For two thousand years, finding areas and finding tangents were thought to be completely separate problems. Then Newton and Leibniz discovered the connection. This single insight, that differentiation and integration are inverse operations, turned calculus from a bag of tricks into a unified theory. It is arguably the most important theorem in mathematics.
Today's hook, When you differentiate $\int_0^x t^2 \, dt$, you get $x^2$ back. Differentiating undoes integrating. But why? And what does that have to do with area? By the end of this lesson, that equation will be the most obvious thing you've ever seen.
0/5QUESTS
1
Orient to the Fundamental Theorem
Work through the visible teaching and complete each embedded check.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
Consider: $\dfrac{d}{dx}\!\left(\displaystyle\int_0^x t^2 \, dt\right) = x^2$. Without any formula what does this equation tell you about the relationship between differentiation and integration?
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The big idea
+5 XP to read
There are two parts to the Fundamental Theorem, and they work together to make a single claim: differentiation and integration are inverse operations. One undoes the other, just like addition and subtraction.
The FTC tells us that every area problem can be solved with an antiderivative, and every antiderivative can be verified by differentiation. Part 2 is the computational engine: $\int_a^b f(x)\,dx = F(b) - F(a)$. Part 1 is the conceptual bridge: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$.
$$\int_a^b f(x)\,dx = F(b) - F(a)$$
FTC Part 2
$\int_a^b f(x)\,dx = F(b) - F(a)$. Find an antiderivative, evaluate at both endpoints, subtract.
FTC Part 1
$\dfrac{d}{dx}\!\int_a^x f(t)\,dt = f(x)$. Differentiating an integral returns the original function.
Chain rule extension
$\dfrac{d}{dx}\!\int_a^{u(x)} f(t)\,dt = f(u(x)) \cdot u'(x)$. Always multiply by the derivative of the upper limit.
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What you'll master
Know
Key facts
FTC Part 1: $\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)$
FTC Part 2: $\int_a^b f(x)\,dx = F(b) - F(a)$
Differentiation and integration are inverse operations
Understand
Concepts
Why the FTC connects the two branches of calculus
How antiderivatives solve area problems
The historical significance of the theorem
Can do
Skills
Apply FTC Part 2 to evaluate definite integrals
Differentiate integral functions using FTC Part 1
Apply the chain rule extension with composite upper limits
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Build the FTC language
Work through the visible teaching and complete each embedded check.
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Key terms
AntiderivativeA function $F$ such that $F'(x) = f(x)$. Also written $\int f(x)\,dx$.
Definite integral$\int_a^b f(x)\,dx$, the net signed area between $f$ and the x-axis on $[a,b]$.
FTC Part 1$\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)$. Differentiation inverts integration.
FTC Part 2$\int_a^b f(x)\,dx = F(b) - F(a)$. Evaluate area via antiderivative endpoints.
Accumulation function$g(x) = \int_a^x f(t)\,dt$, the total accumulated change of $f$ from $a$ to $x$.
Inverse operationsTwo operations that undo each other; differentiation and integration are inverses (up to +C).
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Connect both parts of the FTC
Work through the visible teaching and complete each embedded check.
05
FTC Part 2, the computational engine
core concept
Statement: If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then:
$$\int_a^b f(x)\,dx = F(b) - F(a)$$
What it means: To find the area under $y = f(x)$ from $a$ to $b$, find any antiderivative $F$, evaluate it at the endpoints, and subtract. No summing of infinitely many rectangles required.
Why it matters: Before the FTC, finding areas required Riemann sums, tedious and often impossible. After the FTC, finding areas became as simple as finding antiderivatives.
FTC Part 2: the shaded area equals $F(b) - F(a)$ for any antiderivative $F$.
Example: Find the area under $y = x^2$ from $0$ to $3$.
The same answer comes from Riemann sums, but obtained in seconds rather than hours.
FTC Part 2: $\int_a^b f(x)\,dx = F(b) - F(a)$, where $F' = f$; Notation: write $[F(x)]_a^b$ for "evaluate $F$ at $b$ then subtract $F$ at $a$"
Pause, copy FTC Part 2: $\int_a^b f(x)\,dx = F(b) - F(a)$ where $F' = f$, and the bracket notation $[F(x)]_a^b$ (evaluate at $b$, subtract at $a$) into your book.
Quick check: Using FTC Part 2, which expression correctly evaluates $\int_0^4 x^3\,dx$?
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FTC Part 1, the conceptual bridge
core concept
We just saw FTC Part 2: differentiating the antiderivative and substituting limits evaluates any definite integral. That raises a question: FTC Part 2 uses the antiderivative to evaluate integrals, but is there a reverse statement showing that differentiating an integral recovers the original function? This card answers it → FTC Part 1: $\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)$, meaning differentiation is the exact inverse of integration.
Statement: If $f$ is continuous on $[a, b]$, then the accumulation function $g(x) = \int_a^x f(t)\,dt$ is differentiable and $g'(x) = f(x)$.
$$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$$
What it means: Differentiating an integral gives back the original function. Integration followed by differentiation returns you to where you started.
Accumulation functions in economics. Economists define the capital accumulation function $K(t) = \int_0^t I(s)\,ds$ where $I(s)$ is the investment rate at time $s$. By FTC Part 1, $\dfrac{dK}{dt} = I(t)$, the rate of change of capital equals the investment rate. This is how national accountants track a country's capital stock: the FTC bridges the instantaneous (investment rate) with the cumulative (total capital), the same bridge that connects speed to distance, flow to volume, and growth to size.
Pause, copy FTC Part 1: $\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)$ and its chain rule extension $\dfrac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x)) \cdot u'(x)$ into your book.
Did you get this? True or false: $\dfrac{d}{dx}\!\left(\displaystyle\int_0^{x^2} e^t\,dt\right) = e^{x^2}$.
4
Apply the FTC
Work through the visible teaching and complete each embedded check.
Worked examples · 3 in a row, reveal as you go
PROBLEM 1 · FTC PART 2
Evaluate $\displaystyle\int_0^2 (x^3 + 3x^2 - 2x + 1)\,dx$ using FTC Part 2.
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$F(x) = \dfrac{x^4}{4} + x^3 - x^2 + x$
Find the antiderivative term by term. The constant $C$ is not needed, it cancels on subtraction.
Apply FTC Part 1 with chain rule: $f(u(x)) \cdot u'(x) = \ln(x^2) \cdot 2x$. Simplify using $\ln(x^2) = 2\ln x$ (for $x > 0$).
PROBLEM 3 · CAPITAL ACCUMULATION
If $K(t) = \displaystyle\int_0^t (3s + 2)\,ds$ represents capital accumulation, find $\dfrac{dK}{dt}$ and interpret it.
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By FTC Part 1: $\dfrac{dK}{dt} = 3t + 2$
The upper limit is $t$ (not a composite), so differentiation simply returns the integrand with $s$ replaced by $t$.
2
At time $t$, the investment rate is $I(t) = 3t + 2$ units per period.
Interpretation: $\frac{dK}{dt}$ is the instantaneous rate of capital accumulation, i.e., the investment rate at time $t$. The FTC guarantees this connection between the cumulative stock and the flow rate.
Fill in the blank: $\dfrac{d}{dx}\!\left(\displaystyle\int_0^{3x} e^{t^2}\,dt\right) = e^{(3x)^2} \cdot$ ___
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Avoid FTC traps
Work through the visible teaching and complete each embedded check.
Common errors · the 3 traps that cost marks
Trap 01
Forgetting to subtract $F(a)$
Writing $F(b)$ alone is wrong, you must always evaluate at both limits. Many students evaluate at $b$ and forget to subtract $F(a)$, especially when $a = 0$ and $F(0) = 0$. Write the bracket $[F(x)]_a^b$ to force the habit.
Trap 02
Missing the chain rule factor
$\dfrac{d}{dx}\int_0^{x^2} f(t)\,dt = f(x^2) \cdot 2x$, not $f(x^2)$ alone. Whenever the upper limit is not just $x$, you must multiply by the derivative of the upper limit.
Trap 03
Differentiating the integrand instead
FTC Part 1 says differentiate the integral, not differentiate the integrand. If you differentiate $f(t)$ inside the integral, you get $f'(t)$, which is wrong. The answer is simply $f(x)$, the original function evaluated at the upper limit.
Odd one out: Three of these are correct applications of the FTC. Which one is wrong?
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Practise and reflect
Work through the visible teaching and complete each embedded check.
Earlier you considered: $\dfrac{d}{dx}\!\left(\int_0^x t^2\,dt\right) = x^2$. This is FTC Part 1 in action: differentiating an accumulation function returns the original integrand. It shows that differentiation undoes integration, they are inverse operations, just like addition and subtraction. This inverse relationship is what makes calculus a unified subject rather than a collection of separate techniques.
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7
Complete the short-answer practice
Apply the lesson methods, then compare each response with its comprehensive answer.
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Focus-area checkpoint
checkpoint
Use the visible short-answer practice below during this lesson. After this lesson sequence, use Checkpoint 2 for the focus-area checkpoint.
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Short answer
ApplyBand 43 marks
Q1. Evaluate $\displaystyle\int_0^2 (x^3 + 3x^2 - 2x + 1)\,dx$ using FTC Part 2. Show all working. (3 marks)
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ApplyBand 43 marks
Q2. Find $\dfrac{d}{dx}\!\left(\displaystyle\int_1^{x^2} (t^3 + 1)\,dt\right)$. Show all working. (3 marks)
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AnalyseBand 53 marks
Q3. The Fundamental Theorem of Calculus is often called the most important theorem in mathematics. Explain why it deserves this title by describing: (a) what problem it solved that mathematicians had struggled with for centuries, (b) how it connects the two main branches of calculus, and (c) one real-world application where this connection is essential. (3 marks)
Q2 (3 marks): By FTC Part 1 with chain rule: $\frac{d}{dx}\int_1^{x^2}(t^3+1)\,dt = ((x^2)^3 + 1) \cdot 2x$ [2] $= (x^6 + 1) \cdot 2x = 2x^7 + 2x$ [1].
Q3 (3 marks): (a) Finding areas required Riemann sums, summing infinitely many rectangles, computationally infeasible for most functions [1]. (b) The FTC shows differentiation and integration are inverse operations, connecting slopes of curves with areas under them [1]. (c) In physics, velocity is the derivative of displacement; the FTC guarantees integrating velocity gives displacement, making kinematics consistent [1].
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Review and complete
Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.
01
Retrieve the lesson before you leave
State both parts of the Fundamental Theorem of Calculus.
Explain how differentiation and integration become inverse processes.
Describe when the chain rule is needed with an integral-defined function.
Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.
01
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
