For two thousand years, mathematicians knew how to find areas and how to find tangents — but they believed these were separate, unrelated problems. Then Isaac Newton and Gottfried Leibniz discovered the connection: differentiation and integration are inverse operations. This insight, the Fundamental Theorem of Calculus, is arguably the most important theorem in mathematics. It turned calculus from a collection of tricks into a unified theory — and made modern science possible.
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If $\frac{d}{dx}(\int_0^x t^2 \, dt) = x^2$, what does this tell you about the relationship between differentiation and integration?
Core Content
Statement: If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then:
∫_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.
Why it matters: Before the FTC, finding areas required summing infinitely many rectangles (Riemann sums) — tedious and often impossible. After the FTC, finding areas became as simple as finding antiderivatives.
Example: Find the area under $y = x^2$ from $0$ to $3$.
∫_0^3 x^2 dx = [x^3/3]_0^3 = 27/3 - 0 = 9
This is the same answer we get from Riemann sums — but obtained in seconds rather than hours.
Statement: If $f$ is continuous on $[a, b]$, then the function $g$ defined by:
g(x) = ∫_a^x f(t) dt
is continuous on $[a, b]$, differentiable on $(a, b)$, 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.
Example:
d/dx(∫_0^x t^3 dt) = x^3
d/dx(∫_2^x e^t dt) = e^x
With chain rule: If the upper limit is a function $u(x)$:
d/dx(∫_a^{u(x)} f(t) dt) = f(u(x)) · u'(x)
Find $\frac{d}{dx}\left(\int_1^{x^2} \ln t \, dt\right)$.
The derivative.
By FTC Part 1 with chain rule:
d/dx(∫_1^{x^2} ln(t) dt) = ln(x^2) · d/dx(x^2)
= ln(x^2) · 2x
= 2x · 2ln(x)
= 4x ln(x)
$4x \ln x$ (for $x > 0$).
Find $\frac{d}{dx}\left(\int_0^{3x} e^{t^2} \, dt\right)$.
Answer:
$e^{(3x)^2} \cdot 3 = 3e^{9x^2}$.
$\int_a^b f(x) \, dx = F(b) - F(a)$
$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$
$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u) \cdot u'$
Differentiation and integration are inverses
The equation $\frac{d}{dx}(\int_0^x t^2 \, dt) = x^2$ shows that differentiation undoes integration. This is the essence of FTC Part 1. It means that if we integrate a function and then differentiate the result, we get back the original function. The two operations are inverses — like addition and subtraction, or multiplication and division. This inverse relationship is what makes calculus a unified subject rather than a collection of separate techniques.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Evaluate $\int_0^2 (x^3 + 3x^2 - 2x + 1) \, dx$ using FTC Part 2. Show all working. 3 MARKS
9. Find $\frac{d}{dx}\left(\int_1^{x^2} (t^3 + 1) \, dt\right)$. Show all working. 3 MARKS
10. 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
1. $[\frac{x^4}{4}]_0^4 = 64 - 0 = 64$
2. $\sqrt{x}$ by FTC Part 1.
3. $e^{x^3} \cdot 3x^2 = 3x^2 e^{x^3}$.
4. $[\ln x]_1^e = 1 - 0 = 1$.
1. It connects differentiation and integration, turning hard area problems into easy antiderivative problems.
2. $g(x+h) - g(x) = \int_x^{x+h} f(t) \, dt \approx f(x) \cdot h$, so $\frac{g(x+h) - g(x)}{h} \approx f(x)$.
3. $\frac{dK}{dt} = 3t + 2$. This is the investment rate at time $t$.
Q8 (3 marks): $F(x) = \frac{x^4}{4} + x^3 - x^2 + x$ [1]. $[F(x)]_0^2 = (\frac{16}{4} + 8 - 4 + 2) - (0)$ [1] $= 4 + 8 - 4 + 2 = 10$ [1].
Q9 (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].
Q10 (3 marks): (a) Before the FTC, finding areas required Riemann sums — summing infinitely many rectangles, which was computationally infeasible for most functions [1]. (b) The FTC shows that differentiation and integration are inverse operations, connecting the geometry of curves (tangents/slopes) with the geometry of regions (areas/volumes) [1]. (c) In physics, velocity is the derivative of displacement, and displacement is the integral of velocity. The FTC guarantees that computing displacement from velocity (integration) and then finding velocity from displacement (differentiation) returns the original function, making kinematics consistent [1].
Climb platforms using FTC Part 1 and Part 2, evaluating definite integrals and differentiating integral functions. Pool: lesson 5.
Tick when you've finished all activities and checked your answers.