Antidifferentiation of Polynomial Functions
Learn antidifferentiation for Victorian Year 11 Mathematical Methods (VCAA) — the reverse of differentiation, finding a function back from its gradient function.
You will learn to reverse the power rule term by term, always include the constant of integration, and use a known point to pin down that constant and recover the original polynomial function.
Theory
In Year 11 Mathematical Methods (Unit 2), antidifferentiation reverses differentiation: given the derivative \(f'(x)\), you recover the original function \(f(x)\). For a power \(x^n\) you use the reverse power rule — raise the index by one and divide by the new index — and always add a constant of integration \(+c\). This page shows how to antidifferentiate polynomials term by term, and how a boundary condition pins down the value of \(c\).
An antiderivative (or indefinite integral) of \(f'(x)\) is a function \(f(x)\) whose derivative is \(f'(x)\). Because the derivative of any constant is \(0\), a derivative does not remember the constant term — so there is a whole family of antiderivatives that differ only by a constant.
We record this family with the constant of integration \(+c\). For example, since \(\dfrac{d}{dx}(x^2)=2x\), the general antiderivative of \(2x\) is \(\displaystyle\int 2x\,dx=x^2+c\).
To reverse a power we use the reverse power rule: \(\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+c\) (for \(n\neq-1\)). A polynomial is antidifferentiated one term at a time, keeping the constant multipliers.
Reverse power rule (raise the index, divide by the new index):
Constant and constant-multiple:
Term by term (antidifferentiate each term and add):
How to antidifferentiate a polynomial
- Term by term: split the polynomial into separate power terms.
- Reverse power rule: for each \(ax^n\), raise the index to \(n+1\) and divide by \(n+1\), giving \(\dfrac{a\,x^{n+1}}{n+1}\). A constant \(k\) becomes \(kx\).
- Add \(+c\): write a single constant of integration for the whole answer.
- Apply the condition (if given): substitute the known point into \(f(x)\), solve for \(c\), then write the particular function — and evaluate it where asked.
Set up the integral of \(f'(x)\):
| \(f(x)\) | \(=\) | \(\displaystyle\int (3x^2-4x+1)\,dx\) |
Reverse power rule — each term (raise the index, divide by the new index):
| \(=\) | \(3\times\dfrac{x^{3}}{3}-4\times\dfrac{x^{2}}{2}+1x+c\) |
Simplify the coefficients:
| \(=\) | \(x^3-2x^2+x+c\) |
General antiderivative: \(f(x)=x^3-2x^2+x+c\).
Antidifferentiate to get the general form:
| \(f(x)\) | \(=\) | \(\displaystyle\int (4x-6)\,dx\) |
| \(=\) | \(2x^2-6x+c\) |
Substitute the condition \(f(1)=10\):
| \(2(1)^2-6(1)+c\) | \(=\) | \(10\) |
| \(2-6+c\) | \(=\) | \(10\) |
| \(-4+c\) | \(=\) | \(10\) |
| \(c\) | \(=\) | \(14\) |
So \(c=14\), giving \(f(x)=2x^2-6x+14\).
Antidifferentiate to find the general form:
| \(f(x)\) | \(=\) | \(\displaystyle\int 3x^2\,dx\) |
| \(=\) | \(x^3+c\) |
Find \(c\) using \(f(2)=10\):
| \((2)^3+c\) | \(=\) | \(10\) |
| \(8+c\) | \(=\) | \(10\) |
| \(c\) | \(=\) | \(2\) |
So \(f(x)=x^3+2\).
Evaluate \(f(3)\):
| \(f(3)\) | \(=\) | \((3)^3+2\) |
| \(=\) | \(27+2\) | |
| \(=\) | \(29\) |
\(f(3)=29\).
Antidifferentiate the gradient:
| \(y\) | \(=\) | \(\displaystyle\int (3x^2-2)\,dx\) |
| \(=\) | \(x^3-2x+c\) |
Use the point \((1,\,4)\) to find \(c\):
| \((1)^3-2(1)+c\) | \(=\) | \(4\) |
| \(1-2+c\) | \(=\) | \(4\) |
| \(-1+c\) | \(=\) | \(4\) |
| \(c\) | \(=\) | \(5\) |
Equation of the curve: \(y=x^3-2x+5\).
Common pitfalls
Frequently asked questions
What is antidifferentiation?
It is the reverse of differentiation: given \(f'(x)\), you find a function \(f(x)\) whose derivative is \(f'(x)\).
What is the reverse power rule?
To antidifferentiate \(x^n\), raise the index by one and divide by the new index: \(\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+c\) (for \(n\neq-1\)).
Why do we add a constant of integration?
The derivative of any constant is \(0\), so infinitely many functions share the same derivative. The \(+c\) records that whole family.
How do I antidifferentiate a polynomial?
Antidifferentiate each term separately with the reverse power rule, keep the coefficients, and add a single \(+c\) at the end.
How do I find the value of c?
Substitute the given point (a boundary condition such as \(f(1)=10\)) into the antiderivative \(f(x)\) and solve the resulting equation for \(c\).