Evaluating the Constant of Integration

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Primitive Functions: Evaluating the Constant

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Key Questions

  • If #F(x)# is an antiderivative of a function #f(x)#, that is,

    #F'(x)=f(x)#,

    then

    #G(x)=F(x)+C#, where #C# is any constant,

    is also an antiderivative of #f(x)# since

    #G'(x)=[F(x)+C]'=F'(x)=f(x)#.

    Hence, there are a family of functions (only differ by a constant) that are antiderivatives of #f(x)#. In order to include all antiderivatives of #f(x)#, the constant of integration #C# is used for indefinite integrals.

    #int f(x)dx=F(x)+C#

    The importance of #C# is that it allows us to express the general form of antiderivatives.


    I hope that this was helpful.

  • If #G(x)# is an antiderivative of #f(x)# and #G(x_0)=y_0#, then

    #G(x)=int f(x)dx=F(x)+C#,

    where #F(x)# is any antiderivative of #f(x)#.

    Since

    #G(x_0)=F(x_0)+C=y_0#,

    we have

    #C=y_0-F(x_0)#.


    I hope that this was helpful.

  • Normally, we want this integral function to be specified with a capital #f#, so that we can specify the antiderivative as #f(x)#.

    However, using your variable naming, let's say that #F(x)# is the antiderivative of #f'(x)#, then by the Net Change Theorem, we have:

    #f(x)=F(x)+C#

    Therefore, the constant of integration is:

    #C=f(x)-F(x)#
    #=f(2)-F(2)#
    #=1-F(2)#

    This is a simple answer, however for many students, it is very difficult to this this abstractly. So, let's look at a concrete example:

    #F(x)=x^3# to match your variables
    #F'(x)=f'(x)=3x^2# to match your variables
    #f(x)=int 3x^2 dx#
    #=x^3+C#
    #=F(x)+C#

    Now, substitute the given values:

    #f(2)=x^3+C=1#
    #2^3+C=1#
    #F(2)+C=1#
    #C=1-F(2)#

    So, if an abstract problem makes it difficult for you to find a solution, start with a concrete one to help you find a pattern.

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