Anannt lesson · Unit 6: Integration and Accumulation of Change

Accumulation functions and the FTC

Define $A(x)=\int_a^x f(t)\,dt$ , predict whether $A$ increases when $f$ is negative, and differentiate accumulation with a variable upper limit, including $x^2$ .

About 45 minutes. Reading this page is exposure, not mastery.

Anannt Concept Lens

The idea in the form we want you to carry into an unfamiliar problem — Meera Krishnan’s faculty note.

Fix a continuous function $f$ and a starting input $a$ . Define $A(x)=\int_a^x f(t)\,dt$ . Then $A(x)$ is the signed accumulation of $f$ from $a$ to $x$ . If $f$ is negative on an interval, you are accumulating negative contributions, so $A$ decreases there even if a geometric-area picture looks “positive” after taking absolute values. The Fundamental Theorem says $A'(x)=f(x)$ : the rate at which accumulation grows is the current height of $f$ .

If the upper limit is not $x$ but a function $u(x)$ , the chain rule returns. $\frac{d}{dx}\int_a^{x^2} f(t)\,dt = f(x^2)\cdot 2x$ . The $f(x^2)$ factor is the FTC; the $2x$ is the chain. Feedback in this lesson treats a sign error (believing negative $f$ still raises $A$ ) as a different misconception from a missing chain factor.

Signature activity — predict accumulation before the reveal

Let $f(t)=t-2$ and $A(x)=\int_0^x f(t)\,dt$. On $(0,2)$, $f$ is negative. Before the graph of $A$ appears, decide whether $A$ is rising.

Worked example

Let $f(t)=t-2$ and $A(x)=\int_0^x f(t)\,dt$ . (1) For $0<x<2$ , is $A$ increasing? (2) Compute $A'(x)$ . (3) Compute $\frac{d}{dx}\int_0^{x^2} f(t)\,dt$ .

On $(0,2)$ , $f(t)<0$ , so the integrand is negative. $A(x)$ is adding negative signed area, so $A$ is decreasing. (A prediction is required before looking at the graph of $A$ .)
Directly, $A(x)=\int_0^x(t-2)\,dt=\frac{x^2}{2}-2x$ , so $A'(x)=x-2=f(x)$ , matching the FTC.
Let $u=x^2$ . Then $\frac{d}{dx}\int_0^{u(x)} f= f(u) u' = (x^2-2)(2x)$ .

$A$ decreases while $f$ is negative. $A'(x)=x-2$ . The $x^2$ limit produces an extra $2x$ . Missing $2x$ is not the same error as the sign misconception.

Contrasting non-example

Claiming $A$ must rise because “area is positive.” Signed accumulation can fall. Another non-example: $\frac{d}{dx}\int_0^{x^2} f(t)\,dt = f(x^2)$ with no chain factor. A third: differentiating under the integral by replacing $t$ with $x$ inside $f$ while keeping the limits, which is not the FTC statement.

Anannt Error Clinic

The mix-up we see most often, named kindly, with what to try instead.

Error Clinic — two named misconceptions.

Sign misconception: “ $A$ increases whenever there is area under a curve.” If $f$ is below the axis, $A$ is heading down. The lab asks you to predict this before the $A$ graph appears.

Missing chain factor: students remember $A'(x)=f(x)$ and then replace $x$ by $x^2$ without multiplying by $2x$ . The later table/graph check uses the same skill with a different representation, so a memorised formula for this one prompt will not transfer automatically.

Anannt Method Choice

If the upper limit is $x$ , FTC gives $f(x)$ (lower limit constant). If the upper limit is $u(x)$ , multiply by $u'(x)$ . If the lower limit is the variable, reverse the sign. If both limits vary, split at a constant.

Before this idea, I would usually check Signed accumulation versus geometric area, Chain rule. You can still open the lesson — the risk is building on a gap, not a locked door. A short bridge is safer if those skills are still unknown.

Completion so far: opened. Opening this page is exposure. Independent demonstration is a later, separate bar.

Explain why — Ask Anannt

If a stem, graph, or key looks ambiguous, use Report ambiguous mathematics on the item. Faculty keep a correction trail; we do not silently rewrite your past attempts.

After this lesson

Stay honest with the independent check if you can. When you are done — stuck or not — tell us where you are. Parent WhatsApp is required on the next page. Under 13: a parent should finish that form.