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What is Calculus II

the Fundamental Theorem
of Calculus?

The bridge connecting differentiation and integration — proving they are inverse operations and making definite integrals computable in seconds.

Simple Definition

Before this theorem, differentiation and integration looked like two completely separate inventions. Derivatives measure rates of change; integrals measure accumulated area. Why would those be related?

The Fundamental Theorem of Calculus (FTC) reveals they are exact inverses of each other — just as addition and subtraction are inverses, or multiplication and division. This single insight transformed calculus from a collection of clever techniques into a unified, powerful theory.

It has two parts. Part 1 says that differentiation undoes integration. Part 2 gives you a practical formula: to compute a definite integral, find any antiderivative and evaluate it at the two endpoints.

Intuition

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Speed and distance — the perfect analogy

If $v(t)$ is your speed at time $t$, the integral $\int_a^b v(t)\,dt$ gives your total distance travelled from time $a$ to $b$. But distance is also position $s(t)$ — and $v(t) = s'(t)$. So the integral of the derivative gives back the original quantity's net change: $s(b) - s(a)$. That's the FTC in your car every day.

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Two operations that cancel

Differentiate then integrate: you get back (almost) what you started with. Integrate then differentiate: you get back exactly what you started with. The FTC makes this precise — and it's why antiderivatives are the key to computing areas. Without the FTC, you'd have to compute Riemann sums by hand for every integral.

Two Parts of the Theorem

Part 1

Differentiation of an Integral

$$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$$

If you define a function by integrating $f$ from a fixed point $a$ to a variable upper limit $x$, then differentiating that function gives you back $f(x)$. Integration and differentiation cancel out.

Part 2

Evaluating Definite Integrals

$$\int_a^b f(x)\,dx = F(b) - F(a)$$

If $F$ is any antiderivative of $f$ (i.e. $F' = f$), then the definite integral from $a$ to $b$ equals $F(b) - F(a)$. No Riemann sums needed — just plug in the endpoints.

Visual — Accumulation Function

Upper limit b = 3.0

Shaded area = ∫₀ᵇ sin(x) dx = F(b) − F(0). Drag slider to change b.

How to Use Part 2 — Step by Step

To evaluate $\displaystyle\int_1^3 x^2\,dx$:

Find an antiderivative $F(x)$

Use the power rule in reverse: $F(x) = \dfrac{x^3}{3}$. (No need to add $+C$ — it cancels.)

Write bracket notation

$\Big[\dfrac{x^3}{3}\Big]_1^3$

Evaluate at upper bound

$F(3) = \dfrac{27}{3} = 9$

Subtract lower bound value

$F(1) = \dfrac{1}{3}$ → Answer: $9 - \dfrac{1}{3} = \dfrac{26}{3}$

Quick Examples

Example 1 — Basic

$\displaystyle\int_0^2 3x^2\,dx$

$F(x)=x^3$
$[x^3]_0^2 = 8 - 0 = 8$

Example 2 — Trig

$\displaystyle\int_0^{\pi} \sin x\,dx$

$F(x)=-\cos x$
$-\cos\pi+\cos 0 = 1+1 = 2$

Example 3 — Part 1

$\dfrac{d}{dx}\displaystyle\int_0^x e^{t^2}\,dt$

By FTC Part 1:
$= e^{x^2}$ directly

Example 4 — Chain Rule + FTC

$\dfrac{d}{dx}\displaystyle\int_0^{x^2} \cos t\,dt$

FTC + chain rule:
$= \cos(x^2) \cdot 2x$

Common Mistakes

✗ Subtracting in the wrong order

It's always $F(b) - F(a)$ — upper bound minus lower bound. Writing $F(a) - F(b)$ flips the sign of the entire answer. Double-check which limit is $a$ and which is $b$.

✗ Forgetting the chain rule in FTC Part 1

If the upper limit is $g(x)$ instead of $x$, you must multiply by $g'(x)$. $\frac{d}{dx}\int_0^{x^2} f(t)\,dt = f(x^2) \cdot 2x$, not just $f(x^2)$.

✗ Adding +C to a definite integral

The constant $C$ cancels when you subtract $F(a)$ from $F(b)$. You only need $+C$ for indefinite integrals. Writing $[F(x)+C]_a^b$ is harmless but redundant and confusing.

Related Concepts

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