Calculus AB · Integration and Accumulation of Change · 14 min read · Updated 2026-05-10
Fundamental Theorem of Calculus and accumulation functions — AP Calculus AB
AP Calculus AB · Integration and Accumulation of Change · 14 min read
1. Accumulation Functions★★☆☆☆⏱ 3 min
An accumulation function is a function whose value equals the net signed area accumulated under another function from a fixed starting bound to a variable input. The standard form is:
A(x) = \int_{a}^{x} f(t) dt
Here, $a$ is a fixed constant, $f$ is continuous on the interval containing $a$ and $x$, and $t$ is a dummy variable: a placeholder that does not appear in the final output of $A(x)$. Common variations include variable lower bounds or two variable bounds, which can be rewritten using the property $\int_{b}^{a} f(t) dt = -\int_{a}^{b} f(t) dt$.
2. FTC Part 1: Derivative of Accumulation Functions★★★☆☆⏱ 4 min
FTC Part 1 formalizes the inverse relationship between integration and differentiation, telling us how to differentiate an accumulation function. The basic statement of the theorem is:
For AP exam purposes, we extend this rule to variable bounds using the chain rule, resulting in three common cases:
If $A(x) = \int_{a}^{u(x)} f(t) dt$, then $A'(x) = f(u(x)) \cdot u'(x)$
If $A(x) = \int_{v(x)}^{a} f(t) dt$, then $A'(x) = -f(v(x)) \cdot v'(x)$
If $A(x) = \int_{v(x)}^{u(x)} f(t) dt$, then $A'(x) = f(u(x))u'(x) - f(v(x))v'(x)$
3. FTC Part 2: Evaluating Definite Integrals★★☆☆☆⏱ 3 min
FTC Part 2 gives a method to calculate the exact value of a definite integral using antiderivatives, instead of approximating with Riemann sums. The formal statement is:
A key requirement for this theorem to hold is that $f$ is continuous over the entire interval $[a,b]$ — it does not apply to integrands with discontinuities inside the interval.
4. AP Style Practice Problems★★★☆☆⏱ 4 min
Common Pitfalls
Why: Students forget the upper bound is a function of $x$, so they skip the chain rule step.
Why: Students forget flipping the limits of integration introduces a negative sign.
Why: Students mix up the order of subtraction when recalling FTC Part 2.
Why: Students reuse the output variable by habit, leading to confusion when taking derivatives.
Why: Students forget FTC only applies when the integrand is continuous on the entire interval; $\frac{1}{x^2}$ has an infinite discontinuity at $x=0$ inside $[-1,2]$.
Why: Students confuse constant definite integrals with accumulation functions that have a variable bound.