Specific anti-derivatives
Form the family and use each condition on the quantity it describes
\[C=F(h)-P(h)\]
- Form the family first Find a primitive with zero constant term, then add the arbitrary constant \(\displaystyle F=P+C\)
- Substitute into the function A supplied function value belongs to the function, not its derivative. Use it to solve for the constant \(\displaystyle F(h)=y,\quad P(h)+C=y\)
- Use stationary information on the gradient At a stationary input, the gradient is zero. Determine an unknown gradient parameter before forming the family \(\displaystyle F'(h)=0\)
- 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 the derivative and supplied value, then evaluate it at three
\[F'(x)=2x-3,\quad F(2)=4\]
- \(\displaystyle F(x)=x^2-3x+C\)Form the complete family
- \(\displaystyle 4-6+C=4\quad\Rightarrow\quad C=6\)Substitute into F, not its gradient
- \(\displaystyle F(x)=x^2-3x+6\)State the recovered function
- \(\displaystyle F(3)=9-9+6=6\)Evaluate the unique recovered function
Common mistakes
Boundary applied to the gradient
✕Apply the function value to the derivative✓Apply the boundary value to the original function
The condition describes the original functionStationary always means turning
✕Classify every gradient zero as a maximum or minimum✓Check the gradient sign on both sides
A double gradient root can give stationary inflectionDifficulty levels
introInitial conditions
standardNonzero boundary inputs for cubic/quartic functions
challengeA stationary condition fixes a gradient parameter and then C
examPreserved senior boundary cases and derivative/value verification
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 11, VCE Units 1 & 2, Mathematical Methods, chapter 14: Anti-differentiation and integrals