Include the integration constant

Raise the curve, keep the slope

The top graph shows \(f(x) = 3 - x\). The teal curve below is \(F(x) = 3x - \tfrac{x^2}{2}\), and its slope at every \(x\) is \(3 - x\), so \(F\) is an antiderivative of \(f\). Orange segments on the teal curve show these slopes.

Slide C to raise the teal curve by 2. The segments move with the curve and still lie along it, because a shift upward does not tilt a curve.

The raised curve \(F(x) + 2\) is also an antiderivative of \(f\), and so is every faint curve. All of them have the same slope at each \(x\) and differ only in height.

Two antiderivatives differ by a constant

The teal curve and the pale curve are both antiderivatives of the curve on top. The vertical orange bracket between them shows their gap at one value of x.

Move the bracket along the curves. The gap has the same length at every x.

This happens because the slope of the difference of the two curves is the difference of their slopes. Both slopes equal the height of the top curve, so the difference has slope 0 at every x.

A curve with slope 0 on an interval is flat, so the difference is a constant. Any two antiderivatives of the same function differ by a constant.

Notation

1. If \(F\) is an antiderivative of \(f\), then \(F + C\) is one too, for every constant \(C\), because the derivative of \(C\) is 0.

   \[\frac{d}{dx}\big(\cb{F(x)} + C\big) = \cy{f(x)}\]

2. On an interval, every antiderivative of \(f\) has this form. The collection of all of them is the family of antiderivatives, written as the indefinite integral of \(f\).

   \[\int f(x)\,dx = \cb{F(x)} + C\]

   The letter \(C\) stands for any real number.

3. Write + C in every indefinite integral. Without it, the answer is one curve and not the whole family.

4. Add C once, at the end. Separate constants for separate terms combine into one.

   \[\begin{aligned} &(x^2 + C_1) + (3x + C_2) \\ &\quad = x^2 + 3x + C \end{aligned}\]

   Here \(C = C_1 + C_2\).

5. A definite integral is a number, so it has no C. The constant cancels between the bounds.

   \[\begin{aligned} &\big(F(b) + C\big) - \big(F(a) + C\big) \\ &\quad = F(b) - F(a) \end{aligned}\]

When it fails

The function \(\tfrac{1}{x^2}\) is not defined at 0, so its graph has two pieces, one on each side. The antiderivative \(-\tfrac1x\) has two pieces too.

Lift the right piece and leave the left piece. Both pieces still have slope \(\tfrac{1}{x^2}\). With a gap in the domain, the family needs one constant for each piece: \(-\tfrac1x + C_1\) for \(x < 0\), and \(-\tfrac1x + C_2\) for \(x > 0\).

Example 1: A family from a sum

Find all antiderivatives of \(f(x) = 6x - 4\).

1. Find an antiderivative of each term: \(3x^2\) for \(6x\), and \(-4x\) for \(-4\).

2. Add the constant C.

   \[\int (6x - 4)\,dx = \cb{3x^2 - 4x} + C\]

3. Differentiate to confirm the answer. The derivative of C is 0, so the result is 6x − 4 for every C.

   \[\frac{d}{dx}\big(3x^2 - 4x + C\big) = 6x - 4\]

4. Each value of C is one curve in the picture, and all the curves have slope 6x − 4 at each x.

Example 2: The missing constant

Find all antiderivatives of \(f(x) = 12x^2 + 2\).

1. Find an antiderivative of each term: \(4x^3\) for \(12x^2\), and \(2x\) for 2.

2. A tempting wrong move: State the answer without a constant.

   \[\int (12x^2 + 2)\,dx = 4x^3 + 2x\]

   The orange segments fit every faint curve, so each of them has slope \(12x^2 + 2\), for example \(4x^3 + 2x + 3\). An answer without \(C\) names only one of them, the red curve.

3. Add the constant C. It stands for every shift of the curve.

   \[\int (12x^2 + 2)\,dx = \cb{4x^3 + 2x} + C\]

4. One constant is enough. The sum \((4x^3 + C_1) + (2x + C_2)\) equals \(4x^3 + 2x + C\), with \(C = C_1 + C_2\).

Practice

Find all antiderivatives of \(f(x) = 10x^4 - 3\). Then find the gap between the curve with \(C = 5\) and the curve with \(C = -2\).

The two curves are shifted copies of one another. Slide the orange bracket along them. Its length is the gap, and it is the same at every \(x\).

Questions

The top graph shows \(f(x) = x^2\). The pale curve through the origin is \(F(x) = \tfrac{x^3}{3}\). The teal curve is \(F(x) + C\) for one value of \(C\).

1. What is \(C\) for the teal curve?

   • 2
   • −2
   • 11
   • 0
   Answer

   2

2. What is the slope of the teal curve at \(x = 2\)?

   • 4
   • 6
   • 2
   • \(\tfrac{14}{3}\)
   Answer

   4

3. Which answer is complete for \(\displaystyle\int (8x^3 - 5)\,dx\)?

   • \(2x^4 - 5x + C\)
   • \(2x^4 - 5x\)
   • \(2x^4 - 5 + C\)
   • \(24x^2 + C\)
   Answer

   \(2x^4 - 5x + C\)

Summary

You add + C once, at the end, to write all the antiderivatives of a function.

This works because two antiderivatives of the same function differ by a function with slope 0, which is a constant.

Every indefinite integral has a C, and no definite integral does, because the constant cancels between the bounds.

This is one of five skills in topic 6.8 of Which & Why: AP Calculus.