How do you find the constant of integration for #intf'(x)dx# if #f(2)=1#?

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Answer 1 · 2014-08-19T18:42:34

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.