Area function

Sweep a vertical line across a curve and watch the area under it grow — or shrink, if the curve dips below the axis. That sweeping accumulation is captured by a single object: the area function $A(x) = \int_a^x f(t)\,dt$ , a function of the moving right-hand boundary $x$ , built out of a definite integral. It sits inside the CBSE Class 12 Integrals chapter as the bridge between definite integrals (fixed-limit numbers) and the Fundamental Theorem of Calculus, which is the article's real payoff: differentiating $A(x)$ hands you back $f(x)$ . This article builds the area function from scratch, computes it explicitly for two standard examples, and pins down the traps — sign behaviour, the dummy variable, and the fact that $A(x)$ is not always increasing.

The core insight

The area function $A(x) = \int_a^x f(t)\,dt$ measures signed accumulated area up to a moving point $x$ , satisfies $A(a)=0$ and $A(b)=\int_a^b f(t)\,dt$ , and its behaviour (increasing, decreasing, or flat) tracks the sign of $f$ at each $x$ .

The core idea

For $f$ continuous on $[a,b]$ , define

A(x) = \int_a^x f(t)\,dt, \qquad a \le x \le b

where: $t$ is the dummy variable of integration (it never appears in the final expression), $x$ is the real variable the area function depends on, and $a$ is the fixed lower limit chosen when $A$ is defined. Three properties follow immediately from the definition, and a CBSE student is expected to quote and use all three:

Starting value. $A(a) = \int_a^a f(t)\,dt = 0$ — no area has accumulated at the very first instant, whatever $f$ is.
Endpoint value. $A(b) = \int_a^b f(t)\,dt$ — evaluating the area function at the far end of the interval recovers the ordinary definite integral over the whole interval. This is the link that makes $A(x)$ useful rather than a curiosity.
Monotonicity tracks the sign of $f$ . If $f(t) \ge 0$ throughout $[a,b]$ , each additional sliver of width $dx$ contributes a non-negative amount $f(x)\,dx$ , so $A(x)$ is non-decreasing. Where $f$ turns negative, that sliver subtracts, and $A(x)$ decreases — the integral is a signed area, not a geometric (always-positive) one.

The deeper reason the area function matters: it turns out to be an anti-derivative of $f$ itself, so $A'(x) = f(x)$ — the first Fundamental Theorem of Calculus (a separate article in this wiki). Practically, this means $A(x)$ is almost never computed by a genuine limiting process of thin rectangles; any anti-derivative found by standard techniques (substitution, parts, standard forms) gives $A(x)$ directly, with the constant of integration pinned down by forcing $A(a)=0$ .

Figure

Two graphs stacked, top a straight line through the origin labelled f of t equals 2t with a shaded triangular region up to x, bottom a parabola labelled A of x equals x squared showing the accumulated area growing as x increases

Derivation AThe area function of f(t) = 2t, built directly from a limit of thin strips

The goal is to construct $A(x) = \int_0^x 2t\,dt$ from first principles — as a genuine limit of a sum of thin rectangular strips, not by quoting an anti-derivative — and show the result equals both the anti-derivative shortcut and the elementary geometric (triangle) area, so that all three routes to the area function are seen to agree for this concrete case.

Step 1. Split $[0,x]$ into $n$ equal strips of width $\Delta t = x/n$ , with right-endpoints $t_k = k\Delta t$ for $k=1,\dots,n$ . On each strip approximate the area by a thin rectangle of height $f(t_k) = 2t_k$ and width $\Delta t$ , so the total approximate area is $S_n = \sum_{k=1}^{n} f(t_k)\,\Delta t = \sum_{k=1}^n 2\!\left(\frac{kx}{n}\right)\!\frac{x}{n} = \frac{2x^2}{n^2}\sum_{k=1}^n k$ .

Step 2. Use the standard sum $\sum_{k=1}^n k = \dfrac{n(n+1)}{2}$ , so $S_n = \dfrac{2x^2}{n^2}\cdot\dfrac{n(n+1)}{2} = x^2\!\left(\dfrac{n+1}{n}\right) = x^2\!\left(1+\dfrac1n\right)$ .

Step 3. Let the number of strips grow without bound: $A(x) = \displaystyle\lim_{n\to\infty} S_n = \lim_{n\to\infty} x^2\!\left(1+\frac1n\right) = x^2$ , since $1/n \to 0$ . This is the genuine limiting definition of the definite integral, computed without ever invoking an anti-derivative.

Step 4. Cross-check against the anti-derivative shortcut: $F(t)=t^2$ is an anti-derivative of $2t$ , so $A(x) = \big[t^2\big]_0^x = x^2 - 0 = x^2$ — identical to Step 3. Cross-check against geometry: $y=2t$ from $t=0$ to $t=x$ is a line through the origin of slope $2$ , reaching height $2x$ at $t=x$ , so the shaded region is a right triangle of base $x$ and height $2x$ , giving area $\tfrac12(x)(2x)=x^2$ . All three methods — Riemann limit, anti-derivative, elementary geometry — agree exactly, confirming $A(x)=x^2$ .

Worked example

For $f(t)=t^2-4$ on $[0,x]$ , find $A(x) = \displaystyle\int_0^x (t^2-4)\,dt$ , evaluate $A(1)$ , $A(2)$ , $A(3)$ , and check the sign pattern against $A'(x)=f(x)$ .

Step 1. Find an anti-derivative: $F(t) = \dfrac{t^3}{3}-4t$ . So $A(x) = \left[\dfrac{t^3}{3}-4t\right]_0^x = \dfrac{x^3}{3}-4x$ .

Step 2. Evaluate at the three points: $A(1) = \dfrac13-4=-\dfrac{11}{3}\approx-3.67$ , $\ A(2)=\dfrac83-8=-\dfrac{16}{3}\approx-5.33$ , $\ A(3)=9-12=-3$ .

Step 3. Read the sign pattern. $f(t)=t^2-4$ is negative for $0\le t<2$ and positive for $t>2$ , since $t^2-4=0$ at $t=2$ . Between $x=1$ and $x=2$ , $A(x)$ falls from $-3.67$ to $-5.33$ because $f$ is negative there (area is being subtracted); between $x=2$ and $x=3$ , $A(x)$ rises from $-5.33$ to $-3$ because $f$ has turned positive. This is $A(x)$ decreasing, then increasing — not monotonic — exactly as the FTC relation $A'(x)=f(x)=x^2-4$ predicts: $A'(x)<0$ for $x<2$ and $A'(x)>0$ for $x>2$ .

Step 4. Check $A(2)$ is the local minimum: $A'(2) = 2^2-4=0$ , confirming $x=2$ is a stationary point of $A(x)$ , matching the numbers above where $A$ stops falling and starts rising right around $x=2$ . The area function is $A(x)=\dfrac{x^3}{3}-4x$ , with $A(1)\approx-3.67$ , $A(2)\approx-5.33$ , and $A(3)=-3$ .

Competitive stretch

For the same $f(t)=t^2-4$ , find the positive $x$ at which $A(x)=0$ again (other than $x=0$ ). Set $\dfrac{x^3}{3}-4x=0 \Rightarrow x\!\left(\dfrac{x^2}{3}-4\right)=0$ , giving $x=0$ or $x^2=12$ , so $x=2\sqrt3\approx3.46$ . Since $A$ is negative throughout $(0,2\sqrt3)$ and $A(0)=0$ , the accumulated negative area up to $t=2$ is exactly cancelled by the positive area gained between $t=2$ and $t=2\sqrt3$ .

Common traps

Assuming the area function always increases — it only increases where $f(x)\ge0$ ; where $f$ is negative, $A(x)$ decreases, since the integral is a signed area, not an unsigned geometric one.
Confusing the dummy variable $t$ with the outer variable $x$ — writing $A(x)=\int_a^x f(x)\,dx$ reuses the same letter for both and is a common notational slip.
Treating $A(x)$ as ordinary (unsigned) geometric area when $f$ dips below the axis — it does not measure that; a true unsigned area needs the integral split at the sign change and the negative piece's absolute value taken separately.
Forgetting that $A(a)=0$ always, regardless of $f$ — a fast sanity check whenever $A(x)$ is computed explicitly.

Ask the tutor: want to see the area function plotted alongside its generating curve, or try computing $A(x)$ for a new function step by step?

Why does A(x) sometimes decrease if an integral is supposed to give area?

Because a definite integral gives signed area, not the everyday geometric kind. When $f(t)$ is negative, each thin strip $f(x)\,dx$ contributes a negative amount, so the running total $A(x)$ falls as $x$ increases through that region. It only behaves like ordinary, always-growing area when $f(t)\ge0$ throughout. If you actually want the unsigned area swept by a curve that dips below the axis, split the integral at the sign change and take the absolute value of the negative piece separately — that technique appears in Applications of Integrals.

Is the letter t inside the integral the same as the x outside?

No, and mixing them up is the single most common slip with area functions. The $t$ is a dummy variable — it is integrated away and never appears in the final formula for $A(x)$ . The $x$ in the upper limit is the real, independent variable that $A$ depends on. Writing $f(t)$ inside and $f(x)$ as the upper-limit function keeps the two roles visually distinct; if you write $\int_a^x f(x)\,dx$ with the same letter both places, the expression becomes ambiguous about which $x$ is fixed and which is varying.

Do I need to take an actual limit of rectangles every time I compute an area function?

No — that limiting process (shown once in the derivation above) is how the definite integral is defined, but you almost never redo it by hand. Once you know an anti-derivative $F(t)$ of $f(t)$ , the Fundamental Theorem of Calculus guarantees $A(x) = F(x) - F(a)$ , so ordinary anti-derivative techniques (substitution, integration by parts, standard forms) compute $A(x)$ directly. The limit-of-strips construction is worth understanding once so you trust why the shortcut works, not something to repeat on every problem.

Area function

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads