Reversing differentiation — recovering functions from their rates of change, building the family of antiderivatives, and preparing the foundation for the Fundamental Theorem of Calculus
Chapter 9 introduces integration as the reverse process of differentiation. You have spent the previous five chapters learning to differentiate functions — to find rates of change, slopes, and instantaneous behavior. This chapter turns that process around and asks a different question: if you know the rate of change of a quantity, can you recover the quantity itself?
The answer is yes — but with an important qualification. When you reverse differentiation, you do not get a single function back. You get a family of functions, all differing by a constant. That constant, called the constant of integration, is not a technicality. It carries real meaning: it represents the unknown initial condition, the starting position, the baseline value, or some other quantity that must be determined from additional information.
This chapter builds carefully toward that understanding. You will learn what antiderivatives are, how to write and interpret indefinite integral notation, how to apply basic integration rules, how to solve initial value problems, and how to use antiderivatives to model motion, economics, and science. The chapter also distinguishes clearly between finding a family of antiderivatives and finding accumulated area — that distinction will matter deeply in Chapters 10 and 11.
A car is moving along a straight road. You know the car's velocity at every moment in time, given by the function v(t) = 3t² − 4t + 1, where v is measured in meters per second and t in seconds. You want to know the car's position at time t = 5 seconds. But to find position, you need to know where the car started — you need the position at t = 0. This problem has two parts: first, find a function whose derivative is v(t) — that is an antiderivative. Second, use the starting position to determine which antiderivative is the right one — that is an initial value problem. Both parts are central to Chapter 9.
Every time you know a rate of change and need to recover the underlying quantity, you are doing integration. A physicist knows the velocity of a falling object and wants the position. An economist knows marginal cost and wants total cost. A biologist knows the rate of population growth and wants the population itself. An engineer knows the rate of water flow and wants the total volume that has flowed.
In each case, the reverse of differentiation — finding an antiderivative — is the tool that connects rate of change back to the accumulated quantity. Integration is not only the inverse of differentiation in a formal algebraic sense. It is the conceptual inverse in a physical, scientific, and economic sense as well.
Differentiation answers the question: given a function f(x), what is its rate of change? Integration answers a different question: given a function f(x) that represents a rate of change, what function has f(x) as its derivative? That second question is the one this chapter addresses.
A function F(x) is an antiderivative of f(x) on an interval if F′(x) = f(x) for every x in that interval. In other words, F is an antiderivative of f if differentiating F produces f.
Let f(x) = 2x. You already know that d/dx[x²] = 2x. So F(x) = x² is an antiderivative of f(x) = 2x. But is it the only one? Check: d/dx[x² + 1] = 2x. So x² + 1 is also an antiderivative. Check: d/dx[x² − 7] = 2x. So x² − 7 is also an antiderivative. Check: d/dx[x² + 100] = 2x. So x² + 100 is also an antiderivative.
This pattern reveals a fundamental fact. Adding any constant to an antiderivative produces another antiderivative, because the derivative of a constant is zero.
If F(x) is an antiderivative of f(x) on an interval, then the most general antiderivative of f(x) on that interval is
F(x) + C
where C is any constant, called the constant of integration.
Think of antiderivatives the way you think of families of parallel lines in algebra. The line y = 3x + 5 and the line y = 3x − 2 have the same slope but different intercepts. In the same way, F(x) + 2 and F(x) − 9 have the same derivative but different vertical positions. Integration produces the slope structure; the constant of integration provides the vertical position.
Draw three curves on the same coordinate plane: y = x² + 5, y = x², and y = x² − 3. Label each. Draw tangent lines at the same x-value on each curve. Show that all three tangent lines have the same slope. Note that the curves are vertical translations of each other. Each one is a valid antiderivative of f(x) = 2x.
Because integration reverses differentiation, you can always check an antiderivative by differentiating it and confirming that you recover the original function. If you differentiate your answer and get back f(x), your antiderivative is correct. If you do not, something went wrong and you must find the error.
F′(x) = d/dx [(1/4)x⁴ − 3x² + 5x + C]
= (1/4)·4x³ − 3·2x + 5·1 + 0
= x³ − 6x + 5
Every differentiation rule has a corresponding integration rule. Learning to see integration as the reverse of differentiation — not as a mysterious new process — is one of the most productive habits a calculus student can build.
| Differentiation | Integration |
|---|---|
| d/dx[x^(n+1)] = (n+1)xⁿ | ∫ xⁿ dx = x^(n+1)/(n+1) + C |
| d/dx[eˣ] = eˣ | ∫ eˣ dx = eˣ + C |
| d/dx[sin x] = cos x | ∫ cos x dx = sin x + C |
| d/dx[−cos x] = sin x | ∫ sin x dx = −cos x + C |
| d/dx[ln|x|] = 1/x | ∫ (1/x) dx = ln|x| + C |
The notation for an indefinite integral uses the integral sign ∫ — an elongated S standing for "sum," a choice that connects to the accumulated-area interpretation developed in Chapter 10. The indefinite integral of f(x) with respect to x is written as ∫ f(x) dx, and represents the most general antiderivative of f(x).
Every part of the notation carries meaning. The ∫ is the integral sign — it signals that you are finding an antiderivative. The f(x) is the integrand — the function being integrated. The dx is the differential — it tells you which variable you are integrating with respect to.
∫ f(x) dx = F(x) + C
where F′(x) = f(x) and C is the constant of integration. The result of an indefinite integral is a family of functions, not a single function.
The dx in the integral notation tells you the variable of integration. When you write ∫ 3t² dt, the variable is t, not x. The answer will be a function of t. Writing ∫ 3t² dx would be a different and generally incorrect expression for this context.
∫ 4x³ dx = x⁴ + CEvery time you differentiate a constant, the constant vanishes. When you reverse differentiation, you cannot recover a constant that has already vanished — unless you know something else about the situation. This is why integration always produces a family of functions. The constant of integration, C, is a placeholder for whatever constant was present before differentiation but cannot be recovered from the derivative alone.
Geometrically, the antiderivatives of f(x) are a family of vertically shifted curves. They all have the same shape — the same slopes at every x — but they pass through different points on the y-axis.
Students sometimes write ∫ f(x) dx = F(x), omitting +C, because they found one correct antiderivative and stopped. This is incomplete. It describes one curve in a family, not the family itself. A missing +C is not a formatting error — it represents a conceptual mistake: it implies you believe there is only one antiderivative, when in fact there are infinitely many.
For any real number n ≠ −1,
∫ xⁿ dx = x^(n+1) / (n+1) + C
The exponent increases by 1, and then you divide by the new exponent. The restriction n ≠ −1 is required because n + 1 = 0 would produce division by zero. The integral of x⁻¹ = 1/x is handled separately using logarithms.
(a) ∫ x⁵ dx = x⁶/6 + C
(b) ∫ x^(1/2) dx = x^(3/2) / (3/2) + C = (2/3)x^(3/2) + C
(c) ∫ x⁻³ dx = x⁻² / (−2) + C = −1/(2x²) + C
∫ 3x^(−1/2) dx = 3 · x^(1/2) / (1/2) + C = 6x^(1/2) + C = 6√x + C
Constant Multiple Rule: ∫ k·f(x) dx = k ∫ f(x) dx
Sum and Difference Rule: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
Integral of a Constant: ∫ k dx = kx + C
Integral of 1/x: ∫ (1/x) dx = ln|x| + C (x ≠ 0)
∫ (5x⁴ − 3x² + 7x − 2) dx
= 5·(x⁵/5) − 3·(x³/3) + 7·(x²/2) − 2x + C
= x⁵ − x³ + (7/2)x² − 2x + C
When you integrate a sum term by term, each term would produce its own constant. But since C₁ + C₂ + C₃ is itself just one arbitrary constant, combine all of them into a single +C at the end. Never write multiple separate +C values in one expression.
The natural exponential function eˣ has the remarkable property that it is its own derivative: d/dx[eˣ] = eˣ. This means it is also its own antiderivative — up to the constant of integration.
∫ eˣ dx = eˣ + C
∫ aˣ dx = aˣ / ln a + C (a > 0, a ≠ 1)
∫ e^(kx) dx = e^(kx) / k + C (k ≠ 0)
(a) ∫ (3eˣ − 2x) dx = 3eˣ − x² + C
(b) ∫ 2ˣ dx = 2ˣ / ln 2 + C
(c) ∫ e^(5x) dx = e^(5x) / 5 + C
The formula ∫ e^(kx) dx = e^(kx)/k + C works only when the exponent is exactly kx — a linear expression in x with no additional functions. For more complex exponents, such as e^(x²), this formula does not apply and the integral requires more advanced techniques covered in Calculus II.
| Differentiation | Integration |
|---|---|
| d/dx[sin x] = cos x | ∫ cos x dx = sin x + C |
| d/dx[−cos x] = sin x | ∫ sin x dx = −cos x + C |
| d/dx[tan x] = sec²x | ∫ sec²x dx = tan x + C |
| d/dx[−cot x] = csc²x | ∫ csc²x dx = −cot x + C |
| d/dx[sec x] = sec x tan x | ∫ sec x tan x dx = sec x + C |
| d/dx[−csc x] = csc x cot x | ∫ csc x cot x dx = −csc x + C |
Integrating sine gives negative cosine; integrating the cosecant-related functions also introduces a negative sign. This mirrors the pattern in the derivative rules and can always be confirmed by differentiating your answer.
(a) ∫ (3cos x − 2sin x) dx
= 3 sin x − 2(−cos x) + C
= 3 sin x + 2 cos x + C
(b) ∫ cos(4x) dx = sin(4x) / 4 + C
When the argument of a trig function is kx, the chain rule introduces a factor of k that must be compensated:
∫ cos(kx) dx = sin(kx)/k + C
∫ sin(kx) dx = −cos(kx)/k + C
An initial value problem consists of two parts: (1) a differential equation, which in this chapter means a statement that y′ = f(x), and (2) an initial condition, which is a known value of the function at a specific input — typically y(x₀) = y₀. The goal is to find the unique function y = F(x) that satisfies both.
Step 1. Find the general antiderivative F(x) + C.
Step 2. Use the initial condition to solve for C, producing the unique particular solution.
y = ∫ (6x² − 4x + 1) dx = 2x³ − 2x² + x + C
y(0) = 2(0)³ − 2(0)² + (0) + C = 3
C = 3
P(t) = ∫ 200e^(0.05t) dt
= 200 · e^(0.05t)/0.05 + C
= 4000e^(0.05t) + C
P(0) = 4000e⁰ + C = 4000
4000 + C = 4000 → C = 0
f(x) = ∫ 3x² dx = x³ + Cf(1) = (1)³ + C = 7 → C = 6
In motion problems, s(t) is the position of an object at time t, v(t) = s′(t) is the velocity, and a(t) = v′(t) = s″(t) is the acceleration. Differentiation moves down this chain from position to acceleration. Integration moves back up.
Differentiate: s(t) → v(t) → a(t)
Integrate: a(t) → v(t) → s(t)
Each integration step introduces a new constant, which must be determined from an initial condition.
If s(t) is in meters and t is in seconds, then v(t) is in m/s and a(t) is in m/s². Integrating a(t) in m/s² with respect to t in seconds gives a result in m/s — velocity. The units confirm the operation makes sense. Always check units as a form of error detection.
v(t) = ∫ (12t − 6) dt = 6t² − 6t + Cv(0) = 0 − 0 + C = 4 → C = 4s(t) = ∫ (3t² − 4t + 1) dt = t³ − 2t² + t + Cs(0) = 0 − 0 + 0 + C = 5 → C = 5v(t) = ∫ −9.8 dt = −9.8t + C₁
Apply v(0) = 15: C₁ = 15
v(t) = −9.8t + 15
s(t) = ∫ (−9.8t + 15) dt = −4.9t² + 15t + C₂
Apply s(0) = 2: C₂ = 2
A differential equation is an equation that involves an unknown function and one or more of its derivatives. Every problem in this chapter of the form y′ = f(x) is technically a simple differential equation. The equation y′ = f(x) says: find a function y whose derivative is f(x). The solution is the family of antiderivatives y = F(x) + C.
Most laws of physics, economics, biology, and engineering are naturally expressed as relationships between quantities and their rates of change. Newton's second law — force equals mass times acceleration — gives a differential equation for velocity. A savings account with continuous compounding satisfies dA/dt = rA. The rate of cooling of a hot object is proportional to the temperature difference from its surroundings. All of these are differential equations. Chapter 13 will develop this theory systematically.
y = ∫ (x² + 3) dx = x³/3 + 3x + C
y(2) = 8/3 + 6 + C = 10
8/3 + C = 4 → C = 4 − 8/3 = 4/3
W(t) = ∫ (2t + 3) dt = t² + 3t + CW(0) = 0 + 0 + C = 10 → C = 10In economics, marginal cost MC(x) is the rate of change of total cost with respect to the number of items produced: MC(x) = C′(x). To recover total cost from marginal cost, you integrate. The constant of integration represents the fixed cost — the cost incurred even when production is zero.
C(x) = ∫ (0.6x² − 4x + 50) dx
= 0.2x³ − 2x² + 50x + K
C(0) = 0 − 0 + 0 + K = 2000 → K = 2000
The +C is not optional. It represents a family of functions, not a single one.
Division by zero is undefined. The power rule does not apply when n = −1.
Differentiating e^(3x) produces a factor of 3 from the chain rule. Reversing this requires dividing by 3. Always check by differentiating your answer.
Check: d/dx[−cos x] = sin x. The negative sign is real and necessary.
Integration does not distribute over multiplication. Verify all integral steps by checking the derivative of your answer.
The answer must be in terms of the variable of integration indicated by the differential dt or dx.
Finding F(x) + C is the first step, not the final answer. An answer of F(x) + C when a particular solution is requested is incomplete.
When integrating a sum, all individual constants combine into one. Write a single +C at the end.
This chapter introduced antiderivatives and indefinite integrals. The indefinite integral ∫ f(x) dx = F(x) + C produces a family of functions — not a number. But many of the most important applications of integration produce numbers: the area under a curve between two values, the total distance traveled during a time interval, the net change in a population over five years.
To compute these quantities, you need the definite integral, which integrates over a specific interval from x = a to x = b and produces a numerical result. Chapter 10 will build the definite integral from the ground up — starting with rectangles, moving to Riemann sums, and arriving at the precise definition of accumulated area as a limit.
The connection between the work of this chapter and the Fundamental Theorem of Calculus is this:
∫ab f(x) dx = F(b) − F(a)
where F is any antiderivative of f. The constant of integration disappears when you subtract F(a) from F(b), which is why definite integrals produce definite numbers rather than families. Understanding what antiderivatives are and how to find them — the work of this chapter — is the foundation. Everything in Chapters 10 and 11 builds on it.
Find the general antiderivative. Include +C. Check each answer by differentiation.
Use the instructions and your knowledge of antiderivative geometry to respond.
Each problem contains an error. Identify it and write the correct solution.
What is missing? Write the correct answer.
Identify the error. What is the correct formula?
What chain-rule factor is missing? Write the correct answer and verify by differentiating.
What sign error was made? Verify the correct answer by differentiation.
Two errors appear here. Identify both. What is the correct particular solution?
Why is this approach invalid? What is the correct method and answer?
∫ x⁷ dx∫ (2x^(1/2) − 5x + 3) dx∫ e^(6x) dx∫ (cos x − 1/x) dx∫ (4x³ − 2x + e^x + sin x) dxy′ = 4x³ − 2x, y(0) = 1dy/dx = eˣ + 3, y(0) = −2f″(x) = 10x + 2, f′(0) = −3, f(1) = 4a(t) = −10 m/s², initial velocity v(0) = 20 m/s, and initial position s(0) = 0. Find s(t) and determine when the particle momentarily stops.a(t) = 30 − 4t m/s². It starts from rest at the launch pad. Find v(t) and s(t).MC(x) = 2x + 30 dollars per unit, with fixed cost $800. Write the total cost function. What is the cost of producing 20 units?r(t) = 50 − 2t thousand gallons per day. At t = 0, the reservoir holds 400 thousand gallons. Find W(t) and determine when the inflow rate drops to zero.∫ sin(3x) dx = cos(3x) + C. Identify the two errors and write the correct answer.f′(x) = x², f(2) = 5, and concludes f(x) = x³/3. What error was made? What is the correct particular solution?Chapter 9 gave you the tools to reverse differentiation. You can now find antiderivatives, evaluate indefinite integrals, apply integration rules, and solve initial value problems. But antiderivatives alone do not tell you the area under a curve, the net displacement over an interval, or the total accumulated change between two specific moments in time. For that, you need the definite integral.
Chapter 10 builds the definite integral from the ground up — starting with rectangles, moving to Riemann sums, and arriving at the precise definition of accumulated area as a limit. You will see that computing accumulated quantities is not just a geometric problem. It is the same problem as finding antiderivatives, and the connection between the two is the most important theorem in calculus. The work of Chapter 9 is the essential preparation.