Antidifferentiating Rational Powers
Extend antidifferentiation to rational powers for Victorian Year 11 Mathematical Methods (VCAA) — reversing the power rule for negative and fractional indices, including surds and reciprocals.
You will learn to rewrite roots and reciprocals as powers, add one to the index and divide, avoid the special excluded case, and use a boundary condition to find the constant of integration.
Theory
In Year 11 Mathematical Methods (Unit 2), anti-differentiation reverses differentiation. For a power function you add one to the index and divide by the new index: \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+c\), valid for any rational \(n\neq-1\). This page shows how to antidifferentiate surds and reciprocal powers by first writing them as \(x^{n}\), how to handle coefficients and sums term by term, and how a given point fixes the constant \(c\).
Anti-differentiation is the reverse of differentiation. If \(f'(x)\) is the derivative (or gradient function), then an antiderivative \(f(x)\) is a function whose derivative is \(f'(x)\). Because the derivative of a constant is \(0\), every antiderivative carries an arbitrary constant of integration \(+c\).
For a power function \(x^{n}\), antidifferentiating raises the index by one and divides by the new index. This works for any rational \(n\) — positive, negative or fractional — with the single exception \(n=-1\), for which the rule would divide by zero.
The key skill is rewriting surds and reciprocals as powers first: \(\sqrt{x}=x^{1/2}\), \(\dfrac{1}{\sqrt{x}}=x^{-1/2}\), \(\sqrt[3]{x}=x^{1/3}\) and \(\dfrac{1}{x^{k}}=x^{-k}\). Then the power rule applies directly.
The power rule for anti-differentiation, for rational \(n\neq-1\):
A constant coefficient is kept out the front, and sums are done term by term:
Useful rewrites before applying the rule:
| Surd / fraction form | Power form \(x^{n}\) | Antiderivative |
|---|---|---|
| \(\sqrt{x}\) | \(x^{1/2}\) | \(\dfrac{2}{3}x^{3/2}+c\) |
| \(\dfrac{1}{\sqrt{x}}\) | \(x^{-1/2}\) | \(2\sqrt{x}+c\) |
| \(\sqrt[3]{x}\) | \(x^{1/3}\) | \(\dfrac{3}{4}x^{4/3}+c\) |
| \(\dfrac{1}{x^{2}}\) | \(x^{-2}\) | \(-\dfrac{1}{x}+c\) |
How to antidifferentiate a rational power
- Rewrite: turn every surd and fraction into \(x^{n}\) form (\(\sqrt{x}=x^{1/2}\), \(\dfrac{1}{x^{k}}=x^{-k}\)), splitting or expanding first if needed.
- Add one to the index: the new power is \(n+1\).
- Divide by the new index: write \(\dfrac{x^{n+1}}{n+1}\); dividing by a fraction means multiplying by its reciprocal (e.g. \(\div\tfrac{3}{2}=\times\tfrac{2}{3}\)).
- Keep coefficients, add \(+c\): multiply through by any constant and never forget the constant of integration.
- Use a condition (if given): substitute the known point to solve for \(c\), then answer the question.
Write the surd as a power: \(\sqrt{x}=x^{1/2}\).
Add one to the index and divide by the new index:
| \(f(x)\) | \(=\) | \(\int x^{1/2}\,dx\) |
| \(=\) | \(\dfrac{x^{1/2+1}}{1/2+1}+c\) | |
| \(=\) | \(\dfrac{x^{3/2}}{3/2}+c\) | |
| \(=\) | \(\dfrac{2}{3}x^{3/2}+c\) |
\(f(x)=\dfrac{2}{3}x^{3/2}+c\).
The index is \(n=-2\ (\neq-1)\), so the power rule applies. Keep the coefficient out the front.
Add one to the index, divide by the new index:
| \(f(x)\) | \(=\) | \(\int 3x^{-2}\,dx\) |
| \(=\) | \(3\cdot\dfrac{x^{-2+1}}{-2+1}+c\) | |
| \(=\) | \(3\cdot\dfrac{x^{-1}}{-1}+c\) | |
| \(=\) | \(-3x^{-1}+c\) | |
| \(=\) | \(-\dfrac{3}{x}+c\) |
\(f(x)=-\dfrac{3}{x}+c\).
Split the fraction into powers of \(x\):
| \(f'(x)\) | \(=\) | \(\dfrac{x^{2}}{x^{2}}+\dfrac{1}{x^{2}}\) |
| \(=\) | \(1+x^{-2}\) |
Antidifferentiate term by term:
| \(f(x)\) | \(=\) | \(\int\left(1+x^{-2}\right)dx\) |
| \(=\) | \(x+\dfrac{x^{-1}}{-1}+c\) | |
| \(=\) | \(x-\dfrac{1}{x}+c\) |
\(f(x)=x-\dfrac{1}{x}+c\).
The total cost is the antiderivative of the marginal cost. Write \(\dfrac{4}{\sqrt{x}}=4x^{-1/2}\).
Antidifferentiate the marginal cost:
| \(C(x)\) | \(=\) | \(\int 4x^{-1/2}\,dx\) |
| \(=\) | \(4\cdot\dfrac{x^{1/2}}{1/2}+c\) | |
| \(=\) | \(8x^{1/2}+c\) | |
| \(=\) | \(8\sqrt{x}+c\) |
Use \(C(4)=50\) to find \(c\):
| \(50\) | \(=\) | \(8\sqrt{4}+c\) |
| \(50\) | \(=\) | \(16+c\) |
| \(c\) | \(=\) | \(34\) |
Evaluate \(C(9)\):
| \(C(9)\) | \(=\) | \(8\sqrt{9}+34\) |
| \(=\) | \(24+34\) | |
| \(=\) | \(58\) |
The total cost of producing \(9\) items is \(\$58\).
Common pitfalls
Frequently asked questions
How do you antidifferentiate a power of x?
Add one to the index and divide by the new index: \(\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+c\), for any rational \(n\neq-1\).
How do you antidifferentiate a surd like the square root of x?
Rewrite it as a power first: \(\sqrt{x}=x^{1/2}\), then apply the rule to get \(\dfrac{2}{3}x^{3/2}+c\).
What do you do with a fraction like 1 over root x?
Write it as a negative fractional power, \(\dfrac{1}{\sqrt{x}}=x^{-1/2}\), then antidifferentiate to get \(2\sqrt{x}+c\).
Why can't you use the rule when n equals negative one?
The new index is \(n+1=0\), and you cannot divide by \(0\). So \(x^{-1}=\dfrac{1}{x}\) is the one power the rule does not cover.
Why is there a plus c, and how do you find its value?
Any two antiderivatives differ by a constant, so \(+c\) covers them all. A given point (an initial condition) lets you substitute and solve for \(c\).