The fundamental theorem
The bottom curve $F$ is the running total of the top curve $f$. Pick any $x$. The slope of $F$ at that $x$ — where can you read it off the $f$ graph?
At $x = 1.57$ the bar reads $2.200$ and the tangent slope $2.200$ — rise $2.20$ over run $1$. At $x = 4.71$ the bar shrinks to $0.200$ and the tangent flattens to $0.200$. The slope stat is computed from the $F$ curve alone — and it lands on the height stat at every stop.
The match is forced. Nudge the sweep from $x$ to $x + h$: the total gains a thin strip of width $h$ and height essentially $f(x)$ — a continuous $f$ barely moves that far. So $\frac{F(x+h) - F(x)}{h} \approx f(x)$, and $h \to 0$ gives Part 1 of the Fundamental Theorem — the slope of the total is the height of the rate:
Part 2 makes it a computing device. Call $G$ an antiderivative of $f$ when $G' = f$. $F$ is one; any two differ by a constant, which cancels in differences, so $\int_a^b f(x)\,dx = G(b) - G(a)$. Here $G(x) = 1.2x - \cos x$ gives $\int_0^5 (\sin x + 1.2)\,dx = (6 - \cos 5) - (-1) = 7 - \cos 5 \approx 6.716$ — the number lesson 1's rectangles could only chase.
The F′(x) slope stat is computed from the bottom curve alone, yet it matches f(x) height at every $x$. Which statement is being demonstrated?
Use Part 2 with the antiderivative $G(x) = 1.2x - \cos x$ to evaluate $\int_0^5 (\sin x + 1.2)\,dx$.
Transfer: define $h(x) = \int_0^x \cos(t^2)\,dt$. No elementary formula for $h$ exists. What is $h'(x)$?
One fact, two faces. Part 1: $\frac{d}{dx}\int_0^x f(t)\,dt = f(x)$ — slope of the total equals height of the rate. Part 2: $\int_a^b f(x)\,dx = G(b) - G(a)$ for any antiderivative — two evaluations replace infinitely many rectangles.