The fundamental theorem of calculus links the two halves of calculus. In Calculus 1 you used it; now we can prove it, using bounds for integrals and the definition of the derivative. The proof also shows exactly what can go wrong at a jump.
🔑 Key idea
Let $f$ be integrable on $[a, b]$ and $F(x) = \int_a^x f(t)\,dt$. Then $F$ is continuous, indeed $|F(x) - F(y)| \le M|x - y|$ when $|f| \le M$. If $f$ is continuous at $c$, then $F'(c) = f(c)$. At a jump of $f$, $F$ has a corner: its one-sided derivatives are the one-sided values of $f$.
💡 Why it works
Continuity. $|F(x) - F(y)| = \left|\int_y^x f\right| \le M|x - y|$. Derivative. For $h \ne 0$, $$\frac{F(c + h) - F(c)}{h} - f(c) = \frac{1}{h}\int_c^{c+h} \big(f(t) - f(c)\big)\,dt.$$ Let $\varepsilon > 0$. Continuity at $c$ gives $\delta$ with $|f(t) - f(c)| < \varepsilon$ for $|t - c| < \delta$. For $0 < |h| < \delta$, the integrand is less than $\varepsilon$ in size over an interval of length $|h|$, so the right side is at most $\frac{1}{|h|}\cdot\varepsilon|h| = \varepsilon$. So the difference quotient tends to $f(c)$. $\blacksquare$ The other half, $\int_a^b g' = g(b) - g(a)$, follows from the mean value theorem applied on each piece of a partition.
The area function of a step function: $F$ is made of straight pieces, with a corner where $f$ jumps.
The two halves of the theorem fit together like this. Part 1 (above) says that integrating and then differentiating gets you back where you started, at points of continuity. Part 2 says that differentiating and then integrating also gets you back: if $g$ is differentiable and $g'$ is integrable, then for any partition the mean value theorem gives $g(x_i) - g(x_{i-1}) = g'(c_i)\Delta x_i$, so $g(b) - g(a) = \sum g'(c_i)\Delta x_i$, a Riemann sum for $\int_a^b g'$. It lies between the lower and upper sums for every partition, so it equals the integral.
Worked examples
✏️ Worked example: The area function of a step function
Let $f(t) = 1$ for $t < 2$ and $f(t) = 3$ for $t \ge 2$, and $F(x) = \int_0^x f$. Find $F(3)$ and the one-sided derivatives of $F$ at $2$.
For $x > 2$: $F(x) = 1 \cdot 2 + 3(x - 2)$, so $F(3) = 5$.
Just left of $2$, $F$ has slope $1$; just right, slope $3$.
Answer: $F(3) = 5$; left derivative $1$, right derivative $3$, so $F'(2)$ does not exist.
✏️ Worked example: The chain rule with the FTC
Let $G(x) = \int_1^{x^{2}} (t^{2} + 3)\,dt$. Find $G'(2)$.
$G(x) = F(x^{2})$ with $F' = t^{2} + 3$, so $G'(x) = \left((x^{2})^{2} + 3\right)\cdot 2x$.
Answer: $G'(2) = (16 + 3) \cdot 4 = 76$.
⚠️ Common mistake
"$F' = f$ everywhere." The FTC gives $F'(c) = f(c)$ at points where $f$ is continuous. At a jump of $f$, the difference quotients of $F$ from the two sides tend to the two different one-sided values, so $F$ is not differentiable there. The graph of $F$ is still continuous: integration smooths jumps into corners.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.