Area under simple curves

A definite integral is not just a number that pops out of the Fundamental Theorem of Calculus — it is literally the area of a region, and that geometric reading is the whole point of this chapter. Once you can evaluate $\int_a^b f(x)\,dx$ , the natural next question is: what shape does this number measure? For CBSE Class 12, the answer is restricted to friendly, standard-form curves — a straight line, a circle $x^2+y^2=r^2$ , a parabola like $y^2=4ax$ , or an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ — bounded by the coordinate axes or a pair of vertical or horizontal lines. The method is always the same three moves: sketch the curve, write $y$ as a function of $x$ (or $x$ as a function of $y$ ), and integrate between the correct limits. This article expands that method with a full derivation, one worked example carried through to a competitive-level extension, and the traps examiners love to set.

The core insight

The area bounded by a curve $y=f(x)$ , the $x$ -axis, and the lines $x=a,\,x=b$ is the definite integral $A=\left|\int_a^b y\,dx\right|$ , and the same idea integrated with respect to $y$ handles regions bounded by the $y$ -axis.

The core idea

Area as a definite integral. If $y=f(x)\ge 0$ is continuous on $[a,b]$ , the region trapped between the curve, the $x$ -axis, and the vertical lines $x=a$ and $x=b$ has area

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

where: $f(x)$ is the curve's equation solved for $y$ , and $a,b$ are the $x$ -limits of the region (often supplied by where the curve crosses an axis, if no explicit line is given). This is not a coincidence or an analogy — it is the Fundamental Theorem of Calculus read geometrically: the antiderivative's change over $[a,b]$ equals the signed area swept by the curve.

Signed area versus true area. A definite integral is signed: if the curve dips below the $x$ -axis on part of $[a,b]$ , that portion contributes negatively and could partly cancel area above the axis. Since a physical area is never negative, the syllabus convention is

A=\left|\int_a^b y\,dx\right|,

where: the absolute value is taken only after splitting the interval at any axis-crossing, so each piece is made positive before being added — you cannot simply drop the sign on one combined integral if the curve crosses the axis inside $[a,b]$ .

Integrating along $y$ instead. When a region is bounded by the $y$ -axis and the lines $y=c,\,y=d$ , express $x=g(y)$ and integrate the other way:

A=\int_c^d x\,dy.

Standard-curve shortcuts. Because CBSE restricts this topic to circles, parabolas, and ellipses in standard position, two results are worth knowing by heart, both obtained by taking four times the first-quadrant quarter using symmetry: $A_{\text{circle}}=4\int_0^r\sqrt{r^2-x^2}\,dx=\pi r^2$ , and $A_{\text{ellipse}}=4\int_0^a b\sqrt{1-\frac{x^2}{a^2}}\,dx=\pi ab$ . For a parabola such as $y^2=4ax$ , take the upper branch $y=2\sqrt{ax}$ before integrating, since the full curve is not a function of $x$ alone.

Figure

Shaded region under the parabola y squared equals 4x between x equals 0 and x equals 4 in the first quadrant

Derivation AThe area formula and the circle's standard result

The goal is to justify, from first principles of the definite integral as a limit of a Riemann sum, why the area of a region under a curve equals $\int_a^b y\,dx$ , and then apply that single formula to obtain the full-circle area $\pi r^2$ as a concrete standard-curve result used throughout this chapter.

Step 1. Split $[a,b]$ into $n$ thin strips of width $\Delta x=\frac{b-a}{n}$ . In each strip, approximate the region under the curve by a thin rectangle of height $y=f(x_i)$ and width $\Delta x$ , so its area is $f(x_i)\,\Delta x$ . Summing over all $n$ strips gives an approximate total area $\sum_{i=1}^{n}f(x_i)\,\Delta x$ .

Step 2. As $n\to\infty$ , the strips become infinitesimally thin and the rectangle approximation becomes exact, because the error in each strip (the sliver between the rectangle top and the curve) shrinks faster than the strip width. By definition, this limit is the definite integral: $\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i)\,\Delta x=\int_a^b f(x)\,dx$ . This is exactly the area under the curve, so $A=\int_a^b y\,dx$ is not an assumption but the definition of the definite integral applied geometrically.

Step 3. Apply this to the standard circle $x^2+y^2=r^2$ . In the first quadrant, $y=\sqrt{r^2-x^2}$ , and by the symmetry of the circle about both axes, the full area is four times the first-quadrant piece from $x=0$ to $x=r$ : $A_{\text{circle}}=4\int_0^r\sqrt{r^2-x^2}\,dx$ . Using the standard antiderivative $\int\sqrt{r^2-x^2}\,dx=\frac{x}{2}\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\!\left(\frac{x}{r}\right)+C$ , evaluate from $0$ to $r$ : at $x=r$ the first term vanishes since $\sqrt{r^2-r^2}=0$ , leaving $\frac{r^2}{2}\sin^{-1}(1)=\frac{r^2}{2}\cdot\frac{\pi}{2}=\frac{\pi r^2}{4}$ ; at $x=0$ both terms are zero.

Step 4. Multiply by the symmetry factor: $A_{\text{circle}}=4\times\frac{\pi r^2}{4}=\pi r^2$ , the familiar formula, now obtained purely from integration rather than assumed. This confirms the machine works: build the integral from the region's boundary equations, use symmetry to cut the work to one quadrant, then evaluate with the Fundamental Theorem of Calculus.

Worked example

Problem. Find the area of the region bounded by the parabola $y^2=4x$ , the $x$ -axis, and the line $x=4$ , in the first quadrant.

Step 1. Sketch the curve: $y^2=4x$ opens rightward with vertex at the origin, and in the first quadrant $y=2\sqrt{x}\ge 0$ .

Step 2. Set up the integral for the strip from $x=0$ to $x=4$ , under the curve and above the $x$ -axis: $A=\int_0^4 2\sqrt{x}\,dx$ .

Step 3. Integrate: $A=2\int_0^4 x^{1/2}\,dx=2\left[\dfrac{x^{3/2}}{3/2}\right]_0^4=\dfrac{4}{3}\left[x^{3/2}\right]_0^4=\dfrac{4}{3}\left(4^{3/2}-0\right)$ .

Step 4. Evaluate: $4^{3/2}=(\sqrt{4})^3=2^3=8$ , so $A=\dfrac{4}{3}\times 8=\mathbf{\dfrac{32}{3}\ \text{square units}}$ .

Competitive stretch

Now extend the same method to a circle. Find the first-quadrant area of $x^2+y^2=16$ between $x=0$ and $x=4$ . Here $y=\sqrt{16-x^2}$ , so $A=\int_0^4\sqrt{16-x^2}\,dx=\left[\dfrac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\!\left(\dfrac{x}{4}\right)\right]_0^4$ . At $x=4$ : $0+8\sin^{-1}(1)=8\cdot\frac{\pi}{2}=4\pi$ . At $x=0$ : both terms are $0$ . So $A=4\pi$ square units — a quarter of the full circle, since $\frac14\pi r^2=\frac14\pi(4)^2=4\pi$ checks out exactly, confirming the derivation above.

Common traps

Forgetting the absolute value. If the curve crosses the $x$ -axis within the limits, split the integral at the crossing point and take the modulus of each piece before adding — a single unsplit integral can silently cancel area.
Wrong variable of integration. Regions that are naturally described by $x=g(y)$ (sideways parabolas, vertical-strip problems) need $\int x\,dy$ ; forcing $\int y\,dx$ instead makes the setup awkward or wrong.
Skipping the sketch. Without even a rough graph it is easy to integrate over the wrong interval or pick up an extra or missing piece of the region.
Mixing up circle and ellipse branches. $y=\pm\sqrt{r^2-x^2}$ has two branches; always confirm which one (usually $y\ge0$ ) bounds the enclosed region before integrating.

Ask the tutor: Want to work through one more area problem together, such as a parabola opening along the y-axis, or a region bounded between two curves?

Why do we take the absolute value instead of just integrating straight through?

Because a definite integral is signed — it measures area above the x-axis as positive and area below as negative, and if you integrate straight through a region that dips below the axis, the below-axis part subtracts from the total instead of adding to it. A true geometric area can never be negative, so the correct method is to first find where the curve crosses the x-axis, split the integral there, and add the absolute value of each piece. Skipping this step is the single most common way marks are lost in this chapter.

How do I decide whether to integrate with respect to x or with respect to y?

Look at which variable the curve is more naturally solved for, and which axis bounds the region. If the region is bounded by the x-axis and vertical lines x equals a and x equals b, write y as a function of x and use A equals the integral of y dx. If instead the region is bounded by the y-axis and horizontal lines y equals c and y equals d, solve for x as a function of y and use A equals the integral of x dy. Sketching the region first almost always makes the right choice obvious.

Do I need to memorise the circle and ellipse area formulas, or derive them every time?

For CBSE exams, know both the formula (pi r squared for a circle, pi a b for an ellipse) and how to derive it quickly using the quarter-region symmetry trick, because examiners sometimes ask you to derive rather than just quote. The derivation itself is short: write the first-quadrant equation, integrate using the standard result for the square-root form, then multiply by four. Practising that derivation once or twice makes it fast enough to reproduce under exam pressure.

Area under simple curves

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads