Topic 6.4 The Fundamental Theorem of Calculus and Accumulation Functions: differentiate accumulation functions using the first part of the Fundamental Theorem of Calculus.

What this topic is asking

The College Board (Topic 6.4) defines the accumulation function $g(x) = \int_a^x f(t)\,dt$ and gives the first part of the Fundamental Theorem of Calculus (FTC): its derivative is just the integrand evaluated at the upper limit. With a variable upper limit, the chain rule is needed.

The Fundamental Theorem, first part

The dummy variable $t$ inside the integral is replaced by the upper limit $x$ when differentiating. The lower limit $a$, being constant, contributes nothing to the derivative.

The chain rule for variable limits

A worked derivative of an accumulation function

Why the lower limit does not matter

Changing the constant lower limit shifts an accumulation function by a constant (the fixed area between the two lower limits), and the derivative of a constant is zero. So $\int_0^x f(t)\,dt$ and $\int_5^x f(t)\,dt$ have the same derivative $f(x)$; they differ only by a constant vertical shift. This is why the FTC's first part ignores the lower limit. Recognizing this prevents the error of trying to "use" the lower limit when differentiating, and it foreshadows why antiderivatives carry an arbitrary constant.

Accumulation functions as new functions

The deeper point is that the accumulation function $g(x) = \int_a^x f(t)\,dt$ defines a function even when $f$ has no elementary antiderivative. Functions like $\int_0^x e^{-t^2}\,dt$ cannot be written in closed form, yet the FTC still tells you the derivative exactly: $e^{-x^2}$. This lets the exam ask about the increasing/decreasing and concavity behavior of $g$ using the sign of $f$ (which is $g'$) and the sign of $f'$ (which is $g''$), connecting Unit 6 back to the analytical tools of Unit 5. The accumulation function is a function whose derivative you know even when you cannot write the function itself.

Handling a variable lower limit

When the lower limit varies and the upper limit is constant, flip the integral first. Since $\int_{v(x)}^{a} f(t)\,dt = -\int_a^{v(x)} f(t)\,dt$, differentiating gives $-f(v(x))\cdot v'(x)$: the integrand at the lower limit, times the derivative of the lower limit, with a minus sign from the reversal. If both limits vary, split the integral at any convenient constant $c$ into $\int_{v(x)}^{c} + \int_c^{u(x)}$ and differentiate each piece, producing $f(u(x))u'(x) - f(v(x))v'(x)$. Recognizing that a varying lower limit carries a minus sign, and that two varying limits give a difference of two chain-rule terms, covers every version of the FTC derivative the AB exam can pose.

A check using the second part of the FTC

You can confirm the first-part result against the evaluation form of the theorem. If $F$ is an antiderivative of $f$, then $g(x) = \int_a^x f(t)\,dt = F(x) - F(a)$, and differentiating this gives $g'(x) = F'(x) = f(x)$, since $F(a)$ is constant. This is exactly the first part of the FTC, derived from the second, and it makes the constant-lower-limit fact visible: $-F(a)$ disappears on differentiation. Seeing the two parts of the theorem reproduce each other reinforces that differentiation and integration are inverse operations, which is the single idea underlying all of Unit 6.