The Fundamental Theorem of Calculus
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.
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?
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: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$.
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
Concepts
- Why the FTC connects the two branches of calculus
- How antiderivatives solve area problems
- The historical significance of the theorem
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
Statement: If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then:
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 $\int_0^4 x^3\,dx$?
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 $g'(x) = f(x)$.
What it means: Differentiating an integral gives back the original function. Integration followed by differentiation returns you to where you started.
Simple examples:
With chain rule — composite upper limit: If the upper limit is $u(x)$ instead of $x$:
FTC Part 1: $\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)$; Chain rule extension: $\dfrac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x)) \cdot u'(x)$
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: $\dfrac{d}{dx}\!\left(\displaystyle\int_0^{x^2} e^t\,dt\right) = e^{x^2}$.
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle\int_0^2 (x^3 + 3x^2 - 2x + 1)\,dx$ using FTC Part 2.
Find $\dfrac{d}{dx}\!\left(\displaystyle\int_1^{x^2} \ln t\,dt\right)$.
If $K(t) = \displaystyle\int_0^t (3s + 2)\,ds$ represents capital accumulation, find $\dfrac{dK}{dt}$ and interpret it.
Fill in the blank: $\dfrac{d}{dx}\!\left(\displaystyle\int_0^{3x} e^{t^2}\,dt\right) = e^{(3x)^2} \cdot$ ___
Common errors · the 3 traps that cost marks
Odd one out: Three of these are correct applications of the FTC. Which one is wrong?
Quick-fire practice · 4 problems
Evaluate $\displaystyle\int_0^4 x^3\,dx$
Find $\dfrac{d}{dx}\!\left(\displaystyle\int_2^x \sqrt{t}\,dt\right)$
Find $\dfrac{d}{dx}\!\left(\displaystyle\int_0^{x^3} e^t\,dt\right)$
Evaluate $\displaystyle\int_1^e \frac{1}{x}\,dx$
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $\displaystyle\int_0^2 (x^3 + 3x^2 - 2x + 1)\,dx$ using FTC Part 2. Show all working. (3 marks)
Q2. Find $\dfrac{d}{dx}\!\left(\displaystyle\int_1^{x^2} (t^3 + 1)\,dt\right)$. Show all working. (3 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)
Comprehensive answers (click to reveal)
Drill 1: $\left[\frac{x^4}{4}\right]_0^4 = 64$ · 2: $\sqrt{x}$ (FTC Part 1) · 3: $e^{x^3} \cdot 3x^2$ · 4: $[\ln x]_1^e = 1$
Q1 (3 marks): $F(x) = \frac{x^4}{4} + x^3 - x^2 + x$ [1]. $F(2) = \frac{16}{4} + 8 - 4 + 2 = 4 + 8 - 4 + 2 = 10$ [1]. $F(0) = 0$, so answer is $10$ [1].
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].
Five timed questions on the FTC. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
Enter the arenaClimb platforms by evaluating definite integrals and differentiating integral functions using both parts of the FTC.
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