Fundamental Theorem of Calculus

The fundamental theorem of calculus links the concept of differentiation with that of integration.

Statement

!!! theorem "Fundamental Theorem of Calculus (Part 1)" {#thm:ftc1} If \(f\) is continuous on \([a,b]\) and \(F\) is an antiderivative of \(f\), then:

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

!!! theorem "Fundamental Theorem of Calculus (Part 2)" {#thm:ftc2} If \(f\) is continuous on \([a,b]\), then the function \(g\) defined by:

$$ g(x) = \int_a^x f(t) , dt $$

is continuous on $[a,b]$ and differentiable on $(a,b)$, with $g'(x) = f(x)$.

Proof

Proof of Theorem 1

By Theorem ?, we know \(g'(x) = f(x)\), so \(g\) is an antiderivative of \(f\). Let \(F\) be any antiderivative of \(f\), so \(F(x) = g(x) + C\). Then:

\[ F(b) - F(a) = (g(b) + C) - (g(a) + C) = g(b) - g(a) \]

\[ g(b) - g(a) = \int_a^b f(t) , dt - \int_a^a f(t) , dt = \int_a^b f(t) , dt \]

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

Proof of Theorem 2

For \(x \in (a,b)\):

\[ g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{\int_x^{x+h} f(t) , dt}{h} \]

By the mean value theorem for integrals, there exists \(c \in [x, x+h]\) such that:

\[ \int_x^{x+h} f(t) , dt = f(c) \cdot h \]

Therefore \(g'(x) = \lim_{h \to 0} f(c) = f(x)\), by continuity of \(f\).

Key Equation

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

Applications

Computing a Definite Integral

\[ \int_0^1 x^2 , dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]

Relationship with Area

The integral \(\int_a^b f(x) , dx\) represents the net signed area under the curve \(y = f(x)\) from \(x = a\) to \(x = b\).

Limit of a Function — Foundation for continuity
Continuity — Required for FTC
Derivative — Inverse operation
Mean Value Theorem — Used in proof of FTC part 2