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Antiderivatives and Basic Integrals · Tutorial 1054 of 1495

Integrating Sums And Constant Multiples

Apply the power rule term by term to sums and constant multiples, then verify each antiderivative by differentiating.

Beginner 8 min read

What You'll Learn

  • Integrate each term of a polynomial separately.
  • Carry constant coefficients through the power rule correctly.
  • Apply linearity to sums, differences, and constant multiples.
  • Write the full family of antiderivatives with one arbitrary constant.
  • Check an answer by differentiating it.
  • Identify and correct common coefficient and integration-constant errors.

From One Power to a Whole Expression

Finding an antiderivative of a single power is only the first step in integrating polynomial expressions. A polynomial may contain several powers of \(x\), each with its own coefficient, and those terms may be added or subtracted. The practical method is to handle each term separately, keeping its coefficient attached to it throughout the calculation.

For a nonnegative integer \(n\), the power rule for antiderivatives gives an antiderivative of \(x^n\) by increasing the exponent by one and dividing by the new exponent. A constant multiple of \(x^n\) keeps that constant multiple. The same process applies term by term to a finite sum.

Rule: If \(a_0,a_1,\ldots,a_m\) are real constants and \(m\) is a nonnegative integer, then \(\int \left(a_0+a_1x+\cdots+a_mx^m\right)\,dx =\frac{a_0}{1}x+\frac{a_1}{2}x^2+\cdots+\frac{a_m}{m+1}x^{m+1}+C\). Here \(C\) is an arbitrary real constant. The rule applies to a finite sum of nonnegative integer powers on any interval in \(\mathbb R\).

How to Handle Each Term

For each term \(ax^n\), where \(a\) is a constant and \(n\) is a nonnegative integer, apply the power rule to the power and retain the coefficient:

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

Thus, the coefficient is divided by \(n+1\), while the exponent increases from \(n\) to \(n+1\). The terms can then be added back together. Subtraction causes no change in the method: a negative coefficient simply stays negative during the calculation.

The result of an indefinite integral is a family of functions, not a single function. After integrating all the terms, include one arbitrary constant \(C\). Writing separate constants for each term is unnecessary, since adding several constants produces just one constant.

Worked Example: Integrating a Three-Term Polynomial

Find all antiderivatives of \(6x^2+4x+5\).

First, separate the terms and keep each coefficient with its power:

$$ \int(6x^2+4x+5)\,dx = \int 6x^2\,dx+\int 4x\,dx+\int 5\,dx. $$

For \(6x^2\), increase the exponent \(2\) to \(3\) and divide the coefficient by \(3\):

$$ \int 6x^2\,dx = \frac{6}{2+1}x^{2+1} = \frac{6}{3}x^3 = 2x^3. $$

For \(4x=4x^1\), increase the exponent \(1\) to \(2\) and divide the coefficient by \(2\):

$$ \int 4x\,dx = \frac{4}{1+1}x^{1+1} = \frac{4}{2}x^2 = 2x^2. $$

The constant \(5\) is \(5x^0\), so its antiderivative is \(5x\):

$$ \int 5\,dx = \frac{5}{0+1}x^{0+1} = 5x. $$

Combining the terms and adding the arbitrary constant gives

$$ \int(6x^2+4x+5)\,dx = 2x^3+2x^2+5x+C. $$

Check by differentiating the result:

$$ \frac{d}{dx}(2x^3+2x^2+5x+C) = 6x^2+4x+5. $$

This is the original integrand, so the antiderivative is verified.

Why the Term-by-Term Rule Works

The method follows from two facts: the power rule for differentiating \(x^{n+1}\), and the linearity of differentiation. Specifically, differentiation distributes over finite sums and constant multiples. We can therefore check the proposed antiderivative for a general polynomial directly.

Justification: Let \(p(x)=\sum_{k=0}^{m}a_kx^k\), where \(m\) is a nonnegative integer and each \(a_k\) is real. Define \(F(x)=\sum_{k=0}^{m}\frac{a_k}{k+1}x^{k+1}\). Differentiating each term gives \(F'(x)=\sum_{k=0}^{m}\frac{a_k}{k+1}(k+1)x^k=\sum_{k=0}^{m}a_kx^k=p(x)\). Adding a constant \(C\) does not change the derivative, so every function \(F(x)+C\) is an antiderivative. Conversely, on an interval, any two antiderivatives of the same function differ by a constant. Thus the displayed formula gives the full family of antiderivatives.

This verification explains why it is legitimate to work one term at a time: after differentiation, each term returns to its matching term in the original polynomial. The formula requires a finite sum and nonnegative integer powers in this tutorial’s setting. Each exponent \(k\) then becomes \(k+1\), which is nonzero, so the division in the rule is defined.

Coefficients, Signs, and Common Factors

Coefficients need not be positive integers. A negative coefficient remains negative, and a fractional coefficient is divided by the new exponent in the same way. If an expression has a common constant factor, you may either keep it attached to each term or factor it outside the integral. Both approaches give the same result, provided the factor is included exactly once.

Worked Example: A Difference with a Negative Coefficient

Find all antiderivatives of \(8x^3-6x^2+2x\).

Integrate the three terms separately. The first term gives

$$ \int 8x^3\,dx = \frac{8}{3+1}x^{3+1} = \frac{8}{4}x^4 = 2x^4. $$

The second term has coefficient \(-6\). Keep its sign while increasing the exponent and dividing by \(3\):

$$ \int -6x^2\,dx = \frac{-6}{2+1}x^{2+1} = \frac{-6}{3}x^3 = -2x^3. $$

For the third term, increase the exponent \(1\) to \(2\) and divide \(2\) by \(2\):

$$ \int 2x\,dx = \frac{2}{1+1}x^{1+1} = \frac{2}{2}x^2 = x^2. $$

Putting the results together gives

$$ \int(8x^3-6x^2+2x)\,dx = 2x^4-2x^3+x^2+C. $$

Differentiate to check every term, including its sign:

$$ \frac{d}{dx}(2x^4-2x^3+x^2+C) = 8x^3-6x^2+2x. $$

The derivative matches the original expression, so the calculation is correct.

When a number multiplies a sum, it can be distributed first or kept as a common factor during integration. In either case, the power rule applies to each power, and the constant multiplier must remain part of the calculation.

Worked Example: Keeping a Constant Factor

Find all antiderivatives of \(3(2x^2-x+4)\).

One approach is to distribute the \(3\) before integrating:

$$ 3(2x^2-x+4) = 6x^2-3x+12. $$

Now integrate each term:

$$ \begin{aligned} \int 6x^2\,dx &=\frac{6}{3}x^3=2x^3,\\ \int -3x\,dx &=\frac{-3}{2}x^2=-\frac{3}{2}x^2,\\ \int 12\,dx &=\frac{12}{1}x=12x. \end{aligned} $$

Therefore, the family of antiderivatives is

$$ 2x^3-\frac{3}{2}x^2+12x+C. $$

Alternatively, keep the factor \(3\) outside and integrate the terms inside:

$$ 3\left(x^3-\frac{1}{2}x^2+4x\right)+C = 3x^3-\frac{3}{2}x^2+12x+C. $$

Notice that this last expression differs from the first by more than a constant: the \(x^3\) coefficient would be \(3\), not \(2\). The error is that the antiderivative of \(2x^2\) inside the parentheses is \( \frac{2}{3}x^3\), not \(x^3\). Carrying out the inside calculation correctly gives

$$ 3\left(\frac{2}{3}x^3-\frac{1}{2}x^2+4x\right)+C = 2x^3-\frac{3}{2}x^2+12x+C. $$

Check the correct result by differentiation:

$$ \frac{d}{dx}\left(2x^3-\frac{3}{2}x^2+12x+C\right) = 6x^2-3x+12 = 3(2x^2-x+4). $$

The original expression is recovered.

Common Mistakes and Their Corrections

A frequent error is to increase the power but forget to divide the coefficient by the new exponent. For example, the following calculation is wrong:

$$ \int 6x^2\,dx \ne 6x^3. $$

Differentiating the proposed answer shows the problem: \(\frac{d}{dx}(6x^3)=18x^2\), which is not \(6x^2\). The correction is to divide \(6\) by \(2+1=3\):

$$ \int 6x^2\,dx = \frac{6}{3}x^3 = 2x^3, \qquad \frac{d}{dx}(2x^3)=6x^2. $$

Another common error is to omit \(C\). Without it, you have found one antiderivative, but not the entire family. For instance, \(2x^3+2x^2+5x\) is an antiderivative of \(6x^2+4x+5\), but so is \(2x^3+2x^2+5x+7\). The notation \(+C\) represents all such constant differences.

Takeaway: For every term \(ax^n\), use \(\frac{a}{n+1}x^{n+1}\). Preserve the sign and coefficient, combine the resulting terms, include one arbitrary constant, and differentiate the result to check it.

Check Your Understanding

For each question, apply the rule term by term and check your proposed antiderivative by differentiating.

  1. Find all antiderivatives of \(5x^2+3x+1\).
  2. Find all antiderivatives of \(9x^3-8x\).
  3. Find all antiderivatives of \(4(3x^2+2x-1)\). Show the distributed integrand before integrating.
  4. A student writes \(\int 10x^4\,dx=10x^5+C\). Identify the error and give the corrected family of antiderivatives.
  5. Explain why integrating a finite sum term by term is valid by describing what happens when the proposed result is differentiated.

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