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Antiderivatives and Substitution · Tutorial 489 of 999

Termwise Antiderivatives of Polynomial Expressions

Apply the power rule term by term to polynomial expressions, then verify the resulting antiderivative family by differentiating.

Intermediate 9 min read

What You'll Learn

  • Recognize a polynomial as a finite sum of nonnegative integer powers.
  • Integrate each polynomial term with the power rule for antiderivatives.
  • Combine the separate integration constants into one arbitrary constant.
  • Handle constant terms and missing powers correctly.
  • Check a proposed polynomial antiderivative by differentiating it.
  • Identify common errors involving exponents, coefficients, and constants.

Why Integrate a Polynomial Term by Term?

A polynomial may contain several powers of its variable, but its terms can be handled separately. This works because the derivative of a sum is the sum of the derivatives. To find an antiderivative, we can reverse that process: apply the power rule to each term, then combine the results into one family.

This method is useful because it turns one expression into a short sequence of familiar power-rule steps. It also makes the result easy to check: differentiate the proposed antiderivative and confirm that each original term returns.

Polynomials and the Power Rule

A polynomial in \(x\) is a finite sum of terms whose powers of \(x\) are nonnegative integers. A general polynomial can be written as

$$ P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0, $$

where the coefficients \(a_0,a_1,\ldots,a_n\) are real numbers and \(n\) is a nonnegative integer. Some coefficients may be zero, so a polynomial does not have to contain every power between \(0\) and \(n\). A constant is also a polynomial: it is the case in which there are no positive powers of \(x\).

For a nonnegative integer \(k\), the power rule for antiderivatives says that

$$ \int x^k\,dx=\frac{x^{k+1}}{k+1}+C. $$

Here \(C\) is an arbitrary real constant. The rule applies because \(k+1\ne 0\). For a term with coefficient \(a\), the constant multiple rule gives

$$ \int ax^k\,dx = a\int x^k\,dx = \frac{a}{k+1}x^{k+1}+C. $$

Thus, each term keeps its coefficient, its exponent increases by one, and its coefficient is divided by the new exponent.

Definition: A termwise antiderivative of a polynomial is formed by finding an antiderivative of each term and adding the results. The full family includes one arbitrary constant.

The Termwise Rule

For a polynomial written in powers of \(x\), the termwise rule is

$$ \int \left(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\right)\,dx = \frac{a_n}{n+1}x^{n+1} +\frac{a_{n-1}}{n}x^n +\cdots +\frac{a_1}{2}x^2 +a_0x+C. $$

For the constant term \(a_0\), the power is \(x^0\). Applying the same rule gives \(a_0x^{0+1}/(0+1)=a_0x\). This is why a constant term becomes a linear term.

There is only one arbitrary constant in the final family. If each term is integrated with its own constant, their sum is still just one arbitrary constant: for example, \(C_1+C_2+C_3\) can be renamed \(C\). Keeping a single \(C\) makes the family clear without changing its meaning.

Why the Rule Works

The justification follows from the derivative power rule and the linearity of differentiation. For each term \(a_kx^k\), the power rule gives an antiderivative \(\frac{a_k}{k+1}x^{k+1}\). Its derivative is

$$ \frac{d}{dx}\left(\frac{a_k}{k+1}x^{k+1}\right) = \frac{a_k}{k+1}(k+1)x^k = a_kx^k. $$

Now add these antiderivatives for all the terms. Because the derivative of a finite sum is the sum of the derivatives, the derivative of the resulting expression is the sum of the original terms, which is \(P(x)\). Adding a constant does not change that derivative.

Theorem: If \(P(x)=\sum_{k=0}^{n}a_kx^k\) is a polynomial, then its complete family of antiderivatives on the real line is $$ F(x)=\sum_{k=0}^{n}\frac{a_k}{k+1}x^{k+1}+C, $$ where \(C\) is any real number.

To check the formula directly, differentiate the sum:

$$ F'(x) = \sum_{k=0}^{n} \frac{a_k}{k+1}(k+1)x^k = \sum_{k=0}^{n}a_kx^k = P(x). $$

So the displayed expression is an antiderivative. It also gives the complete family: any two antiderivatives of the same function on an interval differ by a constant. In this case, their difference has derivative zero, so it is constant; therefore no other forms need to be added to the family.

Worked Examples

Worked Example: Integrating a Polynomial with Four Terms

Find the general antiderivative of \(6x^3-4x^2+5x-8\).

Separate the terms and apply the power rule to each one:

$$ \begin{aligned} \int (6x^3-4x^2+5x-8)\,dx &=6\int x^3\,dx-4\int x^2\,dx+5\int x\,dx-8\int 1\,dx\\ &=6\left(\frac{x^4}{4}\right)-4\left(\frac{x^3}{3}\right) +5\left(\frac{x^2}{2}\right)-8x+C\\ &=\frac{3}{2}x^4-\frac{4}{3}x^3+\frac{5}{2}x^2-8x+C. \end{aligned} $$

Check by differentiating each term:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{3}{2}x^4-\frac{4}{3}x^3+\frac{5}{2}x^2-8x+C\right) &=6x^3-4x^2+5x-8. \end{aligned} $$

The derivative matches the original polynomial, so the antiderivative family is correct.

Worked Example: Integrating with a Fractional Coefficient

Find the general antiderivative of \(-\frac{3}{2}x^5+4x^2-7\).

Integrate each term. For the constant \(-7\), use \(-7=-7x^0\):

$$ \begin{aligned} \int \left(-\frac{3}{2}x^5+4x^2-7\right)\,dx &=-\frac{3}{2}\left(\frac{x^6}{6}\right) +4\left(\frac{x^3}{3}\right)-7x+C\\ &=-\frac{3}{12}x^6+\frac{4}{3}x^3-7x+C\\ &=-\frac{1}{4}x^6+\frac{4}{3}x^3-7x+C. \end{aligned} $$

Differentiate the result to check each coefficient and power:

$$ \begin{aligned} \frac{d}{dx}\left(-\frac{1}{4}x^6+\frac{4}{3}x^3-7x+C\right) &=-\frac{6}{4}x^5+\frac{12}{3}x^2-7\\ &=-\frac{3}{2}x^5+4x^2-7. \end{aligned} $$

This is the original integrand, confirming the answer.

Worked Example: A Polynomial with Missing Powers

Find the general antiderivative of \(2x^4-6x+9\). The \(x^3\) and \(x^2\) terms are absent; they can be viewed as terms with coefficient zero.

Integrate the terms that are present:

$$ \begin{aligned} \int (2x^4-6x+9)\,dx &=2\left(\frac{x^5}{5}\right)-6\left(\frac{x^2}{2}\right)+9x+C\\ &=\frac{2}{5}x^5-3x^2+9x+C. \end{aligned} $$

Verify by differentiation:

$$ \begin{aligned} \frac{d}{dx}\left(\frac{2}{5}x^5-3x^2+9x+C\right) &=\frac{10}{5}x^4-6x+9\\ &=2x^4-6x+9. \end{aligned} $$

The missing powers cause no extra terms: their coefficients are zero, and their contributions to the antiderivative are zero.

A Reliable Termwise Process

When integrating a polynomial, keep the coefficient, exponent, and constant term visible at every step. The following sequence prevents most arithmetic errors:

1
Identify the terms.
Read the expression as a sum or difference of constant multiples of powers of the variable.
2
Apply the power rule to each term.
Increase each exponent by one and divide its coefficient by that new exponent. A constant term becomes a multiple of \(x\).
3
Combine the results.
Write the termwise antiderivatives as one polynomial and include one arbitrary constant \(C\).
4
Differentiate to check.
Confirm that every resulting term matches the corresponding term in the original polynomial.

Common Mistakes and Their Corrections

A frequent error is to increase the exponent but forget to divide by the new exponent. For example, an incorrect calculation might say

$$ \int 3x^2\,dx=3x^3+C. $$

This is wrong because differentiating \(3x^3\) gives \(9x^2\), not \(3x^2\). The correction is to increase the power from \(2\) to \(3\) and divide the coefficient by \(3\):

$$ \int 3x^2\,dx = 3\left(\frac{x^3}{3}\right)+C = x^3+C, \qquad \frac{d}{dx}(x^3+C)=3x^2. $$

Another error is to treat a constant as though its antiderivative were still constant. A constant \(a\) is \(ax^0\), so its antiderivative is \(ax\), since \(\frac{d}{dx}(ax)=a\). Also avoid writing a separate arbitrary constant for every term in the final result. If several constants appear during the calculation, their sum is one arbitrary constant, so the answer should end with a single \(+C\).

Why This Method Matters

Termwise integration is a direct use of the linearity of the derivative in reverse. It gives an efficient method for a broad class of expressions and a clear way to check the result. The essential verification is not that the final polynomial “looks right,” but that differentiating it recovers every coefficient and power of the original polynomial.

For polynomial expressions, the method always applies: there are finitely many terms, and each term has a nonnegative integer power. The exponent increases by one, the coefficient is divided by that new exponent, and the one arbitrary constant accounts for the full family of antiderivatives.

Check Your Understanding

For each question, apply the termwise rule and use differentiation to check your result.

  1. Find the general antiderivative of \(5x^3-2x+4\).
  2. What is the antiderivative of the constant polynomial \(6\)?
  3. Find the general antiderivative of \(-4x^4+3x^2-1\).
  4. Explain why the final family contains one arbitrary constant rather than a separate constant for every term.
  5. A student claims \(\int 2x^3\,dx=2x^4+C\). Identify the error and correct the calculation.

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