1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Area between curves

  1. Apr 18, 2006 #1
    Write, but do not evaluate the integral that will give the area between [tex] y = cos x [/tex] and [tex] y = x/2 - 1 [/tex], bounded on the left by the y-axis

    I've sketched the graphs, so I know that [tex] y = cos x [/tex] is above [tex] y = x/2 - 1 [/tex], so the indefinite integral to solve would be [tex] \int (cos x) - (x/2 -1) dx [/tex]

    I know the lower bound is zero, since it's bordered by the y-axis, and I know that to find the upper bound I need to find the point of intersection of the two curves.

    The professon told us to use "technology", which usually means Mathematica. I can't seem to get Mathematica to solve the equation [tex] cos x = x/2 - 1 [/tex]

    Any advice on either how to get Mathematica to solve such an equation, or another method of finding the point of intersection?

  2. jcsd
  3. Apr 18, 2006 #2
    Using my calculator I get that their intersection is at [tex]x\approx1.646[/tex].
  4. Apr 18, 2006 #3
    How did you manipulate the equation to calculate the answer? Or did you just use Newton's method?
    Last edited: Apr 19, 2006
  5. Apr 19, 2006 #4
    I just used my calculator. I don't believe that this can be solved for explicitly. Newton's Method would work, but I graphed it on my TI-89 and found the intersection point.
  6. Apr 19, 2006 #5
    Thanks for the help. I'm still getting used to the idea that most equations are unsolvable.
  7. Apr 20, 2006 #6
    You can graph it on any graphing calculator and use the ISECT (intersect) function to find where they interstect, and that's your x value solution.

    So you'd have:
    y1 = cosx
    y2= x/2 -1
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Area between curves