Antiderivatives and the Constant of Integration

10 min read

Textbook: Logan, A First Course in Differential Equations, 3rd ed. • Chapter: 1 • Section: 2

Quick Reference


Textbook	Logan, A First Course in Differential Equations, 3rd ed.
Chapter.Section.Subsection	1.2 Antiderivatives (review of the calculus prerequisite)
Pages	16-21
Course	MATH347

Page numbers are faculty-verified against Logan, 3rd ed.

Try This First

A small balloon is being inflated. At time $t$ (in seconds) the air is going in at a rate of $g(t) = 6t$ cubic centimeters per second, and at $t = 0$ the balloon already holds $4$ cubic centimeters.

Before reading any further, try to answer two things on paper:

What kind of object is the volume $V$? Is it a single number, or is it something that changes as $t$ moves forward?
If you only knew the rate $6t$ and nothing about the starting amount, could you pin down the volume exactly? What would still be missing?

Hold onto your answers. Turning a known rate back into the original quantity is the work ahead, along with the one piece of information that the rate alone can never tell you.

Notice (open after you have tried it)

The volume is a process, not a fixed number: it takes an input time $t$ and returns an output volume $V(t)$. The rate $6t$ tells you how fast the output is changing, but it does not tell you where the output started. Two balloons with the very same inflation rate can hold different amounts at every instant if they began with different amounts. That missing starting information is the constant of integration.

Building the Idea From a Rate

A differential equations course begins with a simple reversal. In a calculus course the usual question is: given a function, what is its derivative? Here the question runs the other way: given the rate of change, what was the original function?

Think of the output as a quantity that moves as the input moves. If you are told the speed of a car at every instant, you know how its distance is changing, but you do not yet know where the car started. The starting position is a separate fact. Recovering the position from the speed is exactly the work of an antiderivative, and the unknown starting value is exactly the reason a constant appears.

This is the most basic differential equation, written $x' = g(t)$. It says: the unknown function $x = x(t)$ has a known derivative $g(t)$. Solving it means finding every function whose derivative is $g(t)$.

Prerequisite Hub

Builds on (cross-course)

The foundations come from second-quarter calculus rather than from an earlier MATH347 topic.

Skill	Course	Why it is needed
Antiderivative concept	MATH162	An antiderivative of $g$ is a function whose derivative is $g$; this page reuses that idea and attaches a constant to it.
Indefinite integral	MATH162	The notation $\int g(t)\,dt$ and the basic power, exponential, and trig integral rules are assumed.

Unlocks (later in MATH347)

Skill	What it adds
`antiderivatives-ivp`	Pins the constant $C$ down using an initial condition, turning the general solution into one particular curve.
`separable-de`	Generalizes “integrate both sides” to equations where the rate depends on the unknown itself.

The Official Definition

Antiderivative as the solution of $x' = g(t)$.

To solve the differential equation $x' = g(t)$ for the unknown $x = x(t)$, where $g(t)$ is a given continuous function, write

$$x(t) = \int g(t)\,dt + C,$$

an arbitrary constant of integration. This one-parameter family of curves are the integral curves of the differential equation and represent the general solution.

Source: Logan, A First Course in Differential Equations, 3rd ed., §1.2 Antiderivatives, p.16, Eqs. (1.7)-(1.8).

Constant of integration / one-parameter family.

The antiderivative is unique only up to an additive constant; the indefinite integral $\int g(t)\,dt + C$ describes a whole family of antiderivatives, one curve for each value of $C$.

Source: Logan, 3rd ed., §1.2, p.16.

Reading the definition with two representations

The two ideas below describe the same family of solutions, one in symbols and one as a picture.

Representation	What it says
Algebraic	$x(t) = \int g(t)\,dt + C$, with $C$ free to be any real number.
Graphical	A stack of parallel curves, called integral curves, each one shifted vertically from the next. Choosing $C$ selects one curve from the stack.

Try to translate between them: if the picture shows two integral curves that are always $3$ units apart vertically, what is the difference between their $C$ values? (It is $3$, because shifting $C$ shifts every output by the same amount.)

Key Theorem: Fundamental Theorem of Calculus

Fundamental Theorem of Calculus (variable upper limit).

If $g(t)$ is a continuous function, the derivative of an integral with variable upper limit is

$$\frac{d}{dt}\int_a^t g(s)\,ds = g(t),$$

where the lower limit $a$ is any number.

Source: Logan, 3rd ed., Theorem 1.12, p.18.

This theorem is the guarantee that the recipe in the definition actually works. It says that differentiating the integral gives back $g$, so the function built from an integral really is an antiderivative of $g$. The choice of lower limit $a$ only changes the result by a constant, which is one more way to see where the constant of integration comes from.

Worked Examples

Example 1: Recover the function from its rate (the balloon)

Return to the opener. The volume satisfies $V'(t) = 6t$.

Predict first. The rate $6t$ grows with time, so the volume should grow faster and faster. The antiderivative of a linear rate is quadratic, so expect a $t^2$ term plus a constant.

Compute.

$$V(t) = \int 6t\,dt + C = 3t^2 + C.$$

Check against the prediction. The result is quadratic in $t$, as expected, and differentiating gives $V'(t) = 6t$, which matches the given rate. The constant $C$ is exactly the missing starting information from the opener.

This is the general solution: a one-parameter family. Every member has the same rate; they differ only in their starting height.

Example 2: One rate, a whole family of curves

Solve $x' = \cos t$.

Predict first. A cosine rate should integrate to a sine, since the derivative of sine is cosine.

Compute.

$$x(t) = \int \cos t\,dt + C = \sin t + C.$$

Check. Differentiate: $\frac{d}{dt}(\sin t + C) = \cos t$, which is the given rate. Correct.

Now read this as a graph. The curves $\sin t$, $\sin t + 1$, and $\sin t - 2$ are three members of the family, each a vertical shift of the others. Picking $C$ picks one curve out of infinitely many.

Example 3: A rate with several terms

Solve $x' = 4t^3 - 2t + 5$.

Predict first. Integrate term by term; each power rises by one degree.

Compute.

$$x(t) = \int (4t^3 - 2t + 5)\,dt + C = t^4 - t^2 + 5t + C.$$

Check. Differentiate term by term: $\frac{d}{dt}(t^4 - t^2 + 5t + C) = 4t^3 - 2t + 5$. Each term matches, so the answer is correct. The single constant $C$ covers all the term-by-term constants at once, because the sum of arbitrary constants is again an arbitrary constant.

Example 4: Why only one constant

A student integrates $x' = 2t$ and writes $x(t) = t^2 + C_1 + C_2$ after adding a constant twice. Is anything wrong?

There is nothing mathematically false, but the two constants collapse into one. The sum $C_1 + C_2$ is itself a single arbitrary real number, so writing $x(t) = t^2 + C$ captures the same family with no information lost. A first-order rate equation produces exactly one free constant, and that one constant is what an initial condition will later determine.

Common Misconceptions

Common misconception

the constant of integration is optional or can be dropped. The constant is not decoration; it carries real information. Without it, $\int 6t\,dt = 3t^2$ describes a single curve and silently assumes the quantity started at $0$ when $t = 0$. The balloon in the opener started at $4$, so its true volume is $3t^2 + 4$, a different curve entirely. Predict-then-check: differentiate both $3t^2$ and $3t^2 + 4$. Both give $6t$, which is why the rate alone cannot decide between them, and why the constant must be kept until an initial condition fixes it.

Common misconception

the rate $g(t)$ already determines the function completely. A rate describes how a quantity is changing, not how much there is. Two functions with the same derivative differ by a constant (this is what “antiderivative up to a constant” means). Test it on a small case: $f(t) = t^2$ and $h(t) = t^2 + 10$ have the identical derivative $2t$, yet $f(1) = 1$ while $h(1) = 11$. Knowing the rate $2t$ does not tell you which function you have until you also know one value.

Practice Problems

Level 1 Read the family

Write the general solution of $x' = 3t^2$.

Thought Process

Antidifferentiate $3t^2$ using the power rule, then attach the constant of integration. Do not drop the constant.

Show Answer

$$x(t) = \int 3t^2\,dt + C = t^3 + C.$$

Differentiating $t^3 + C$ returns $3t^2$, confirming the answer.

Level 2 Several terms

Find the general solution of $x' = 6t^2 - 4t + 7$.

Thought Process

Integrate each term separately. The constants from each term combine into one constant $C$.

Show Answer

$$x(t) = \int (6t^2 - 4t + 7)\,dt + C = 2t^3 - 2t^2 + 7t + C.$$

Check by differentiating: $6t^2 - 4t + 7$. The answer matches the given rate.

Level 3 A non-polynomial rate

Solve $x' = e^{2t}$.

Thought Process

The antiderivative of $e^{kt}$ is $\frac{1}{k}e^{kt}$. Predict that a factor of $\frac{1}{2}$ appears because the chain rule would otherwise produce an extra factor of $2$.

Show Answer

$$x(t) = \int e^{2t}\,dt + C = \frac{1}{2}e^{2t} + C.$$

Check by differentiating: $\frac{d}{dt}\left(\frac{1}{2}e^{2t} + C\right) = \frac{1}{2}\cdot 2 e^{2t} = e^{2t}$. Correct.

Level 4 Match a curve from the family

The general solution of $x' = 2t$ is $x(t) = t^2 + C$. Which value of $C$ gives the curve passing through the point $(1, 5)$? Name the misconception a student avoids by doing this step.

Thought Process

Substitute $t = 1$ and $x = 5$ into $t^2 + C$ and solve for $C$. This selects one curve out of the whole family.

Show Answer

Substitute the point: $5 = (1)^2 + C$, so $C = 4$. The curve is $x(t) = t^2 + 4$.

This step shows why the constant cannot be dropped. The rate $2t$ alone leaves $C$ undetermined; only a known value (here the point $(1,5)$) selects a single curve. A student who omitted $C$ would have no way to make the function pass through the given point. Fixing the constant from a known value is the work of the next node, initial value problems.

Level 5 Why differentiation and integration line up

Define $A(t) = \int_0^t (3s^2 + 1)\,ds$. Without first evaluating the integral, find $A'(t)$ and explain which theorem justifies the answer. Then evaluate $A(t)$ explicitly and confirm that $A'(t)$ matches.

Thought Process

The Fundamental Theorem of Calculus with a variable upper limit gives the derivative directly: differentiating an integral with upper limit $t$ returns the integrand evaluated at $t$. After that, compute the integral by hand and differentiate to confirm.

Show Answer

By the theorem. By the Fundamental Theorem of Calculus (variable upper limit, Logan Theorem 1.12, p.18), since $3s^2 + 1$ is continuous,

$$A'(t) = \frac{d}{dt}\int_0^t (3s^2 + 1)\,ds = 3t^2 + 1.$$

By direct evaluation. Compute the integral:

$$A(t) = \int_0^t (3s^2 + 1)\,ds = \left[ s^3 + s \right]_0^t = (t^3 + t) - (0 + 0) = t^3 + t.$$

Differentiating: $A'(t) = 3t^2 + 1$, which matches the theorem exactly.

Why this matters. This is the proof, in one example, that building a function as an integral really does produce an antiderivative. Notice that choosing the lower limit $0$ fixed a particular member of the family; a different lower limit would shift $A$ by a constant, which is yet another way to see the constant of integration appear.

Mastery Checklist

Novice (Level 1-2):

I can state that solving $x' = g(t)$ means finding every function whose derivative is $g(t)$.
I can antidifferentiate polynomial rates term by term and attach a single constant $C$.
I can explain why the constant of integration cannot be dropped.

Competent (Level 3-4):

I can integrate exponential and trigonometric rates correctly, including the chain-rule factor.
I can describe the general solution as a one-parameter family of integral curves.
I can select one curve from the family using a known value of the function.

Proficient (Level 5):

I can state the Fundamental Theorem of Calculus for a variable upper limit and explain what it guarantees.
I can connect the choice of lower limit in a definite integral to the constant of integration.
I can verify any proposed solution by differentiating it back to the original rate.

Resources

Resource	Where it points
Logan, 3rd ed. §1.2 Antiderivatives (pp.16-21)	https://www.math.unl.edu/~jlogan1/PDFfiles/New3rdEditionODE.pdf
OpenStax Calculus Volume 2 §4.1 Basics of Differential Equations (general vs particular solution)	https://openstax.org/books/calculus-volume-2/pages/4-1-basics-of-differential-equations

The Logan text is the primary source for this course; the OpenStax section is a free, openly licensed companion that frames the same general-versus-particular distinction in standard calculus language.

Connections

Looking back:

Antiderivative concept (MATH162): an antiderivative of $g$ is a function whose derivative is $g$.
Indefinite integral (MATH162): the notation and the basic integration rules reused on this page.

Looking ahead:

Antiderivatives and initial value problems: an initial condition fixes the constant $C$, turning the general solution into one particular curve.
Separable equations: the same “integrate both sides” move, extended to rates that depend on the unknown function itself.

Real-world connections:

Position from velocity: integrating a known speed recovers distance traveled, but the starting position is the free constant.
Volume from inflow rate: the balloon in the opener is a one-parameter family until the starting amount is known.

Previous	Up	Next
MATH347 Week 0 Index	Skills Index	Antiderivatives and Initial Value Problems

Last updated: 2026-06-16