Reversing differentiation: the general antiderivative and why the constant is essential, indefinite integral notation, the reversed power rule and the basic formulas, initial-value problems, and the family of vertically shifted curves sharing one derivative.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
Antiderivatives
Objectives
Five outcomes. Everything here is Chapter 3 read backwards, plus one constant that changes the whole picture.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 419-430 — the section these objectives are drawn from
Warm-up
Chapter 3 answered one question repeatedly: given a function, what is its derivative? Every rule there was a machine running in one direction.
Discussion prompt
Which function has derivative 3x squared? Is there only one?
Hint: Try the obvious answer, then add something to it.
Answer:
The cube of x differentiates to three x squared, so it is an answer. But so does the cube of x plus 7, and the cube of x minus 100 — the derivative of a constant is zero, so adding any constant changes nothing about the slope.
\[ \frac{d}{dx}\!\left[x^{3}+C\right] = 3x^{2} \quad \text{for every constant } C \]
So the reverse question has infinitely many answers, differing only by a constant. That is not a defect in the question but the honest situation, and this section's job is to describe the whole family rather than pick one arbitrarily.
Concept
An antiderivative of a function is any function whose derivative it is. Because constants differentiate to zero, antiderivatives come in families differing by a constant, and the general antiderivative records all of them at once.
antiderivative — A function whose derivative is the given function. If one antiderivative is known, every other differs from it by a constant, so the general antiderivative is that one plus an arbitrary constant C.
\[ F'(x) = f(x) \;\Longrightarrow\; \text{the general antiderivative is } F(x)+C \]
That every antiderivative differs by a constant is a theorem, not an observation, and it follows from the Mean Value Theorem of Section 4.4: two functions with the same derivative everywhere have a constant difference.
Figure (svg): Differentiation and antidifferentiation as opposite directions along the same arrow
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 419-423
Section
Section 1
Concept
Finding a single antiderivative is the hard part; once one is found, every other is that one plus a constant. The Mean Value Theorem is what guarantees nothing else is hiding.
the general antiderivative — The complete description of every function with the given derivative: one antiderivative plus an arbitrary constant. Omitting the constant reports one member of a family as though it were the whole family.
\[ F' = G' \text{ on an interval} \;\Longrightarrow\; F - G \text{ is constant} \]
The theorem is Section 4.4's corollary. If two functions have the same derivative, their difference has derivative zero, and a function with zero derivative on an interval is constant.
Figure (svg): The family of antiderivatives as vertically shifted copies of one curve
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 419-423 — the general antiderivative
Picture it
Four antiderivatives of the same function.
Figure (svg): The family of antiderivatives as vertically shifted copies of one curve
The curves are vertical translates of one another, so at any input they all have the same slope. The derivative records slope and nothing about height, which is exactly the information the constant restores.
Worked example
Example 4.52. Guess, then check by differentiating.
\[ \text{Find the general antiderivative of } f(x)=3x^{2}. \]
Ask what differentiates to a square
Why: The power rule lowers the exponent by one.
\[ \text{start with } x ^{3} \]
Differentiate the guess
Why: Check.
\[ 3 x ^{2},\text{ correct} \]
Add the arbitrary constant
Why: Constants differentiate to zero.
\[ x ^{3} + C \]
Justify that nothing else works
Why: By the corollary to the Mean Value Theorem.
State
Why: The general antiderivative.
\[ x ^{3} + C \]
Figure (svg): Differentiation and antidifferentiation as opposite directions along the same arrow
\[ \int 3x^{2}\,dx = x^{3}+C \]
Verify: differentiate the answer back
Why: Differentiating x cubed plus C gives 3x squared plus 0, which is the original function — so the answer is correct, and this check works for every antiderivative problem. That is the single most useful feature of the subject: unlike most computations, an antiderivative can always be verified completely, in one line, by running Chapter 3 forwards.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 420-421
Sorting
Differentiate each and compare.
Sort into buckets
Sort each candidate.
The first three are the same family and the last two are outside it. Note that the fifth differs from a valid answer by x rather than by a constant, which is why it fails — only constant differences are allowed.
Worked example
Checkpoint 4.52. What omitting it costs.
\[ \text{What is lost by writing } \int 2x\,dx = x^{2}? \]
Note the answer given is an antiderivative
Why: It differentiates correctly.
List others
Why: Add any constant.
\[ x ^{2} + 5, x ^{2} - 3,\text{ all valid} \]
Identify what was reported
Why: One member of the family.
Consider the consequence
Why: In an initial-value problem.
State
Why: The correct answer.
\[ x ^{2} + C \]
Figure (svg): The family of antiderivatives as vertically shifted copies of one curve
\[ \int 2x\,dx = x^{2}+C \]
Verify: see the consequence in a concrete problem
Why: A problem asking for the curve through the point one comma three with slope twice x has answer x squared plus 2. Starting from x squared with no constant leaves nothing to adjust, and the point cannot be fitted at all. So the constant is not a formality: it is the degree of freedom that the initial condition consumes, and the next idea's problems are unsolvable without it.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 421-422
Trap
\[ \int 2x\,dx = x^{2} \]
Report the antiderivative found
Why: The student stops at the first answer.
The answer differentiates correctly but describes one curve where there is a whole family, and leaves nothing to fit a condition with.
\[ \int 2x\,dx = x^{2}+C \]
Report the general antiderivative
Why: The constant records that every vertical shift works equally well.
The habit worth building is writing the plus C at the same moment as the antiderivative rather than afterwards. Added as a separate step it is forgotten; written together it becomes part of the answer's shape.
Fill the middle
One antiderivative found; the family described.
Fill in the blanks
\int 3x^C\,dx = x^___ + ___
Why: The constant records that every vertical shift of the cube has the same derivative. Without it the answer names one curve rather than the family, and an initial condition would have nothing to act on.
Two truths and a lie
All three are about the family.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. A function with any antiderivative has infinitely many, differing by a constant. That is why the answer to an antidifferentiation question is a family rather than a function, and why the notation carries a plus C.
Prediction
Commit before reasoning.
Predict first
Two functions have the same derivative everywhere on an interval. Why must their difference be constant?
Correct: Their difference has zero derivative, and Section 4.4's corollary makes it constant.
\[ (F-G)' = 0 \text{ on } I \;\Longrightarrow\; F-G \text{ constant on } I \]
Why: Subtracting gives a function whose derivative is zero throughout the interval. The Mean Value Theorem then says that between any two points the function's change equals its derivative somewhere times the gap — which is zero — so the function never changes. That is a proof rather than an appeal to intuition, and it is what guarantees the plus C captures every antiderivative and not merely the obvious ones.
Section
Section 2
Concept
The integral sign followed by the function and a differential denotes the general antiderivative. The differential names the variable, which matters when the expression contains more than one letter.
indefinite integral — Notation for the general antiderivative: an elongated S, the integrand, and a differential naming the variable, with the answer written as an antiderivative plus an arbitrary constant.
\[ \int f(x)\,dx = F(x)+C \]
The word indefinite distinguishes this from the definite integral of Chapter 5, which produces a number rather than a family of functions. The connection between them is the Fundamental Theorem.
Figure (svg): The indefinite integral notation, with each part named
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 421-425 — indefinite integrals
Picture it
Each symbol named.
Figure (svg): The indefinite integral notation, with each part named
The differential is the part most often dropped, and it is the part that says which letter varies. In an expression containing several letters it is the only thing distinguishing quite different questions.
Worked example
Example 4.54. The differential is not decoration.
\[ \text{Evaluate } \int 4t^{3}\,dt \text{ and } \int 4t^{3}\,dx. \]
Read the first
Why: The differential names t.
Apply the reversed power rule
Why: Raise and divide.
\[ t ^{4} + C \]
Read the second
Why: The differential names x.
\[ t ^{3}\text{ is } a\text{ constant here} \]
Antidifferentiate the constant
Why: A constant times x.
\[ 4 t ^{3} x + C \]
Compare
Why: Utterly different answers.
Figure (svg): The indefinite integral notation, with each part named
\[ \int 4t^{3}\,dt = t^{4}+C, \qquad \int 4t^{3}\,dx = 4t^{3}x+C \]
Verify: differentiate both back with respect to the right variable
Why: Differentiating t to the fourth with respect to t gives 4t cubed, correct. Differentiating 4t cubed times x with respect to x treats the t part as a constant coefficient and gives 4t cubed, also correct. The two answers look nothing alike and both are right, which is the clearest possible demonstration that the differential carries real information rather than being punctuation.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 422-423
Fill the middle
The same integrand, integrated with respect to x rather than t.
Fill in the blanks
\int 4t^4t^3 x\,dx = ___ + C
Why: The differential says x is the variable, so the whole coefficient is constant and antidifferentiating gives that constant times x. The same integrand with dt would give t to the fourth instead.
Worked example
Checkpoint 4.54. The rules of Section 3.3, reversed.
\[ \text{Evaluate } \int \left(5x^{4}-3\cos x\right)dx. \]
Split across the difference
Why: Reversing the sum rule.
\[ \int 5 x ^{4} \,dx - \int 3 \cos x \,dx \]
Pull the constants out
Why: Reversing the constant-multiple rule.
\[ 5 \int x ^{4} \,dx - 3 \int \cos x \,dx \]
Antidifferentiate each
Why: Power rule and the sine formula.
\[ 5(x ^{5} / 5) - 3(\sin x) \]
Simplify
Why: The fives cancel.
\[ x ^{5} - 3 \sin x \]
Add one constant
Why: Not one per term.
\[ x ^{5} - 3 \sin x + C \]
Figure (svg): The solution to Worked example the sum and constant-multiple rules shown as a ladder of expressions, one row per legal move
\[ \int\!\left(5x^{4}-3\cos x\right)dx = x^{5}-3\sin x+C \]
Verify: differentiate back, and note the single constant
Why: Differentiating gives 5x to the fourth minus 3 cosine x, matching the integrand. Note that although two terms were integrated, only one constant appears: two arbitrary constants added together are just one arbitrary constant, so writing plus C one plus C two would be redundant rather than more careful. The sum and constant-multiple rules transfer directly from Section 3.3 because differentiation obeys them, and reversing a rule that holds forwards holds backwards.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 423-424
Error analysis
A student integrates a product the way sums are integrated.
Annotate
On: \( \int x\cos x\,dx = \left(\tfrac{x^{2}}{2}\right)(\sin x) + C \)
Sums and constant multiples reverse cleanly because differentiation respects them. Products and quotients do not, and the check that catches every such error is differentiating the answer back.
Matching
Each part of the symbol.
Match the pairs
Why: The third and fourth are the two most often dropped, and both carry real information — one decides which letter varies, the other whether the answer is a function or a family.
Sorting
Ask whether differentiation respects the operation.
Sort into buckets
Sort each rule.
The two invalid ones correspond exactly to the two Chapter 3 rules that were NOT simple: the product and quotient rules were complicated forwards, and they have no simple reverse at all. Techniques for those come later.
Prediction
Commit before reasoning.
Predict first
Integrating a sum of three terms, how many arbitrary constants should the answer carry?
Correct: One.
\[ C_{1}+C_{2}+C_{3} = C \]
Why: Each term contributes an arbitrary constant, but adding arbitrary constants together produces an arbitrary constant — nothing is gained by naming them separately, and the answer is no more general with three letters than with one. Writing several is not more careful but merely more cluttered, and it obscures that the family of antiderivatives has exactly one degree of freedom.
Section
Section 3
Concept
Since differentiating multiplies by the exponent and lowers it, antidifferentiating raises the exponent and divides by the raised value. The rule fails for exactly one exponent, where the division would be by zero.
the reversed power rule — The antiderivative of a power is that power with its exponent raised by one, divided by the new exponent, plus a constant. It holds for every exponent except negative one, where the denominator would vanish.
\[ \int x^{n}\,dx = \frac{x^{n+1}}{n+1}+C, \quad n \ne -1 \]
The excluded exponent is the reciprocal, whose antiderivative is the natural logarithm of the absolute value — a formula that comes from Section 3.9 rather than from the power rule at all.
Figure (svg): The power rule reversed, with the exponent-1 exception
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 421-426 — the power rule for integrals
Picture it
Four examples and the exponent that breaks it.
Figure (svg): The power rule reversed, with the exponent-1 exception
The exception is not arbitrary. Raising negative one by one gives zero, and the rule divides by that — so the formula simply has nothing to say there, and a different one is needed.
Worked example
Example 4.55. The rule applies to any exponent but one.
\[ \text{Evaluate } \int\!\left(x^{5}+\sqrt{x}+\frac{1}{x^{3}}\right)dx. \]
Rewrite everything as a power
Why: Radicals and reciprocals.
\[ x ^{5} + x ^{\frac{1}{2}} + x ^{-3} \]
Apply the rule to the first
Why: Raise to 6, divide by 6.
\[ x ^{6} / 6 \]
Apply it to the second
Why: Raise to 3/2, divide by 3/2.
\[ (\frac{2}{3}) x ^{\frac{3}{2}} \]
Apply it to the third
Why: Raise to -2, divide by -2.
\[ -x ^{-2} / 2 \]
Add the constant
Why: One for the whole expression.
\[ +C \]
Figure (svg): The power rule reversed, with the exponent-1 exception
\[ \frac{x^{6}}{6}+\frac{2}{3}x^{3/2}-\frac{1}{2x^{2}}+C \]
Verify: differentiate each term back
Why: The first gives x to the fifth; the second gives x to the one half, which is the square root; the third, written as negative one half x to the negative two, gives x to the negative three, which is the reciprocal cube. All three match the integrand. Note that rewriting radicals and reciprocals as powers before starting is what makes the rule applicable at all — attempting it on a square root written as a radical usually goes wrong.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 424-425
Fill the middle
The power rule reversed on a fourth power.
Fill in the blanks
\int x^5\,dx = \frac5}}}___}+C
Why: The exponent rises from 4 to 5 and the answer is divided by 5. Differentiating brings the 5 down and lowers the exponent again, recovering x to the fourth.
Worked example
Checkpoint 4.55. Where the rule has nothing to say.
\[ \text{Why does the power rule fail for } \int \frac{1}{x}\,dx, \text{ and what is the answer?} \]
Write the integrand as a power
Why: The reciprocal.
\[ x ^{-1} \]
Attempt the rule
Why: Raise the exponent.
\[ x ^{0} / 0 \]
Identify the failure
Why: Division by zero.
Recall Section 3.9
Why: The derivative of the logarithm.
\[ d / \,dx [\ln x] = \frac{1}{x} \]
Account for negative inputs
Why: Absolute value inside.
\[ \ln | x | + C \]
Figure (svg): The solution to Worked example the excluded exponent shown as a ladder of expressions, one row per legal move
\[ \int \frac{1}{x}\,dx = \ln|x|+C \]
Verify: check the absolute value is needed
Why: For negative x the logarithm of x is undefined, but the reciprocal is perfectly well defined there — so an antiderivative must exist. Differentiating the logarithm of negative x by the chain rule gives one over negative x times negative one, which is one over x, correct. The absolute value covers both cases in one formula. Note that this is the one antiderivative in the section that does not come from reversing a power rule at all: it comes from Section 3.9's logarithm.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 425-426
Trap
\[ \int x^{4}\,dx = 4x^{3} + C \]
Differentiate instead of antidifferentiate
Why: The student runs Chapter 3's rule forwards out of habit.
\[ \frac{d}{dx}\!\left[4x^{3}\right] = 12x^{2} \ne x^{4} \]
The check exposes it instantly: differentiating the answer must return the integrand, and here it does not come close.
\[ \int x^{4}\,dx = \frac{x^{5}}{5} + C \]
Raise the exponent and divide by the new one
Why: The reverse of lowering it and multiplying.
Because the reverse operation can always be checked by the forward one, this is a mistake that never needs to survive. Differentiating the answer takes one line and settles the matter completely — no other topic in the course offers a check that cheap and that complete.
Sorting
The power rule, or something else?
Sort into buckets
Sort each integrand.
The second and fifth are the same function written two ways, which is worth noticing: the exception is easy to miss when the reciprocal is written as a fraction rather than as a power. Rewriting everything as a power first makes the exception visible.
Two truths and a lie
All three are about the power rule.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Antidifferentiation RAISES the exponent; lowering it is what differentiation does. Running the rule in the wrong direction is the commonest slip here, and differentiating the answer back catches it in one line.
Prediction
Commit before reasoning.
Predict first
The power rule fails for exactly one exponent. Which, and why?
Correct: Negative one, because the rule would divide by zero.
\[ n=-1: \; \frac{x^{0}}{0} \text{ undefined} \;\Longrightarrow\; \ln|x|+C \]
Why: For every other exponent, raising by one gives something non-zero to divide by. Only negative one raised by one gives zero, and the formula then reads x to the zero over zero, which is meaningless. The exception is structural rather than a special case to memorise, and the missing antiderivative comes from Section 3.9's logarithm instead.
Section
Section 4
Concept
Every derivative proved in Chapter 3 becomes an antiderivative formula when read backwards. There is nothing new to prove; the only new work is recognising which formula applies.
the basic antiderivatives — Reversals of the standard derivatives: the exponential is its own antiderivative, sine antidifferentiates to minus cosine, cosine to sine, the squared secant to tangent, and the standard rational expression to the inverse tangent.
\[ \int e^{x}dx = e^{x}+C, \quad \int\cos x\,dx = \sin x+C \]
The sines and cosines are where signs go wrong. Differentiating sine gives cosine with no sign change, so antidifferentiating cosine gives sine; differentiating cosine introduces a minus, so antidifferentiating sine gives MINUS cosine.
Figure (svg): The basic antiderivative formulas, each read backwards from Chapter 3
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 422-428 — table of antiderivatives
Picture it
Seven formulas, each a Chapter 3 result read backwards.
Figure (svg): The basic antiderivative formulas, each read backwards from Chapter 3
Nothing here is new mathematics. Each line was proved in Chapter 3 and is being read right to left, which is why the table needs recognising rather than memorising afresh.
Worked example
Example 4.56. Watch the signs.
\[ \text{Evaluate } \int\!\left(2\sin x+e^{x}-\sec^{2}x\right)dx. \]
Split the sum and pull out constants
Why: Reversing Section 3.3.
\[ 2 \int \sin x \,dx + \int e ^{x} \,dx - \int \sec ^{2} x \,dx \]
Antidifferentiate the sine
Why: Differentiating cosine gives minus sine, so reverse the sign.
\[ -\cos x \]
Antidifferentiate the exponential
Why: It is its own derivative.
\[ e ^{x} \]
Antidifferentiate the squared secant
Why: It is the derivative of tangent.
Assemble
Why: With one constant.
\[ -2 \cos x + e ^{x} - \tan x + C \]
Figure (svg): The basic antiderivative formulas, each read backwards from Chapter 3
\[ -2\cos x+e^{x}-\tan x+C \]
Verify: differentiate back term by term
Why: Differentiating minus 2 cosine gives plus 2 sine; the exponential gives itself; minus tangent gives minus the squared secant. All three match the integrand exactly, including the signs — which is the check worth doing every time on trigonometric antiderivatives, since the sine and cosine signs are reversed more often than any other error in this section.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 426-427
Fill the middle
Antidifferentiating a sine.
Fill in the blanks
\int \sin x\,dx = -\cos x + C
Why: Since the derivative of cosine is minus sine, the antiderivative of sine must be minus cosine — differentiating it returns plus sine. Cosine, by contrast, antidifferentiates to plain sine with no sign change.
Worked example
Checkpoint 4.56. The integrand rewritten first.
\[ \text{Evaluate } \int \frac{3}{1+x^{2}}\,dx \text{ and } \int \frac{x^{2}+1}{x}\,dx. \]
Recognise the first
Why: The derivative of the inverse tangent.
\[ 3 \arctan x + C \]
Examine the second
Why: A quotient with no direct formula.
Divide term by term
Why: Split the fraction.
\[ x + \frac{1}{x} \]
Antidifferentiate each
Why: Power rule and the logarithm.
\[ x ^{2} / 2 + \ln | x | \]
Add the constant
Why: One for the expression.
\[ +C \]
Figure (svg): The solution to Worked example recognising a disguised formula shown as a ladder of expressions, one row per legal move
\[ 3\arctan x+C, \qquad \frac{x^{2}}{2}+\ln|x|+C \]
Verify: differentiate the second back and note the technique
Why: Differentiating gives x plus one over x, and recombining over a common denominator returns x squared plus one over x, the original integrand. The technique — rewriting an integrand into a form the table covers — is the main skill of the section and the main one of Chapter 5's later methods. There is no quotient rule to reverse, so rewriting is not a shortcut but the only route available.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 427-428
Error analysis
A student antidifferentiates a sine.
Annotate
On: \( \int \sin x\,dx = \cos x + C \)
This is the most frequent error in the section and the easiest to catch. Differentiating the answer takes one line and reveals a sign error as clearly as any other kind.
Matching
Chapter 3's table, reversed.
Match the pairs
Why: The first two are the pair that trips people: cosine goes to sine cleanly, while sine picks up a minus. The other two are direct reversals of derivatives proved in Sections 3.5 and 3.7 respectively.
Sorting
Ask whether the integrand appears in the table as written.
Sort into buckets
Sort each integrand.
The fifth needs only a constant pulled out, which the constant-multiple rule allows. Rewriting is the section's main skill and the main one of every later integration technique too, since the table is short and integrands are not.
Prediction
Commit before reasoning.
Predict first
Which function is unchanged by antidifferentiation, apart from the constant?
Correct: The natural exponential.
\[ \int e^{x}dx = e^{x}+C \]
Why: Section 3.9 proved the exponential is its own derivative, so reading that backwards makes it its own antiderivative, up to the constant. No other elementary function has this property, and it is what makes the exponential the natural solution to equations where a quantity's rate of change is proportional to itself — the growth and decay models of the next chapter.
Section
Section 5
Concept
A differential equation gives the derivative and an initial condition gives one point on the curve. Antidifferentiating produces the family, and the condition determines the constant, selecting a single function.
initial-value problem — A differential equation together with a condition specifying the function's value at one input. The antiderivative supplies the family and the condition supplies the constant.
\[ \frac{dy}{dx}=f(x), \; y(x_{0})=y_{0} \;\Longrightarrow\; y = F(x) + (y_{0}-F(x_{0})) \]
This is where the arbitrary constant earns its place. Without it there would be nothing to adjust, and the condition could not be satisfied except by luck.
Figure (svg): An initial condition selecting one curve from the family
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 428-433 — initial-value problems
Picture it
The family, with the condition marking one member.
Figure (svg): An initial condition selecting one curve from the family
Every curve shown has the correct derivative and only one passes through the marked point. The condition consumes exactly the one degree of freedom that the constant provided.
Worked example
Example 4.57. Family first, then the constant.
\[ \text{Solve } \frac{dy}{dx}=2x \text{ with } y(1)=3. \]
Antidifferentiate
Why: The general antiderivative.
\[ y = x ^{2} + C \]
Impose the condition
Why: Substitute the given point.
\[ 3 = 1 + C \]
Solve for the constant
Why: One equation, one unknown.
\[ C = 2 \]
Write the particular solution
Why: Substitute back.
\[ y = x ^{2} + 2 \]
Check
Why: The condition and the derivative.
\[ y(1) = 3\text{ and y' } = 2 x \]
Figure (svg): An initial condition selecting one curve from the family
\[ y = x^{2}+2 \]
Verify: confirm both requirements separately
Why: Differentiating gives 2x, matching the differential equation; substituting 1 gives 1 plus 2, which is 3, matching the condition. A solution must satisfy both, and checking them separately catches the two distinct ways such a problem goes wrong — an antidifferentiation slip, or an arithmetic slip in solving for the constant.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 429-430
Ranking
Solving an initial-value problem.
Put in order
Why: Step a must come first because the constant does not exist until then. Step e checks the two independent requirements — the right derivative and the right value — which fail in different ways.
Worked example
Example 4.58. Motion under gravity needs two conditions.
\[ \text{A ball is thrown upward at } 20 \text{ m/s from a height of } 2 \text{ m. Find its height.} \]
Start from the acceleration
Why: Gravity alone, downward.
\[ a(t) = -9.8 \]
Antidifferentiate once
Why: With a constant.
\[ v(t) = -9.8 t + C 1 \]
Use the initial velocity
Why: At time zero the speed is 20 upward.
\[ C 1 = 20 \]
Antidifferentiate again
Why: With a second constant.
\[ s(t) = -4.9 t ^{2} + 20 t + C 2 \]
Use the initial height
Why: At time zero the height is 2.
\[ C 2 = 2 \]
Figure (svg): Antidifferentiating twice to recover position from acceleration
\[ s(t) = -4.9t^{2}+20t+2 \]
Verify: identify the constants physically and check a consequence
Why: The two constants are not bookkeeping: the first is the launch speed and the second the launch height, both supplied by the physical setup. As a check, the ball's highest point occurs where the velocity vanishes, at t about 2.04 seconds, giving a height of about 22.4 metres — which is sensible for a 20 metre-per-second throw from 2 metres. Note this reverses Section 3.4's chain from position to velocity to acceleration, running it upward instead.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 430-432
Trap
\[ \frac{dy}{dx}=2x, \; y(1)=3 \]
Substitute the condition into the derivative
Why: The student uses the given point immediately.
\[ 3 = 2(1) \quad \text{(false, and meaningless)} \]
The condition describes the function's value, not its derivative's, so substituting it into the differential equation compares unrelated quantities.
\[ y = x^{2}+C, \; \text{then } 3 = 1+C \;\Longrightarrow\; C=2 \]
Antidifferentiate first, then impose the condition
Why: The condition acts on the antiderivative, where there is a constant to determine.
The order matters because the constant only exists after antidifferentiating. Imposing the condition first has nothing to act on, which is why the resulting equation is nonsense rather than merely wrong.
Fill the middle
The family found; the condition imposed.
Fill in the blanks
y = x^2+C, \; y(1)=3 \;\Longrightarrow\; C = ___
Why: Substituting gives 3 equals 1 plus C, so the constant is 2 and the particular solution is x squared plus 2. That curve is the only member of the family passing through the given point.
Sorting
One per antidifferentiation.
Sort into buckets
Sort each problem by the number of conditions required.
The rule is one condition per antidifferentiation, and it is exactly why projectile problems always supply both a launch speed and a launch height. Missing one leaves a family rather than an answer.
Prediction
Commit before reasoning.
Predict first
Why must the antidifferentiation be done before the initial condition is used?
Correct: Because the constant does not exist until after antidifferentiating.
\[ \text{family first: } y=x^{2}+C, \; \text{then } C=2 \]
Why: The condition gives the function's value at a point, and it is used to solve for the arbitrary constant — which only appears once the family has been written down. Substituting the condition into the differential equation instead compares a function value with a derivative value, which are unrelated quantities, and produces an equation that is simply false. The order is forced by what each step supplies.
Comparison
Fill the blanks. Every antiderivative formula is a derivative read the other way.
Comparison matrix
| Function | Its derivative | Its antiderivative |
|---|---|---|
| x^n | n x^(n-1) | x^(n+1)/(n+1) + C |
| e^x | e^x | e^x + C |
| sin x | cos x | -cos x + C |
| cos x | -sin x | sin x + C |
The third and fourth rows are where signs are lost. Differentiating cosine introduces a minus, so antidifferentiating sine must introduce one too — and differentiating the answer back settles it every time.
Pattern
Given an antiderivative to find.
The final check is the section's great advantage. Unlike most computations, an antiderivative can be verified completely in one line by running the forward rules — so no error here need survive.
Stewart, Calculus: Early Transcendentals 8e, §4.9 Antiderivatives §4.9, pp. 350-357
Check
The power rule reversed.
Check your understanding
What is the general antiderivative of x^4?
Answer: A
Why: Raise the exponent to 5 and divide by 5; differentiating returns x to the fourth.
Check
Signs.
Check your understanding
What is the general antiderivative of sin x?
Answer: A
Why: Differentiating minus cosine gives plus sine, matching the integrand.
Check
An initial-value problem.
Check your understanding
If dy/dx = 2x and y(1) = 3, what is y?
Answer: A
Why: The family is x squared plus C, and the condition gives 3 = 1 + C, so C is 2.
Real world
A car's data recorder samples acceleration but not speed or position. After a collision, an investigator has a complete acceleration record and knows the car was stationary at a measured point five seconds before impact.
Discussion prompt
Explain how position is recovered from acceleration alone, why two pieces of information are needed, and what the constants mean physically.
Hint: Section 3.4 differentiated position twice to get acceleration.
Answer:
Section 3.4 went from position down to velocity to acceleration by differentiating twice. The investigator runs that chain upward, antidifferentiating the acceleration record twice.
\[ a(t) \;\longrightarrow\; v(t)+C_{1} \;\longrightarrow\; s(t)+C_{1}t+C_{2} \]
Each antidifferentiation introduces a constant, so two pieces of information are needed, and the physical setup supplies exactly two: the car was stationary, which fixes the first constant at zero, and it was at a measured point, which fixes the second.
The constants are not bookkeeping. The first is the velocity at the reference time and the second is the position there — the two facts that acceleration alone genuinely cannot contain. A recorder that captured acceleration for a car cruising at 30 metres per second would produce an identical record to one for a stationary car, since neither is accelerating. The acceleration record simply does not distinguish them.
That is the section's central point stated physically: differentiation discards information, and the constant is exactly what was discarded. Recovering it requires evidence from outside the derivative, which is why real accident reconstruction needs a witnessed reference point and not just the data.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does the general antiderivative carry an arbitrary constant?
Correct: Because constants differentiate to zero.
\[ \frac{d}{dx}[C]=0 \;\Longrightarrow\; \text{the reverse cannot recover } C \]
Why: Differentiation destroys constant information: the cube of x and the cube of x plus 7 have identical derivatives, so reversing cannot recover which one was meant. The constant records that whole family honestly rather than picking a member arbitrarily. It is also the degree of freedom that an initial condition consumes, which is why omitting it makes initial-value problems unsolvable rather than merely untidy.
Explain it
They wrote that the antiderivative of 2x is x squared, and cannot see why a plus C is needed.
Discussion prompt
In four sentences or fewer, show them what is missing.
Hint: Ask them for a second answer.
Answer:
Ask them to differentiate x squared plus 7. They get 2x — the same answer, from a different function, so their single answer was not the only one.
The derivative simply does not record height, only slope, so reversing it cannot tell which of infinitely many parallel curves was meant. The plus C names all of them at once, and when a problem gives a point to pass through, that constant is precisely what gets solved for.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the power rule, remember it raises the exponent and that negative one is excluded because the rule would divide by zero. For signs, differentiate your answer back — it exposes a sign error immediately. For rewriting, turn everything into powers or split the fraction before looking for a formula. For initial-value problems, antidifferentiate first and impose the condition second. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw the two-way arrow between a function and its derivative, with a note on the reverse arrow saying it has infinitely many destinations and why. Beneath it, sketch four members of the family of antiderivatives of 2x as parallel curves, mark one input, and note that all four have the same slope there. Below, write the indefinite integral notation with each of its four parts labelled, and give the two answers for the same integrand with dt and with dx. In the middle of the page, write the reversed power rule in a box with its exception beside it and one line explaining why that exponent breaks it. Beside it, list the seven basic antiderivative formulas, circling the two whose signs are most often reversed. In the lower half, solve an initial-value problem completely: family, condition, constant, particular solution, and both checks. At the bottom, draw the acceleration-to-velocity-to-position chain with a constant on each arrow and name each constant physically.
If your family sketch shows curves that are not parallel, look again — differing by a constant is a pure vertical shift, so the curves are congruent and never cross.
Recap
Five things, and every one of them is Chapter 3 read backwards.
| If you see | Then |
|---|---|
| A power other than the reciprocal | Raise the exponent and divide by the new one |
| The reciprocal | The logarithm of the absolute value |
| A product or quotient | Rewrite it: there is no rule to reverse |
| A sine | Minus cosine, not cosine |
| An answer you are unsure of | Differentiate it back |
| A differential equation with a point given | Antidifferentiate first, then fit the constant |
| Two antidifferentiations | Two conditions are needed |
That closes Chapter 4, and with it the derivative. Chapter 5 opens with an apparently unrelated question — the area under a curve — and Section 5.3 proves the astonishing fact that answering it is exactly the antidifferentiation you have just learned.
OpenStax Calculus Volume 1, §4.10 Antiderivatives §4.10, pp. 419-430 — everything on these slides traces back here
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