Question

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

Answer 1

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.