Results 1 to 2 of 2

Math Help - Finding the Area Beneath Trigonometric Curves

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    45

    Finding the Area Beneath Trigonometric Curves

    I know the solution to this particular question, but I'm not too sure of the reasoning behind it. The question asks;

    Two curves, y = \sqrt{3}cosx and y = sinx, are drawn on a graph. The first two intersections to the right of the y-axis are labelled A and B.

    a) Solve the equation \sqrt{3}cosx = sinx to find the x-coordinates of A and B.

    b) Find the area contained between the graphs of y = \sqrt{3}cosx and y = sinx between the points A and B.

    My problem is not in finding the solution, but in how the solution is obtained. For some reason, to find the area, you simply find the integral of sinx - {sqrt}3cosx between the points B ( x = \frac{4\pi}{3}) and A ( x = \frac{\pi}{3}). I would have thought that you would need to find two seperate integrals, one to find the area enclosed between A and \frac{\pi}{2} (i.e. the integral of sinx - \sqrt{3}cosx between x = \frac{\pi}{2} and x = \frac{\pi}{3}) and another to find the area betwee B and \frac{\pi}{2} (i.e. the integral of \sqrt{3}cosx - sinx between x = \frac{4\pi}{3} and x = \frac{\pi}{2}). I would have thought that since part of the graph is below the x-axis then the area would be partly negative and therefore would counterract some of the area enclosed above the x-axis.

    Why is it that you can find the area, which is 4 units^2, using only a single integral, instead of two?
    Last edited by Flay; September 23rd 2008 at 12:07 AM.
    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 Flay View Post
    I know the solution to this particular question, but I'm not too sure of the reasoning behind it. The question asks;

    Two curves, y = \sqrt{3}cosx and y = sinx, are drawn on a graph. The first two intersections to the right of the y-axis are labelled A and B.

    a) Solve the equation \sqrt{3}cosx = sinx to find the x-coordinates of A and B.

    b) Find the area contained between the graphs of y = \sqrt{3}cosx and y = sinx between the points A and B.

    My problem is not in finding the solution, but in how the solution is obtained. For some reason, to find the area, you simply find the integral of sinx - {sqrt}3cosx between the points B ( x = \frac{4\pi}{3}) and A ( x = \frac{\pi}{3}). I would have thought that you would need to find two seperate integrals, one to find the area enclosed between A and \frac{\pi}{2} (i.e. the integral of sinx - \sqrt{3}cosx between x = \frac{\pi}{2} and x = \frac{\pi}{3}) and another to find the area betwee B and \frac{\pi}{2} (i.e. the integral of \sqrt{3}cosx - sinx between x = \frac{4\pi}{3} and x = \frac{\pi}{2}). I would have thought that since part of the graph is below the x-axis then the area would be partly negative and therefore would counterract some of the area enclosed above the x-axis.

    Why is it that you can find the area, which is 4 units^2, using only a single integral, instead of two?
    If f(x) > g(x) over [a, b] then the area bounded by the curves y = f(x) and y = g(x) between x = a and x = b is equal to \int_a^b f(x) - g(x) \, dx.

    Since \sin x > \sqrt{3} \cos x over \left[ \frac{\pi}{3}, \, \frac{4 \pi}{3} \right] the bounded area is \int_{\pi/3}^{4 \pi/3} \sin x - \sqrt{3} \cos x \, dx.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding area between curves?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 22nd 2011, 03:51 PM
  2. finding area enclosed by the curves
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 14th 2011, 06:44 PM
  3. finding the area between two curves
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 18th 2010, 03:29 PM
  4. Finding the area beneath a region
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 13th 2009, 01:12 AM
  5. Finding the Area Between 2 Curves
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 2nd 2008, 10:33 AM

Search Tags


/mathhelpforum @mathhelpforum