Area Under Curves — Why Integration Works

Why can an integral compute area? It’s not obvious that the anti-derivative of a function has anything to do with the area under the curve. Yet it does — and understanding why will make integration problems far more intuitive.

The key insight: area is the limit of infinitely many infinitely thin rectangles. Each rectangle has width $dx$ and height $f(x)$ . The area of one rectangle is $f(x)\,dx$ . Summing all rectangles gives $\int_a^b f(x)\,dx$ .

The Fundamental Theorem of Calculus then tells us this sum equals $F(b) - F(a)$ where $F'(x) = f(x)$ . That’s the connection: differentiation and integration are inverse operations.

In CBSE Class 12, area under curves is a guaranteed 5–6 mark question in boards. JEE also tests this with more complex regions.

Key Terms & Definitions

Definite Integral: $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ , where $F'(x) = f(x)$ .

Area Under Curve: The area bounded between $y = f(x)$ , the x-axis, and the lines $x = a$ and $x = b$ .

A = \int_a^b f(x)\,dx \quad \text{(when } f(x) \geq 0 \text{ on } [a,b]\text{)}

Area Between Two Curves: Area between $y = f(x)$ (upper) and $y = g(x)$ (lower):

A = \int_a^b [f(x) - g(x)]\,dx

Negative Area: If $f(x) < 0$ on $[a,b]$ , the integral gives a negative value. The actual area is the absolute value.

Signed Area: $\int_a^b f(x)\,dx$ counts area above x-axis as positive and below as negative.

The Geometric Intuition

Why Thin Rectangles Work

Divide $[a,b]$ into $n$ equal strips of width $\Delta x = (b-a)/n$ . In each strip, approximate $f(x)$ by a constant (the value at some point $x_i^*$ in the strip). The area of strip $i$ is $f(x_i^*)\Delta x$ .

Total approximate area: $\sum_{i=1}^n f(x_i^*) \Delta x$ (Riemann sum)

As $n \to \infty$ (strips become infinitely thin), this sum converges to the exact area — the definite integral.

This is not circular reasoning. The integral IS defined as this limit. The anti-derivative connection comes from the Fundamental Theorem.

Why F(b) - F(a) Gives Area

Define $A(x)$ = area from $a$ to $x$ under the curve. Then $A(a) = 0$ and $A(b)$ = total area.

The key: $\dfrac{dA}{dx} = f(x)$ . Why? Because adding a tiny strip of width $dx$ and height $f(x)$ adds area $f(x)\,dx$ . So $\dfrac{dA}{dx} = f(x)$ .

This means $A(x)$ IS the antiderivative $F(x)$ (plus a constant). Therefore $A(b) = F(b) - F(a)$ .

Standard Formulas

A = \int_a^b f(x)\,dx \quad (f(x) \geq 0)

If $f(x)$ changes sign, split the integral at the roots.

A = \int_a^b [f(x) - g(x)]\,dx

where $f(x) \geq g(x)$ on $[a,b]$ .

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

Useful when the curve is better expressed as $x = g(y)$ .

Solved Examples

Example 1 — Easy: Area Under Parabola

Find the area enclosed by $y = x^2$ , the x-axis, and the lines $x = 0$ and $x = 3$ .

Solution:

A = \int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = \mathbf{9 \text{ sq units}}

Example 2 — Medium: Area Below x-axis

Find the area enclosed by $y = \sin x$ from $x = 0$ to $x = 2\pi$ .

Careful: $\sin x$ is negative from $\pi$ to $2\pi$ .

A = \int_0^{\pi} \sin x\,dx + \left|\int_{\pi}^{2\pi} \sin x\,dx\right|

= \left[-\cos x\right]_0^{\pi} + \left|\left[-\cos x\right]_{\pi}^{2\pi}\right|

= [(-\cos\pi) - (-\cos 0)] + |[(-\cos 2\pi) - (-\cos\pi)]|

= [1 + 1] + |[-1 - 1]| = 2 + 2 = \mathbf{4 \text{ sq units}}

Note: $\int_0^{2\pi} \sin x\,dx = 0$ (signed area), but actual area = 4.

Example 3 — Medium: Area Between Two Curves

Find the area between $y = x^2$ and $y = x$ (i.e., the area bounded by both curves).

First, find intersections: $x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1$ .

On $[0,1]$ : $x \geq x^2$ (line is above parabola).

A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \mathbf{\frac{1}{6} \text{ sq units}}

Example 4 — Hard (JEE Level): Circle Area

Find the area of the region bounded by the circle $x^2 + y^2 = 4$ .

$y = \sqrt{4 - x^2}$ (upper semicircle). Total area = 2 × area of upper semicircle.

A = 2\int_{-2}^{2} \sqrt{4-x^2}\,dx

Use the formula $\int \sqrt{a^2 - x^2}\,dx = \frac{x\sqrt{a^2-x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$ :

= 2\left[\frac{x\sqrt{4-x^2}}{2} + 2\sin^{-1}\frac{x}{2}\right]_{-2}^{2}

= 2\left[(0 + 2 \cdot \frac{\pi}{2}) - (0 + 2 \cdot (-\frac{\pi}{2}))\right] = 2[π + π] = \mathbf{4\pi \text{ sq units}}

As expected (area of circle = $\pi r^2 = \pi(4) = 4\pi$ ).

Example 5 — Standard Board Question

Find the area bounded by the curve $y = x^3$ , x-axis, and the lines $x = -1$ and $x = 1$ .

Solution: Note: $x^3$ is negative for $x \in [-1, 0]$ and positive for $x \in [0, 1]$ .

A = \left|\int_{-1}^{0} x^3\,dx\right| + \int_0^1 x^3\,dx

= \left|\left[\frac{x^4}{4}\right]_{-1}^{0}\right| + \left[\frac{x^4}{4}\right]_0^1 = \left|0 - \frac{1}{4}\right| + \left(\frac{1}{4} - 0\right) = \frac{1}{4} + \frac{1}{4} = \mathbf{\frac{1}{2} \text{ sq units}}

Exam-Specific Tips

CBSE 2024 frequently asked: Area between $y = x^2$ and $y = \sqrt{x}$ (the parabola and its inverse). Find intersections first (at $x = 0$ and $x = 1$ ), determine which is upper curve ( $\sqrt{x} \geq x^2$ on $[0,1]$ ), then integrate the difference.

JEE Main: Questions on area enclosed by ellipses, circles, or combinations of straight lines and curves. Standard formula for area of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi ab$ . Know this directly.

For area between curves: always sketch (even roughly). Determine which curve is above the other in the relevant interval. The integral should always be (upper curve) − (lower curve), giving a positive result.

Common Mistakes to Avoid

Mistake 1: Not taking absolute value for areas below the x-axis. $\int_0^{2\pi} \sin x\,dx = 0$ , but area = 4. Whenever the curve dips below the x-axis, split the integral at the root(s) and take absolute values.

Mistake 2: Wrong order in “upper − lower.” If $f(x) > g(x)$ , the area between them is $\int(f-g)dx$ , not $\int(g-f)dx$ . If you subtract in the wrong order, you get a negative answer for area — physically meaningless.

Mistake 3: Forgetting to find intersection points. For area between curves, the limits of integration are where the curves intersect. Using $[0,1]$ when the actual intersection is at $x = 0.5$ gives a completely wrong answer.

Mistake 4: Forgetting the factor of 2 for symmetric curves. If a region is symmetric about the y-axis, compute area for $x \geq 0$ and double it. This halves the calculation work — use it.

Practice Questions

Q1. Find area between $y = x$ and $y = x^2$ .

Intersections: $x = x^2 \Rightarrow x = 0, 1$ . On $[0,1]$ , $x \geq x^2$ .

$A = \int_0^1(x-x^2)dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ sq units.

Q2. Find area enclosed by $y = |x|$ and $y = 2$ .

Intersections: $|x| = 2 \Rightarrow x = \pm 2$ . Region: triangle-like shape.

$A = \int_{-2}^{2}(2 - |x|)dx = 2\int_0^2(2-x)dx = 2[2x - \frac{x^2}{2}]_0^2 = 2[4-2] = 4$ sq units.

Q3. Find area bounded by $y = e^x$ , x-axis, $x = 0$ and $x = 1$ .

$A = \int_0^1 e^x dx = [e^x]_0^1 = e - 1$ sq units.

Q4. Area of region bounded by $y^2 = 4x$ and $x = 1$ .

The parabola opens right; at $x = 1$ , $y = \pm 2$ . Area using symmetry:

$A = 2\int_0^1 \sqrt{4x}\,dx = 2\int_0^1 2\sqrt{x}\,dx = 4[\frac{2x^{3/2}}{3}]_0^1 = \frac{8}{3}$ sq units.

FAQs

What does negative area mean geometrically?

A negative integral value means the curve is below the x-axis in that interval. Geometrically, the “signed area” is negative — counting below-axis area as negative and above-axis area as positive. When we say “area” (always positive), we take the absolute value.

Can the area formula be applied horizontally?

Yes. If a curve is expressed as $x = g(y)$ , the area between the curve and the y-axis from $y = c$ to $y = d$ is $\int_c^d g(y)\,dy$ . This is useful for curves like $x = y^2$ where the standard approach is awkward.

What is the area under $f(x) = 0$ ?

Zero. The curve lies on the x-axis — no area is enclosed.

Does a larger integral always mean a larger area?

Not necessarily, because of the signed area issue. The integral $\int_{-1}^{1} x\,dx = 0$ , but the actual area (above + below x-axis) is 1. When computing area, always account for sign changes in the function.

Standard Areas to Memorise

Curve	Bounded by	Area
Circle $x^2+y^2=a^2$	Full circle	$\pi a^2$
Ellipse $x^2/a^2+y^2/b^2=1$	Full ellipse	$\pi ab$
Parabola $y^2=4ax$	Latus rectum ( $x=a$ )	$\frac{8a^2}{3}$
$y=x^2$ and $y=x$	Between curves	$\frac{1}{6}$
$y=\sqrt{x}$ and $y=x^2$	Between curves (0 to 1)	$\frac{1}{3}$

Integration using y-axis — when to use it

When the curve is better expressed as $x = g(y)$ , or when vertical strips would cross the boundary twice, switch to horizontal strips:

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

Example: Area enclosed by $y^2 = 4x$ between $y = 0$ and $y = 4$ :

$x = y^2/4$ . $A = \int_0^4 \frac{y^2}{4}\,dy = \frac{1}{4}\left[\frac{y^3}{3}\right]_0^4 = \frac{1}{4} \times \frac{64}{3} = \frac{16}{3}$

Area between a curve and oblique line

When one boundary is a non-horizontal/non-vertical line (like $y = x + 1$ ), the formula stays the same — just treat it as one of the two curves:

A = \int_a^b |\text{upper}(x) - \text{lower}(x)|\,dx

Find intersections first to set limits.

Additional Solved Example

JEE Main: Find the area enclosed between $y = x^2 + 1$ and $y = 2x + 1$ .

Intersections: $x^2 + 1 = 2x + 1 \implies x^2 - 2x = 0 \implies x(x-2) = 0$ . So $x = 0, 2$ .

On $[0, 2]$ : $2x + 1 \geq x^2 + 1$ (line is above parabola).

$A = \int_0^2 [(2x+1)-(x^2+1)]\,dx = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$

Additional Practice Questions

Q5. Find area enclosed by $y = \ln x$ , x-axis, and $x = e$ .

$A = \int_1^e \ln x\,dx$ . Integration by parts: $\int \ln x\,dx = x\ln x - x + C$ .

$A = [x\ln x - x]_1^e = (e \cdot 1 - e) - (1 \cdot 0 - 1) = 0 + 1 = 1$ sq unit.

Q6. Find the area of the region $\{(x,y): x^2 + y^2 \leq 1 \leq x + y\}$ .

This is the area inside the unit circle but above the line $x + y = 1$ . The line meets the circle at $(1,0)$ and $(0,1)$ . Area = area of quarter circle $-$ area of triangle = $\frac{\pi}{4} - \frac{1}{2}$ .