First Fundamental Theorem of Calculus

Theorem (Fundamental Theorem of Calculus I)

Theorem (FTC I).
Let \(f\) be continuous on \([a,b]\), and define

\[ F(x)\;=\;\int_{a}^{\,x} f(t)\,\mathrm{d}t,\quad x\in[a,b]. \]

Then \(F\) is differentiable on \((a,b)\) and

\[ F'(x) \;=\; f(x) \quad\text{for every }x\in(a,b). \]

Proof. Fundamental Theorem of Calculus I.

Let \(f\) be continuous on \([a,b]\), and define

\[ F(x)\;=\;\int_{a}^{\,x} f(t)\,\mathrm{d}t,\quad x\in[a,b]. \]

We will show that \(F\) is differentiable on \((a,b)\) and that \(F'(x)=f(x)\) for every \(x\in(a,b)\).

To show that \(F\) is differentiable, we need to compute the difference quotient \(\tfrac{F(x+h)-F(x)}{h}\) and take the limit as \(h\to0\).

Fix any point \(x\in(a,b)\) and any \(h\neq0\) such that \(x+h\in(a,b)\). We compute the difference quotient for \(F\) at \(x\):

\[ \frac{F(x+h)-F(x)}{h} \;=\; \frac{1}{h}\Bigl(\int_a^{\,x+h}f(t)\,\mathrm{d}t \;-\;\int_a^{\,x}f(t)\,\mathrm{d}t\Bigr) \;=\; \frac{1}{h}\,\int_{x}^{\,x+h} f(t)\,\mathrm{d}t. \]

Because \(f\) is continuous at \(x\), on the tiny interval \([x,x+h]\) it attains both a minimum \(m_h\) and a maximum \(M_h\) by the Extreme Value Theorem, and these both converge to \(f(x)\) as \(h\to0\).

Thus

\[ m_h \,\le\, \frac{1}{h}\int_{x}^{x+h}f(t)\,\mathrm{d}t \,\le\, M_h, \]

and since both the lower bound \(m_h\to f(x)\) and the upper bound \(M_h\to f(x)\), the Squeeze Theorem gives

\[ \lim_{h\to0}\frac{F(x+h)-F(x)}{h} \;=\; f(x). \]

In other words, \(F'(x)=f(x)\).

Because \(x\) was arbitrary in \((a,b)\), \(F\) is differentiable there with \(F'=f.\)

This completes the proof. ◻