Initial value problems

Given a function, you can find the most general antiderivative, but the constant is arbitrary without more information. 

Example:  DO:  Find the most general antiderivative of $f(x)=5x^2 + 2x - 3$.  (Don't look ahead until finished!)

Solution:  You should have found $$F(x)=\frac{5}{3}x^3 + x^2-3x+C.$$  

DO:  Differentiate to see if this is right.  (In fact, differentiate to check your work every time you antidifferentiate.)

Suppose now we ask you to find the antiderivative $F(x)$ of $f(x)=5x^2+2x-3$ given that $F(3)=30$, which we call an initial condition or initial value. 

Given $F(x)$ that you found above and the initial condition, you can now find the value of $C$ and thus find the specific antiderivative desired.  To do this, plug $3$ into your general antiderivative $F$ above, which you know is equal to $30$.
$$F(3)=\frac{5}{3}*3^3+3^2-3*3+C=5*9+9-9+C=45+C=30 \text{ <--- initial condition}$$
so $C=-15$.  Our specific antiderivative is $$F(x)=\frac{5}{3}x^3 + x^2-3x-15.$$

DO: Check to see that your initial condition holds for $F$, and that you get $f$ when you differentiate.

Such an exercise might also be stated as:

Example:  Find $g(x)$ given that $g'(x)=12x^5 + 2x - 3$ and $g(1)=2$. 

DO:  Think this different notation through carefully.  Then work this slightly different problem yourself before looking further.  Check your work by differentiating.

Solution:  $g(x)=2x^6+x^2-3x+2$.