Differentiating an area AB & BC

Fix a starting point $a$ and let the finish line move. Define

$$g(x) = \int_{a}^{x} f(t)\,dt.$$

For each $x$ this is a number, the signed area collected from $a$ up to $x$, so $g$ is a genuine function of $x$. Three readings of the notation to settle before anything else.

1. The letter inside is not the letter outside. The $t$ is the dummy variable being integrated away; the $x$ is the input to $g$. Writing $\int_{a}^{x} f(x)\,dx$ uses one letter for two jobs and makes the expression unreadable, which is why the convention exists.
2. $g(a) = 0$. An interval of zero width collects zero area, whatever $f$ is doing at $a$.
3. $g$ can be negative. It is a signed area, so it falls wherever $f$ is below the axis.

Now push the finish line a little further, from $x$ to $x + h$. The extra area is a thin strip of width $h$ whose height is about $f(x)$, so $g(x + h) - g(x) \approx f(x)\,h$. Divide by $h$ and let $h \to 0$:

$$g'(x) = f(x).$$

That is the Fundamental Theorem of Calculus, Part 1: if $f$ is continuous, then $g(x) = \int_{a}^{x} f(t)\,dt$ is differentiable and its derivative is $f$ itself. Accumulating and differentiating undo each other, which is the sentence the entire unit exists to establish.

The starting point $a$ has vanished from the conclusion, and that is not an oversight. Moving $a$ shifts every value of $g$ by the same constant, and a constant shift does not change a derivative.

Replace the upper limit $x$ by a function $u(x)$. Now two things move: $x$ moves, and the finish line moves at its own speed, $u'(x)$. The chain rule handles it exactly as it handles any composition:

$$\frac{d}{dx}\int_{a}^{u(x)} f(t)\,dt = f\!\left(u(x)\right)\cdot u'(x).$$

Read the two factors. The height of the new sliver is $f$ evaluated at the finish line, which is $u(x)$ and not $x$. The width of the new sliver is how fast that finish line is travelling, which is $u'(x)$.

So $\frac{d}{dx}\int_{0}^{x^{2}} f(t)\,dt = f(x^{2})\cdot 2x$. When $x$ increases by a little, the upper limit increases by about $2x$ times as much, and the area grows correspondingly faster.

Two observations that make the omission easy to spot.

1. The plain case is the special case. With $u(x) = x$ the factor is $u'(x) = 1$, which is why $g'(x) = f(x)$ looks like it has no chain rule in it. It has one; it is invisible.
2. Check by units or by size. An answer of $f(x^{2})$ alone says the area grows at the same rate whether the finish line is crawling or sprinting, which cannot be right.

Suppose the variable is at the bottom instead:

$$h(x) = \int_{x}^{b} f(t)\,dt.$$

Pushing $x$ to the right removes area from the left end of the region, so $h$ decreases where $f$ is positive. Reversing the limits makes that precise, using the property Topic 6.6 states in general:

$$h(x) = -\int_{b}^{x} f(t)\,dt \quad\implies\quad h'(x) = -f(x).$$

With a function in the lower slot, both adjustments apply at once:

$$\frac{d}{dx}\int_{v(x)}^{b} f(t)\,dt = -f\!\left(v(x)\right)\cdot v'(x).$$

And if both limits move, the integral splits at any convenient fixed point $c$, which turns it into two problems of the kind already solved:

$$\frac{d}{dx}\int_{v(x)}^{u(x)} f(t)\,dt = f\!\left(u(x)\right)u'(x) - f\!\left(v(x)\right)v'(x).$$

The upper limit contributes and the lower limit takes away, with each carrying its own chain-rule factor. A single term is the giveaway that one of the two moving edges was not noticed.

Part 1 says every continuous function has an antiderivative, and hands one over: $\int_{a}^{x} f(t)\,dt$ is a function whose derivative is $f$. That matters even when the antiderivative has no formula. $\int_{0}^{x} e^{-t^{2}}\,dt$ cannot be written with elementary functions, and Part 1 still tells you its derivative exactly: $e^{-x^{2}}$.

Practically, it turns questions about $g$ into questions about $f$, which you can see on a graph:

1. $g$ is increasing where $f$ is positive, since $g' = f$.
2. $g$ has a critical point where $f$ crosses the axis.
3. $g$ is concave up where $f$ is increasing, since $g'' = f'$.

Every increasing, concave and inflection question from Unit 5 comes back here one level down, with $f$ playing the role that $f'$ played there. Topic 6.5 is entirely about reading those off a graph, so the vocabulary is worth re-anchoring now rather than rebuilding later.

One thing Part 1 does not do: it does not evaluate anything. Getting a number out of $\int_{a}^{b} f(t)\,dt$ is Part 2, in Topic 6.7, and it runs the other way, from an antiderivative to a value.