,dx = 0 if  a = b by the fundamental theorem of calculus.
    If you change it to this it should work.
    A = \int_{-\frac{\Pi}{4}}^{\frac{\Pi}{4}}((6 sec\,(x))^2 - 7 cos\,(x))\,dx
    You could also exploit the symmetry to get
    A = 2\int_0^{\frac{\Pi}{4}}((6 sec\,(x))^2 - 7 cos\,(x))\,dx
    Last edited by fgn; December 7th 2006 at 03:05 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2005
    Posts
    172
    Quote Originally Posted by fgn View Post
    If you change it to this it should work.
    A = \int_{-\frac{\Pi}{4}}^{\frac{\Pi}{4}}((6 sec\,(x))^2 - 7 cos\,(x))\,dx
    i understand that part, but not sure how to get the answer from here
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2006
    Posts
    81
    Viet isn't the only one looking for help on this particular problem. I know a couple of others are as well. Can anyone help us through this problem step by step?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    fgn
    fgn is offline
    Newbie
    Joined
    Nov 2006
    Posts
    6
    Quote Originally Posted by viet View Post
    i understand that part, but not sure how to get the answer from here
    So far so good!

    It's good if you try to learn the derivatives of the elementary functions.
    \frac{d}{dx}(tan\,(x)) = \frac{sin^{\,2}(x) + cos^{\,2}(x)}{cos^{\,2}(x)} = sec^{\,2}(x) = tan^{\,2}(x) + 1

    From this you get  \int sec^{\,2}(x)\,dx = tan\,(x) + C

    And your integral becomes
     \int((6 sec\,(x))^2 - 7 cos\,(x))\,dx = 36tan\,(x) - 7sin\,(x) + C

    I hope you can take it from here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of a region
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 28th 2010, 09:06 AM
  2. Area of region
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 27th 2009, 03:46 AM
  3. Area of Region
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 10th 2009, 01:11 PM
  4. Area of a region
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 6th 2009, 01:39 PM
  5. Area of a region
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2006, 11:56 PM

Search Tags


/mathhelpforum @mathhelpforum