Why Verify the Power Rule?
An antiderivative is a function whose derivative is the given function. The power rule for antiderivatives proposes a simple candidate: raise the exponent by one, then divide by the new exponent. Differentiation provides a direct way to check whether that candidate is correct.
This check also explains the rule’s restriction. When the original exponent is negative one, raising it by one gives zero, and division by the new exponent is not defined. That does not prove that \(x^{-1}\) has no antiderivative; it means only that this particular formula cannot provide one.
The Candidate and Its Conditions
For a fixed real number \(n\), consider the function \(x^n\). On the positive real numbers, real powers are defined and the derivative power rule gives \(\frac{d}{dx}x^r=rx^{r-1}\) for any real exponent \(r\). Using that derivative rule, we can test the proposed function \(x^{n+1}/(n+1)\).
The condition \(n\ne -1\) ensures that the denominator \(n+1\) is not zero. The stated domain \(x>0\) ensures that all real powers in the formula are defined using the usual real-valued power function. For some integer exponents, the formula also applies on other intervals where the relevant powers are defined and differentiable. The theorem as stated needs no such extension: it is valid for every real \(n\ne-1\) on \(x>0\).
Deriving the Formula by Differentiation
Keep \(n\) fixed, with \(n\ne-1\), and define \(F(x)=x^{n+1}/(n+1)+C\). The factor \(1/(n+1)\) is constant with respect to \(x\), as is \(C\). Differentiate using the constant-multiple rule, the fact that the derivative of a constant is zero, and the power rule for derivatives:
In the final step, the nonzero factor \(n+1\) cancels with its reciprocal, and \((n+1)-1=n\). Thus the derivative of the proposed function is exactly the original power \(x^n\). This verifies that it is an antiderivative on the stated domain.
The calculation shows why each part of the formula is needed. Raising the exponent from \(n\) to \(n+1\) makes differentiation produce the factor \(n+1\). Dividing by \(n+1\) cancels that factor. The constant \(C\) does not affect the derivative, since \(C'=0\).
Worked Examples
Worked Example: Verify the Rule for \(x^3\)
Here the exponent is \(n=3\), so \(n+1=4\). Since \(3\ne-1\), the rule proposes \(F(x)=\frac{x^4}{4}+C\) on \(x>0\). Differentiate each term:
The derivative is the original function \(x^3\), so \(x^4/4+C\) is verified as an antiderivative. As a direct check, substituting \(n=3\) into the rule gives \(x^{3+1}/(3+1)+C=x^4/4+C\), the same candidate we differentiated.
Worked Example: Verify the Rule for \(x^{1/2}\)
For the real exponent \(n=\frac12\), the denominator in the rule is \(n+1=\frac32\), which is nonzero. Dividing by \(\frac32\) is multiplying by its reciprocal \(\frac23\). The candidate is therefore \(F(x)=\frac23x^{3/2}+C\), for \(x>0\). Differentiate:
The reciprocal factors \(\frac23\) and \(\frac32\) multiply to \(1\), and the exponent reduces from \(\frac32\) to \(\frac12\). This confirms the candidate by differentiation on the positive domain.
Worked Example: Verify the Rule for a Constant Power
Take \(n=0\). The integrand is \(x^0=1\) on \(x>0\), and the formula proposes \(F(x)=\frac{x^{0+1}}{0+1}+C=x+C\). Its derivative is:
Thus the derivative returns the original integrand. The calculation also shows that the formula works when the original exponent is zero: the restriction is not \(n\ne0\), but \(n\ne-1\).
Why the Exponent \(-1\) Is Excluded
If \(n=-1\), then \(n+1=0\). Substitution into the proposed expression would require dividing by zero:
This expression is undefined, so it cannot be a candidate antiderivative. The differentiation verification also depends on cancelling \(n+1\) against \(1/(n+1)\), which is valid only when \(n+1\ne0\). Therefore the power rule for antiderivatives has the explicit restriction \(n\ne-1\). The exceptional case must be handled by a different rule, rather than by substituting \(-1\) into this formula.
It is important to read this restriction accurately. It says that the displayed power formula does not apply when \(n=-1\). It does not say that every other antiderivative method fails, nor that the function \(x^{-1}\) has no antiderivative. It marks the boundary of this particular rule.
A Common Mistake: Multiplying Instead of Dividing
A frequent error is to multiply by the new exponent instead of dividing by it. For example, an incorrect attempt for \(x^3\) might be:
Differentiate this proposed answer to check it. The derivative is \(4(4x^3)+0=16x^3\), not \(x^3\). The proposed function therefore fails the definition of an antiderivative. The factor \(4\) has made the result sixteen times too large.
The correction is to divide by \(4\), not multiply by it. The candidate \(x^4/4+C\) differentiates as \(\frac14(4x^3)+0=x^3\), so it passes the check.
A second error is to apply the formula mechanically at \(n=-1\), perhaps by writing \(x^0/0\). Since division by zero is undefined, this is not a valid expression. The correction is to stop, identify that the exponent is the excluded value, and use a separate method for that case.
How to Check a Proposed Antiderivative
Differentiation makes verification direct. Given a proposed antiderivative, compare its derivative with the original integrand. For the power rule, the main checks are that the exponent increases by exactly one, the coefficient is the reciprocal of that new exponent, and the exponent is not \(-1\).
Write the integrand as \(x^n\) and note the fixed value of \(n\).
Confirm that \(n\ne-1\), so the proposed denominator \(n+1\) is nonzero.
Apply the derivative power rule and the constant-multiple rule, then simplify the coefficient and exponent.
The verification succeeds precisely when the derivative simplifies to \(x^n\).
This procedure matters because a formula can look plausible while still having an incorrect coefficient or exponent. Checking the derivative tests the defining condition itself; it is not merely a comparison of appearances.
Check Your Understanding
Use differentiation to check each claim, and state the relevant restriction where needed.
- For \(n=5\), write the proposed antiderivative of \(x^5\), then differentiate it to verify the result.
- For \(n=\frac23\), what coefficient does the rule give for the proposed antiderivative of \(x^{2/3}\)? Verify the coefficient by differentiating.
- Explain why the factor \(n+1\) appears when differentiating \(x^{n+1}\), and how the proposed antiderivative cancels it.
- What goes wrong if \(n=-1\) is substituted into \(x^{n+1}/(n+1)\)?
- A student proposes \(3x^3+C\) as an antiderivative of \(x^2\). Differentiate the proposal, identify the error, and give the corrected candidate.
- State the domain and exponent conditions in the theorem verified here.
