Symmetrical Areas

Key Questions

  • How do you find the area under the graph of #f(x)=x^3# on the interval #[-1,1]# ?

    Well, we need to be careful when you say "under the graph" since #f(x)=x^3# goes below the x-axis when #x<0#, but you meant the region between the graph and the x-axis, then the area of the region is 1/2.

    Since there are two regions: one from #x=-1# to #x=0# and the other from #x=0# to #x=1#, the area #A# can be found by
    #A=int_{-1}^0(0-x^3)dx+int_0^1(x^3-0)dx#
    by Power Rule,
    #=[-x^4/4]_{-1}^0+[x^4/4]_0^1=1/4+1/4=1/2#

    Wataru · · Sep 11 2014
  • What are the symmetry properties of finding areas using integrals?

    If #f# is an even function (symmetric about the y-axis), then

    #int_{-a}^a f(x) dx=2int_0^a f(x) dx#.

    If #f# is an odd function (symmetric about the origin), then

    #int_{-a}^a f(x) dx=0#.

    Symmetries can be used to simplify computation of definite integrals. Let us look at the following examples.

    Example 1 (Even Function)

    #int_{-1}^1(3x^2+1) dx =2int_0^1(3x^2+1) dx=2[x^3+x]_0^1=2(2-0)=4#

    Example 2 (Odd Function)

    #int_{-pi/3}^{pi/3}{sin theta}/{sqrt{cos^2 theta+1}} d theta=0#

    I hope that this was helpful.

    Wataru · · Sep 28 2014

Questions

Using Integrals to Find Areas and Volumes