Area Under Curves and Area Between Curves | NIMCET

Quick answer: The area between $y = f(x)$ , the x-axis and the lines $x = a$ , $x = b$ is $\int_a^b\lvert f(x)\rvert\,dx$ . The area between two curves $y = f(x)$ and $y = g(x)$ , with $f \ge g$ on [a, b], is $\int_a^b[f(x) - g(x)]\,dx$ , where a and b are usually the intersection points. Standard results: the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ encloses $\pi ab$ , and the region between $y^2 = 4ax$ and its latus rectum has area $\frac{8a^2}{3}$ .

This applies definite integrals and their properties.

Area formulas

Region	Area
Under $y = f(x)$ , $f \ge 0$ , from a to b	$\int_a^b f(x)\,dx$
Curve below the x-axis	$\left\lvert\int_a^b f(x)\,dx\right\rvert$
Curve crossing the x-axis	Split at the roots and add absolute values
Between $x = g(y)$ and the y-axis, from c to d	$\int_c^d\lvert g(y)\rvert\,dy$
Between two curves ( $f \ge g$ )	$\int_a^b[f(x) - g(x)]\,dx$

Method for the area between curves

Step 1: Sketch the curves.

Step 2: Find their intersection points by solving $f(x) = g(x)$ .

Step 3: Decide which curve is on top in each part.

Step 4: Integrate (top − bottom), splitting the interval if the order changes.

Sometimes integrating with respect to y (right − left) is simpler.

Standard areas

Region	Area
Circle $x^2 + y^2 = a^2$	$\pi a^2$
Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	$\pi ab$
$y^2 = 4ax$ and its latus rectum $x = a$	$\frac{8a^2}{3}$
$y^2 = 4ax$ and $x^2 = 4by$	$\frac{16ab}{3}$
$y = x^2$ and $y = x$	$\frac{1}{6}$
One arch of $y = \sin x$ (0 to π)	2

The ellipse and parabola equations are in the ellipse and parabola articles.

Worked examples

Example 1: area under a curve

Find the area under $y = x^2$ from $x = 0$ to $x = 3$ .

$\int_0^3 x^2\,dx = 9$

Example 2: curve crossing the axis

Find the area between $y = \sin x$ and the x-axis from 0 to $2\pi$ .

The curve is below the axis on $(\pi, 2\pi)$ , so the area is $2 + 2 = 4$ (the signed integral would wrongly give 0).

Example 3: between a parabola and a line

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

Step 1: Intersections: $x^2 = 2x$ gives $x = 0, 2$ .

Step 2: On (0, 2), $2x \ge x^2$ :

$\int_0^2(2x - x^2)\,dx = 4 - \frac{8}{3} = \frac{4}{3}$

Example 4: ellipse by integration

Show that the area of $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi ab$ .

Step 1: By symmetry, area $= 4\int_0^a\frac{b}{a}\sqrt{a^2 - x^2}\,dx$ .

Step 2: $\int_0^a\sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{4}$ (a quarter circle).

Step 3: Area $= 4 \cdot \frac{b}{a} \cdot \frac{\pi a^2}{4} = \pi ab$ .

Example 5: with respect to y

Find the area between $x = y^2$ and $x = 4$ .

$\int_{-2}^{2}(4 - y^2)\,dy = 16 - \frac{16}{3} = \frac{32}{3}$

Check: this is $\frac{2}{3}$ of the enclosing rectangle ( $4 \times 4 = 16$ ), as expected for a parabolic segment ✓.

Summary table

Idea	Key result
Area under f	$\int\lvert f\rvert\,dx$
Between curves	$\int(\text{top} - \text{bottom})\,dx$
Ellipse	$\pi ab$
$y = x^2$ and $y = x$	$\frac{1}{6}$
One arch of sin x	2

Exam tips

Always sketch; it shows the intersection points and which curve is on top.
Use symmetry to integrate over half or a quarter of the region.
Choose dy integration when the region is bounded left and right by curves in y.

See all topics in the NIMCET Calculus chapter.

Common mistakes

Integrating a function that goes below the axis without splitting.
Subtracting top from bottom and getting a negative area.
Using the wrong limits instead of the intersection points.
Forgetting the factor 4 when using quarter-region symmetry.

Frequently Asked Questions

How do you find the area between two curves?

Find the intersection points, then integrate (upper curve − lower curve) between them.

What is the area of an ellipse?

πab, for x²/a² + y²/b² = 1.

What is the area between y = x² and y = x?

It is 1/6.

What happens if the curve goes below the x-axis?

That part of the integral is negative, so take the absolute value of each part and add.

What is the area under one arch of y = sin x?

It is 2.