Specific anti-derivatives
Form the family and use each condition on the quantity it describes
- Form the family first Find a primitive with zero constant term, then add an arbitrary constant
- Substitute into the function Apply a supplied function value to the original function, then solve for the additive constant
- Use stationary information on the gradient At a stationary input h, F′(h)=0. Determine an unknown gradient parameter before forming the family
- Verify both conditions Differentiate the chosen function and check its supplied value. Matching only the gradient leaves the constant undetermined
Worked example
Find the function from its derivative and supplied value, then evaluate it at three
- \(\displaystyle F(x)=x^2-3x+C\)Form the complete family
- \(\displaystyle 4-6+C=4\quad\Rightarrow\quad C=6\)Substitute into F, not its gradient
- \(\displaystyle F(3)=9-9+6=6\)Evaluate the unique recovered function
Common mistakes
Difficulty levels
Questions about this topic
Why can C differ between expanded and bracket forms?+
C refers to the displayed base expression. Expanding a bracket may contribute a fixed constant, so compare the complete functions rather than the isolated C values
Why is one boundary value enough?+
All primitives of a polynomial on a connected interval differ by a constant. One function value fixes that constant
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 7: Antiderivatives