Understanding the Area Under the Curve Formula

The area under a curve is the precise measure of the region bounded by a curve and a specified axis, typically the $x$-axis or $y$-axis, between given limits. Computation of such areas relies fundamentally on definite integration, which accumulates the infinitely many infinitesimal strips (rectangular or otherwise) beneath the curve within the range of consideration.

Formulation of the Area Under the Curve With Respect to the $x$-Axis

Let $y = f(x)$ be a continuous, real-valued function defined on the closed interval $[a, b]$. The area, denoted $A$, enclosed between the curve, the $x$-axis, and the vertical lines $x = a$ and $x = b$ is given by

$A = \displaystyle\int_{a}^{b} f(x)\,dx$

If $f(x) \geq 0$ on $[a, b]$, $A$ represents the area of the corresponding region. If $f(x)$ assumes negative values, the signed area is obtained, and the net area bounded between the curve and the $x$-axis is $A = \left| \int_a^b f(x)\,dx \right|$.

Formulation of the Area Under the Curve With Respect to the $y$-Axis

Let $x = g(y)$ be a continuous function defined for $y$ in $[c, d]$. The area bounded by the curve, the $y$-axis, and the horizontal lines $y = c$ and $y = d$ is

$A = \displaystyle\int_{c}^{d} g(y)\,dy$

This formulation is especially employed when the curve is more conveniently expressed as $x$ in terms of $y$.

Area Bounded Between Two Curves

Consider two continuous functions $y_1 = f(x)$ and $y_2 = g(x)$ where $f(x) \geq g(x)$ for all $x \in [a, b]$. The area $A$ between these two curves, from $x = a$ to $x = b$, is evaluated as

$A = \displaystyle\int_{a}^{b} \left[f(x) - g(x)\right]\,dx$

If at some subintervals the roles of the upper and lower functions change, the integration range should be split at intersection points, and appropriate absolute values applied. For further reading, refer to Integral Calculus.

Riemann Sums and the Definite Integral as Area

The process of integration for area calculation originates from the limiting procedure of summing the areas of rectangular strips with infinitesimal widths, known as the Riemann sum. For $y = f(x)$ on $[a, b]$, partition the interval into $n$ subintervals, each of width $\Delta x = \frac{b-a}{n}$. The sum of the areas of these rectangles is

$\displaystyle S_n = \sum_{i=1}^n f(x_i^*)\,\Delta x$

where $x_i^* $ is any sample point in the $i$-th subinterval. The definite integral, and hence the area, is defined as the limit

$A = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*)\,\Delta x = \displaystyle\int_{a}^{b} f(x)\,dx$

Explicit Calculation of the Area Under a Standard Circle

Let the equation of a circle centered at the origin be $x^2 + y^2 = a^2$. Expressing $y$ in terms of $x$, the upper semicircle is $y = \sqrt{a^2 - x^2}$. To calculate the total area, consider the area in the first quadrant and multiply by $4$.

The area in the first quadrant is

$A_{\text{Q1}} = \displaystyle\int_{0}^{a} \sqrt{a^2 - x^2} \,dx$

The total area is $A = 4A_{\text{Q1}}$.

Consider the integral $I = \int_{0}^{a} \sqrt{a^2 - x^2} \,dx$.

To evaluate this, use the substitution $x = a\sin\theta$, giving $dx = a\cos\theta\,d\theta$, and when $x = 0$, $\theta = 0$, when $x = a$, $\theta = \frac{\pi}{2}$.

$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta$

Therefore,

$I = \int_{x=0}^{x=a} \sqrt{a^2 - x^2}\,dx = \int_{\theta=0}^{\theta=\frac{\pi}{2}} a\cos\theta\, (a\cos\theta)\,d\theta$

$= a^2\int_{0}^{\frac{\pi}{2}} \cos^2\theta\, d\theta$

Now, use $\cos^2\theta = \frac{1+\cos2\theta}{2}$, so

$I = a^2\int_{0}^{\frac{\pi}{2}} \frac{1+\cos2\theta}{2}\,d\theta$

$= \frac{a^2}{2}\int_{0}^{\frac{\pi}{2}} (1+\cos2\theta)\,d\theta$

$= \frac{a^2}{2}\left[\int_{0}^{\frac{\pi}{2}} 1 \, d\theta + \int_{0}^{\frac{\pi}{2}} \cos2\theta\,d\theta\right]$

$\int_{0}^{\frac{\pi}{2}} 1\,d\theta = \frac{\pi}{2}$

$\int_{0}^{\frac{\pi}{2}} \cos2\theta\,d\theta = \left.\frac{\sin2\theta}{2}\right|_0^{\frac{\pi}{2}} = \frac{\sin\pi - \sin 0}{2} = 0$

Therefore, $I = \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{a^2\pi}{4}$

Thus, the full area is $A = 4I = a^2\pi$.

Result: The area enclosed by the circle $x^2 + y^2 = a^2$ is $A = \pi a^2$.

Calculation of Area Bounded by a Standard Parabola

Consider the standard parabola $y^2 = 4ax$ for $x \in [0, a]$, taking only the positive root, so $y = 2\sqrt{a x}$.

The area bound by the parabola, $x$-axis, and lines $x=0$ and $x=a$ (the area to the right of the $y$-axis and beneath the curve) is

$A_{\text{right}} = \int_{x=0}^{x=a} 2\sqrt{a x} \,dx$

Expand $2\sqrt{a x} = 2a^{1/2} x^{1/2}$, so

$A_{\text{right}} = 2a^{1/2} \int_{0}^{a} x^{1/2}\,dx$

$\int x^{1/2}\,dx = \frac{2}{3}x^{3/2}$, thus,

$A_{\text{right}} = 2a^{1/2} \cdot \frac{2}{3}x^{3/2} \Big|_{0}^{a}$

$= \frac{4}{3} a^{1/2} (a)^{3/2} - 0 = \frac{4}{3} a^{1/2} a^{3/2}$

$a^{1/2} a^{3/2} = a^{1/2 + 3/2} = a^2$

Thus $A_{\text{right}} = \frac{4}{3}a^2$.

Since the parabola is symmetric about the $x$-axis, the total area between $x=0$, $x=a$ and the curve is $2A_{\text{right}} = \frac{8}{3} a^2$.

Result: The area bounded by $y^2 = 4ax$, $x = 0$, $x = a$, and the $x$-axis is $\dfrac{8 a^2}{3}$.

Calculation of Area Bounded by a Standard Ellipse

For the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, the upper half is $y = b\sqrt{1 - (x/a)^2}$ for $x \in [-a, a]$. The area in the first quadrant is

$A_{\text{Q1}} = \int_{x=0}^{x=a} b \sqrt{1 - \dfrac{x^2}{a^2}}\,dx$

Use the substitution $x = a\sin\theta$, $dx = a\cos\theta d\theta$, $x=0$ implies $\theta=0$, $x=a$ implies $\theta = \frac{\pi}{2}$.

$\sqrt{1-(x/a)^2} = \sqrt{1 - \sin^2\theta} = \cos\theta$

Therefore,

$A_{\text{Q1}} = \int_{\theta=0}^{\pi/2} b\cos\theta \cdot a\cos\theta\,d\theta = ab \int_{0}^{\pi/2} \cos^2\theta\,d\theta$

From above, $\int_{0}^{\pi/2} \cos^2\theta\,d\theta = \dfrac{\pi}{4}$, so

$A_{\text{Q1}} = ab\dfrac{\pi}{4}$

Thus, total area $A = 4A_{\text{Q1}} = ab\pi$.

Result: The area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $A = \pi ab$.

Area Between a Curve and a Line

Given $y_1 = f(x)$ and $y_2 = g(x)$, the area between the curve and the line from $x = a$ to $x = b$ is

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

When the boundary function $g(x)$ is a constant, this provides the area between the curve and a horizontal line. For advanced examples and applications, review Area of a Sector of a Circle Formula.

Worked Example: Area Under a Curve Using Definite Integration

Given: $y = 7 - x^2$ on $x\in [-1, 2]$.

Substitution: $A = \int_{-1}^{2} (7 - x^2)\,dx$

Integrate $7 - x^2$ with respect to $x$:

$\displaystyle\int (7 - x^2)\,dx = 7x - \frac{x^3}{3} + C$

Evaluate the definite integral:

$A = [7x - \frac{x^3}{3}]_{x=-1}^{x=2}$

First, for $x = 2$:

$7(2) - \frac{2^3}{3} = 14 - \frac{8}{3}$

For $x = -1$:

$7(-1) - \frac{(-1)^3}{3} = -7 + \frac{1}{3}$

Subtract:

$A = [14 - \frac{8}{3}] - [-7 + \frac{1}{3}] = 14 - \frac{8}{3} + 7 - \frac{1}{3}$

$= (14 + 7) - \left( \frac{8}{3} + \frac{1}{3}\right ) = 21 - \frac{9}{3} = 21 - 3 = 18$

Final result: The area under the curve is $18$ square units.

Standard Approximate and Numerical Approaches

When the functional form of the curve is complex or its values are known at discrete points, approximate methods such as the trapezoidal rule and Simpson’s rule are applied. In all cases, the aim is to approximate the sum of the infinitesimal areas beneath the curve with the best possible accuracy. For related topics, see Integral Calculus Important Questions.

The Area Under the Curve in Physics and Statistics

The concept of area under a curve extends to many applied contexts. In physics, the area under a velocity-time graph yields the displacement of an object. In statistics, particularly in probability theory, the area under the normal curve within a given interval represents probability. For further conceptual differences, see Difference Between Area and Volume.