Power Rule for Integration
The power rule for integration is a straightforward way to find the indefinite integral of a power of \(x\). To apply it, add 1 to the exponent and divide by the new exponent. The general formula is $$ \int x^\alpha\,dx=\frac{x^{\alpha+1}}{\alpha+1}+c $$ where \(\alpha\) is any real number other than \(-1\), and \(c\) is an arbitrary constant called the constant of integration.
This rule lets me find an antiderivative of functions of the form \(f(x)=x^\alpha\), provided that \(\alpha\neq-1\).
For any real exponent \(\alpha\neq-1\), the formula is valid when \(x>0\).
Note. The power rule is one of the most useful techniques in integral calculus. I think it's worth learning early because it makes many basic integrals easy to evaluate and provides a solid foundation for working with more complicated functions.
How to Use the Power Rule
I'll start with a simple example to show how the rule works.
Consider the indefinite integral of \(f(x)=x^2\).
$$ \int x^2\,dx $$
The exponent is 2. To integrate this function, I add 1 to the exponent, giving 3, and then divide the resulting power by 3.
The calculation is
$$ \int x^2\,dx=\frac{x^{2+1}}{2+1}+c=\frac{x^3}{3}+c $$
I also add an arbitrary real constant $ c $ to represent the entire family of antiderivatives.
Why does this rule work?
Integration and differentiation are closely related. Finding an indefinite integral means finding a function whose derivative is the original integrand.
To check my result, I differentiate the expression I obtained using the power rule for differentiation.
$$ D\left(\frac{x^3}{3}+c\right)=\frac{3x^2}{3}=x^2 $$
The derivative gives me the original integrand, \(x^2\). This confirms that the result is correct.
Special Cases of the Power Rule
The power rule also works when the exponent is not explicitly written or when it is a fraction. The following examples show how to handle these cases.
The Integral of x
First, I'll consider the integral of \(f(x)=x\).
$$ \int x\,dx $$
Although the exponent is not written, it is equal to 1 because \(x=x^1\).
I add 1 to the exponent and divide by the new exponent.
$$ \int x\,dx=\frac{x^2}{2}+c $$
So the antiderivative of \(x\) is \(\frac{x^2}{2}\), plus an arbitrary constant.
The Integral of 1
Next, I'll consider the constant function \(f(x)=1\).
$$ \int 1\,dx $$
Since \(x^0=1\) for \(x\neq0\), I can rewrite the integrand as a power and apply the same rule.
$$ \int x^0\,dx=\frac{x^{0+1}}{0+1}+c = \frac{x^1}{1}+c = x+c $$
Therefore, the indefinite integral of \(f(x)=1\) is
$$ \int 1\,dx=x+c $$
This result is also valid at \(x=0\), since the derivative of \(x+c\) is 1 for every real value of \(x\).
The Integral of a Constant Other Than 1
What happens if the constant is not 1? The method is almost identical.
For example, consider the constant function \(f(x)=5\).
$$ \int 5\,dx $$
I can rewrite 5 as \(5\cdot x^0\), since \(x^0=1\) for \(x\neq0\).
$$ \int 5x^0\,dx $$
Using the constant multiple rule for integration, which follows from the linearity of integration, I can factor the constant 5 out of the integral.
$$ 5\int x^0\,dx $$
Now I apply the power rule.
$$ 5\int x^0\,dx =5\cdot\frac{x^{0+1}}{0+1}+c =5x+c $$
Therefore,
$$ \int 5\,dx=5x+c $$
This result holds for every real value of \(x\), including \(x=0\).
More generally, the indefinite integral of any constant function \(f(x)=k\), where \(k\in\mathbb{R}\), is
$$ \int k\,dx=kx+c $$
The reason is simple: the derivative of \(kx+c\) is always \(k\).
The Integral of the Square Root of x
The power rule also applies to fractional exponents. To see how, I'll calculate the indefinite integral of \(f(x)=\sqrt{x}\).
$$ \int\sqrt{x}\,dx $$
The first step is to rewrite the square root as a power with a fractional exponent.
Recall that
$$ \sqrt{x}=x^{1/2} $$
I can therefore rewrite the integral as
$$ \int x^{1/2}\,dx $$
Now I add 1 to the exponent.
$$ \frac12+1=\frac32 $$
Next, I divide the resulting power by the new exponent.
$$ \int x^{1/2}\,dx=\frac{x^{3/2}}{3/2}+c $$
Dividing by a fraction is equivalent to multiplying by its reciprocal, so I obtain
$$ \int\sqrt{x}\,dx=\frac23x^{3/2}+c $$
I can also express the result using radicals, since $ x^{3/2}=\sqrt{x^3} $.
$$ \int\sqrt{x}\,dx=\frac23\sqrt{x^3}+c $$
The two expressions are equivalent for \(x\geq0\). The antiderivative is differentiable for \(x>0\), and its formula extends continuously to \(x=0\).
Why the Power Rule Does Not Work for an Exponent of -1
There is one important exception to remember: the power rule cannot be used when the exponent is \(-1\).
For example, consider
$$ \int x^{-1}\,dx=\int\frac1x\,dx $$
If I tried to apply the usual power rule, the denominator would become
$$ -1+1=0 $$
Since division by zero is undefined, the formula cannot be used.
Instead, I need the logarithmic integration formula.
$$ \int\frac1x\,dx=\ln|x|+c $$
The result is the natural logarithm of the absolute value of \(x\), plus the constant of integration.
This formula is valid on any interval that does not contain zero, since the function \(1/x\) is undefined at \(x=0\).
Using the Power Rule to Integrate Polynomials
One of the most practical uses of the power rule is finding indefinite integrals of polynomial functions.
A polynomial is a sum of terms, each consisting of a coefficient multiplied by a nonnegative integer power of \(x\). Because integration is linear, I can integrate each term separately and then add the results.
For example, consider the polynomial
$$ f(x)=3x^2+2x+5 $$
I want to calculate its indefinite integral.
$$ \int(3x^2+2x+5)\,dx $$
First, I separate the terms using the sum rule and the constant multiple rule.
$$ \int(3x^2+2x+5)\,dx=3\int x^2\,dx+2\int x\,dx+5\int 1\,dx $$
Next, I apply the power rule to each integral.
$$ \int(3x^2+2x+5)\,dx=3\cdot\frac{x^3}{3}+2\cdot\frac{x^2}{2}+5x+c $$
After simplifying, I obtain
$$ \int(3x^2+2x+5)\,dx=x^3+x^2+5x+c $$
This approach works for any polynomial. I integrate each term using the power rule and combine the results.
Only one constant of integration is needed because the sum of arbitrary constants is itself an arbitrary constant.
Derivation of the Power Rule for Integration
To understand where the formula comes from, I'll derive it using the power rule for differentiation.
Recall that
$$ D(x^n)=nx^{n-1} $$
Since antidifferentiation reverses differentiation, I need to find a function whose derivative is \(x^\alpha\).
Consider the function
$$ F(x)=\frac{x^{\alpha+1}}{\alpha+1} $$
I differentiate it using the power rule.
$$ F'(x)=\frac{1}{\alpha+1}\cdot(\alpha+1)x^\alpha $$
The factors \(\alpha+1\) cancel, leaving
$$ F'(x)=x^\alpha $$
This shows that \(F(x)\) is an antiderivative of \(x^\alpha\).
By adding the arbitrary constant \(c\), I obtain the general formula.
$$ \int x^\alpha\,dx=\frac{x^{\alpha+1}}{\alpha+1}+c $$
The derivation is valid whenever \(\alpha\neq-1\). For an arbitrary real exponent, the formula holds on the interval \(x>0\).
And so on.
