Area under a curve

A city planner staring at a river's meandering bank cannot find its enclosed area with a ruler — the boundary is a curve, not a straight edge. Calculus supplies the tool: chop the region into infinitely many infinitesimally thin vertical strips, add up each strip's tiny area, and the sum collapses into a single definite integral. That is the whole idea behind area under a curve, the classic geometric payoff of the Fundamental Theorem of Calculus. In the ISC (CISCE) Class 12 syllabus this sits in the optional Section-B unit Applications of Integrals, distinct from — but built entirely on — the compulsory Calculus unit's definite-integral machinery. This article rebuilds why the integral equals the area, handles the trap of a curve dipping below the axis, and works the region both ways: as a function of $x$ and as a function of $y$ .

The core insight

The area between a curve $y=f(x)$ , the $x$ -axis, and two vertical lines $x=a,\,x=b$ equals the definite integral $\int_a^b f(x)\,dx$ when $f(x)\ge0$ , and the absolute value of that integral, computed separately on each sign-constant piece, whenever the curve dips below the axis.

The core idea

The basic formula. If $f(x)\ge 0$ throughout $[a,b]$ , the region bounded by $y=f(x)$ , the $x$ -axis, and the lines $x=a$ , $x=b$ has area

A = \int_a^b f(x)\,dx

where: $f(x)$ is the curve's equation, and $a,b$ are the left and right boundary lines with $a<b$ . The integral is evaluated using the Fundamental Theorem of Calculus — find an antiderivative $F(x)$ of $f(x)$ , then compute $F(b)-F(a)$ .

When the curve crosses the axis. A definite integral is a signed quantity: where $f(x)<0$ , the integral comes out negative even though the physical region still has positive area. So if $f$ dips below the axis somewhere inside $[a,b]$ , the correct area is

A = \left|\int_a^b f(x)\,dx\right| \quad \text{(applied separately on each interval where }f\text{ keeps one sign)}

where: the bars mean take the absolute value of the integral computed over each stretch where $f$ does not change sign, then add those absolute values together — never integrate straight across a sign change and hope the total is right.

Swapping the roles of $x$ and $y$ . Some boundaries are more naturally written as $x=g(y)$ — a sideways parabola, say. Bounded by the $y$ -axis and horizontal lines $y=c,\,y=d$ , the same idea gives

A = \int_c^d g(y)\,dy

where: $g(y)$ expresses $x$ in terms of $y$ , and $c,d$ are the lower and upper $y$ -limits. Choosing $x$ - or $y$ -integration is a matter of which variable the boundary is naturally expressed in, not a fixed rule.

Figure

A parabola y equals x squared with the region between the curve, the x axis and the vertical lines x equals 1 and x equals 3 shaded, illustrating the area found by the definite integral

The derivation

Derivation AWhy the definite integral equals the area under a curve

The goal is to show, using nothing beyond the definition of the definite integral as a limit of a sum of thin rectangular strips, that $\int_a^b f(x)\,dx$ genuinely equals the geometric area between $y=f(x)$ , the $x$ -axis, and the lines $x=a,\,x=b$ when $f(x)\ge0$ , and why that same integral turns negative — without the region itself becoming negative — once $f(x)$ dips below the axis.

Step 1. Slice the interval $[a,b]$ into $n$ thin strips of width $\Delta x=(b-a)/n$ , each running from some $x_i$ to $x_i+\Delta x$ . Approximate the curve over each strip by a flat rectangle of height $f(x_i)$ , so the strip's area is $f(x_i)\Delta x$ .

Step 2. Add up all $n$ strip areas: $\sum_{i=1}^{n} f(x_i)\Delta x$ . This sum is only an approximation to the true region, since each rectangle either slightly overshoots or undershoots the curve, but it is a genuine geometric area as long as every $f(x_i)\ge0$ .

Step 3. Let $n\to\infty$ so $\Delta x\to0$ . The rectangles hug the curve ever more closely, and the sum converges — by definition — to the definite integral: $\lim_{n\to\infty}\sum f(x_i)\Delta x = \int_a^b f(x)\,dx$ . Because each term in the sum was a genuine positive area throughout (when $f\ge0$ ), the limit is exactly the region's area.

Step 4. Now suppose $f(x)<0$ over the strip. The "height" $f(x_i)$ is negative, so the strip's signed contribution $f(x_i)\Delta x$ is negative, even though the strip still occupies a physical rectangle of positive area below the axis. The integral therefore reports a negative number for that stretch — not because area can be negative, but because the integral is tracking signed area (area above the axis counted positively, area below counted negatively). Converting back to true geometric area means taking the absolute value of each sign-constant piece before adding, which is exactly the rule stated above. Verified: for $f(x)=x^2$ on $[1,3]$ , both the strip-sum limit and the antiderivative evaluation $\left[\tfrac{x^3}{3}\right]_1^3$ give $\tfrac{26}{3}$ , matching the worked example below.

Worked example

Find the area bounded by the curve $y=\sin x$ , the $x$ -axis, and the lines $x=0$ and $x=2\pi$ .

Step 1. Check the sign of $\sin x$ across the interval: on $[0,\pi]$ , $\sin x\ge0$ ; on $[\pi,2\pi]$ , $\sin x\le0$ . Since the curve changes sign at $x=\pi$ , split the integral there rather than integrating straight across.

Step 2. Integrate the first piece: $A_1 = \int_0^{\pi}\sin x\,dx = \big[-\cos x\big]_0^{\pi} = (-\cos\pi)-(-\cos 0) = 1+1 = 2$ .

Step 3. Integrate the second piece: $A_2 = \int_{\pi}^{2\pi}\sin x\,dx = \big[-\cos x\big]_\pi^{2\pi} = (-\cos2\pi)-(-\cos\pi) = -1-1 = -2$ .

Step 4. Since $A_2$ is negative — the curve lies below the axis on $[\pi,2\pi]$ — take its absolute value before adding: $A = |A_1| + |A_2| = 2+2 = \mathbf{4\ \text{sq. units}}$ . (Had we integrated straight across without splitting, $\int_0^{2\pi}\sin x\,dx = A_1+A_2 = 0$ , the net signed integral — clearly wrong as a total area, since the positive and negative halves cancelled.)

Competitive stretch

The area formula $A=\int_a^b f(x)\,dx$ also matches the earlier polynomial case exactly: for $y=x^2$ bounded by $x=1,\,x=3$ , $A=\int_1^3 x^2\,dx=\left[\tfrac{x^3}{3}\right]_1^3=9-\tfrac13=\tfrac{26}{3}$ sq. units — and swapping to $x=y^2$ bounded by $y=1,\,y=3$ gives the identical number $\tfrac{26}{3}$ , since the two problems are mirror images with $x$ and $y$ relabelled.

Common traps

Integrating straight across a sign change without splitting — this gives the net signed integral, not the total area; always locate zeros of $f$ inside $[a,b]$ first and split there.
Forgetting to take the absolute value of a negative piece — a negative integral value doesn't mean "negative area"; it means the region lies below the axis, and its magnitude is what gets added to the total.
Using the wrong variable of integration for curves more naturally described as $x=g(y)$ — integrate with respect to $y$ using $y$ -limits, rather than forcing an $x$ -based setup.
Treating this as identical to the core Calculus unit — evaluating a definite integral is Calculus-unit technique, but "area bounded by a curve" as an examined application belongs to the separate Applications of Integrals unit; know which unit a paper is testing.

Ask the tutor: Want help spotting exactly where a curve crosses the axis before splitting an area integral? Ask the AI tutor to walk through the setup with your own function.

Why does the definite integral give a negative value when the region I can clearly see has a positive area?

The definite integral is defined as a limit of a sum of thin rectangular strips with height $f(x_i)$ . When the curve lies below the $x$ -axis, $f(x_i)$ is negative, so each strip's signed contribution is negative — the integral is tracking signed area, not the physical area you can measure with a ruler. The fix is simple: wherever the integral comes out negative, that stretch of the curve lies below the axis, and you take the absolute value of that piece before adding it to any other pieces to get the true total area.

How do I know where to split the interval when the curve changes sign?

Set $f(x)=0$ and solve for $x$ within $[a,b]$ — those roots are exactly the points where the curve crosses the axis and the sign of $f$ can flip. For $y=\sin x$ on $[0,2\pi]$ , the zero at $x=\pi$ is the natural split point, giving $[0,\pi]$ where $\sin x\ge0$ and $[\pi,2\pi]$ where $\sin x\le0$ . Always sketch or reason through the sign on each piece before integrating — don't assume a single sign holds across the whole interval just because the endpoints happen to give a positive value.

Do I always have to choose between integrating with respect to x or y, or can I mix them?

For a single region bounded by one curve and vertical or horizontal lines, pick whichever variable the boundary curve is most naturally expressed in and stick with it for that region — integrating with respect to $x$ using $x$ -limits, or with respect to $y$ using $y$ -limits. Mixing, such as using $y$ -limits while integrating a function of $x$ , gives a nonsensical result. More complex regions bounded by two different curves may need the region split into pieces, each handled with its own single, consistent variable choice.