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

Question

6853 views around the world

You can reuse this answer
Creative Commons License

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.

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