How does the accumulation function $A(x)$ generalize the concept of area to signed integrals?

Continuity ensures that as the interval width $h$ approaches zero, the average value of $f$ over $[x, x+h]$ converges to the instantaneous value $f(x)$. Without continuity, the local behavior might oscillate wildly or have jumps, preventing the limit of the difference quotient from settling on a single well-defined value $f(x)$.

Conditions: $A(x) = \int_a^x f(t) dt$; $f$ is continuous at $x$

The fundamental theorem relates accumulation and rate of change by stating that if $F'(x) = f(x)$, then the accumulated area $A(x)$ from a fixed point $a$ to $x$ is given by $A(x) = F(x) - F(a)$. It shows that differentiation and integration are inverse processes: the derivative of the accumulation function $A(x)$ recovers the integrand $f(x)$, and antiderivatives allow the evaluation of definite integrals.

Conditions: The function $f$ is continuous.; $F$ is an antiderivative of $f$ (i.e., $F' = f$).

The derivative $A'(x)$ equals $f(x)$ because moving the endpoint from $x$ to $x+h$ adds a thin strip whose area increment is approximately $f(x)h$. Dividing this increment by $h$ and taking the limit as $h$ approaches zero yields the local average height, which continuity ensures approaches the exact endpoint height $f(x)$.

Conditions: The function $f$ is continuous at $x$.; $A(x)$ is defined as the accumulation of $f$ from a fixed lower limit to $x$.

Continuity ensures that the local average height of the function over a small interval approaches the exact height at the endpoint as the interval width shrinks to zero. This property is crucial for proving that the derivative of the accumulation function $A(x)$ is exactly the integrand $f(x)$, as it allows the replacement of the average value with the point value in the limit.

Conditions: The function $f$ is continuous at the point $x$.; The interval width $h$ approaches zero.