Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.
Fix the left end of an interval and let the right end move. The area accumulated so far is a new function of $x$. The Fundamental Theorem says its rate of change is the height of the curve at the moving edge. This ties integrals to derivatives, and the The Chain Rule extends it.
🔑 Idea clave
Fundamental Theorem of Calculus, Part 1. If $f$ is continuous and $A(x) = \displaystyle\int_a^x f(t)\,dt$, then $A'(x) = f(x)$. With a function as the upper limit: $$\dfrac{d}{dx}\displaystyle\int_a^{u(x)} f(t)\,dt = f\big(u(x)\big)\cdot u'(x)$$.
Top: the area under $f(t) = 2 - t$ from 0 to $x$. Bottom: that area as a function, $A(x)$. The slope of $A$ is always the height of $f$.
💡 Por qué funciona
Increase $x$ by a little, $h$. The extra area is a thin strip from $x$ to $x + h$, almost a rectangle of height $f(x)$ and width $h$: $A(x + h) - A(x) \approx f(x)\,h$. So $\frac{A(x + h) - A(x)}{h} \approx f(x)$, and the approximation becomes exact as $h \to 0$ because $f$ is continuous. Area grows at a rate equal to the height of the leading edge.
Forgetting the chain rule factor when the upper limit is a function of $x$: writing $\cos(x^2)$ instead of $2x\cos(x^2)$ in the second example. Check: if the upper limit is not plain $x$, there must be a factor $u'(x)$ in your answer.
Práctica
Apruebas con 3 respuestas correctas, las 2 últimas seguidas. Cada respuesta correcta da 3 XP.