Area Under a Curve

Area Under a Curve

The concept of finding the area under a curve is a fundamental application of definite integrals. It helps in solving problems related to total accumulation, displacement, probability, economics, and physics.

1. Understanding the Area Under a Curve

Given a function $f(x)$, the area under its graph between two points $x = a$ and $x = b$ can be computed using integration. If $f(x)$ is non-negative on $[a, b]$, the area is given by:

where:

• $a$ and $b$ are the limits of integration.
• $f(x)$ represents the height of the curve at each point.
• $dx$ represents an infinitesimally small width.

If $f(x)$ takes negative values, the integral computes net area, meaning that the area below the x-axis is subtracted from the area above the x-axis.

2. Approximation Using Riemann Sums

Before defining integration, we can approximate the area under a curve using Riemann sums:

1. Divide the interval $[a, b]$ into $n$ subintervals of equal width:

   $$\Delta x = \frac{b-a}{n}$$

2. Choose sample points $x_i^*$ within each subinterval.

3. Compute the sum of the areas of the rectangles:

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

4. As $n \to \infty$, the sum approaches the definite integral.

3. The Definite Integral and the Area

The exact area under $f(x)$ from $x = a$ to $x = b$ is given by:

Case 1: $f(x) \geq 0$

If $f(x)$ is non-negative, the integral directly gives the total area.

Case 2: $f(x)$ is both positive and negative

If $f(x)$ crosses the x-axis, the integral computes net area (positive areas minus negative areas). To find the total area, we must integrate separately where $f(x) \geq 0$ and where $f(x) < 0$, taking absolute values.

where $x = c$ is the root of $f(x)$.

4. Example: Finding Area Under a Curve

Find the area under the curve $f(x) = x^2$ from $x = 0$ to $x = 3$.

1. Compute the integral:

   $$\int_{0}^{3} x^2 \, dx$$

2. Find the antiderivative of $x^2$:

   $$F(x) = \frac{x^3}{3}$$

3. Evaluate from $0$ to $3$:

   $$A = \left[ \frac{x^3}{3} \right]_{0}^{3}$$ $$= \frac{3^3}{3} - \frac{0^3}{3}$$ $$= \frac{27}{3} - 0 = 9$$

Thus, the area under the curve from $x = 0$ to $x = 3$ is 9 square units.

5. Applications of Finding the Area Under a Curve

• Physics: Computing work done, displacement, and electric charge accumulation.
• Economics: Calculating consumer surplus, revenue, and cost functions.
• Biology & Medicine: Finding total drug concentration over time.
• Probability & Statistics: Computing probabilities from probability density functions.

By understanding how to compute areas under curves, we can solve many practical problems in science and engineering.