Area of a Flat Surface

Basic Concepts of Flat Surface Area

In everyday life, we often need to calculate the area of various flat shapes. For shapes with simple forms like squares or triangles, we can use familiar formulas. But what if we want to calculate the area of a region bounded by irregular curves?

Definite integrals provide an elegant solution to this problem. The basic concept of definite integrals stems from Riemann's approach, where we divide a region into small rectangles and then sum their areas.

Imagine we have a function $f(x)$ and want to find the area of the region under the curve from $x = a$ to $x = b$. We can divide the interval $[a, b]$ into $n$ small parts with width $\Delta x$.

Determining Area Using Definite Integrals

To calculate the area of a flat surface using definite integrals, we need to understand several systematic steps:

Identify Integration Limits

The first step is to determine the lower and upper limits of integration. These limits indicate the range of $x$ values that bound the region whose area we want to calculate.

Determine the Integrand Function

The function to be integrated is the function that bounds the region. If the region is above the $x$-axis, then the area of the region is $\int_{a}^{b} f(x) \, dx$.

Evaluate the Integral

After determining the limits and function, we can evaluate the integral using the fundamental theorem of calculus:

where $F(x)$ is the antiderivative of $f(x)$.

Application to Quadratic Functions

Let's apply this concept to a concrete example. Suppose we want to calculate the area of the region bounded by the curve $f(x) = x^2 - 4x$ and the $x$-axis between $x = 1$ and $x = 3$.

Now, try to observe the graph above. The function $f(x) = x^2 - 4x$ turns out to have negative values in the interval $[1, 3]$. We can easily check: when $x = 1$, we get $f(1) = 1 - 4 = -3$. Similarly, when $x = 3$, we get $f(3) = 9 - 12 = -3$.

Well, here's where the uniqueness lies! Since we're looking for area which is always positive, we need to use the absolute value of that function. So our integral becomes:

Let's solve it step by step:

Therefore, the area of that region is $\frac{22}{3}$ square units.

Application to Irrational Functions

Now let's try a slightly more challenging example with an irrational function. We will calculate the area of the region under the curve $f(x) = x\sqrt{x^2 + 5}$ from $x = 0$ to $x = 2$.

For this integral, we need to use the substitution technique. Why? Because there's the form $x\sqrt{x^2 + 5}$ which is quite complex if we solve it directly.

Let's perform substitution with $u = x^2 + 5$. From here, we get the differential $du = 2x \, dx$, which means $x \, dx = \frac{1}{2} du$.

Don't forget to change the integration limits too! When $x = 0$, we get $u = 5$. When $x = 2$, we get $u = 9$.

Now our integral becomes:

Note that $9^{3/2} = (3^2)^{3/2} = 3^3 = 27$ and $5^{3/2} = 5 \cdot \sqrt{5}$.

When using substitution in definite integrals, don't forget to change the integration limits according to the new substitution variable.

Exercises

1. Calculate the area of the region bounded by the curve $y = x^2 + 1$, the $x$-axis, and the lines $x = 0$ and $x = 2$!

2. Determine the area of the region under the curve $y = \frac{1}{x^2 + 1}$ from $x = 0$ to $x = 1$!

3. Calculate the area of the region bounded by the curve $y = 2x - x^2$ and the $x$-axis!

Answer Key

1. First problem with function $y = x^2 + 1$

   Since this function is always positive, we can directly set up the integral:

   $$A = \int_{0}^{2} (x^2 + 1) \, dx$$

   After we integrate and evaluate, we obtain:

   $$A = \left[\frac{x^3}{3} + x\right]_{0}^{2}$$

   $$A = \left(\frac{8}{3} + 2\right) - 0 = \frac{14}{3}$$

   Therefore, the area of that region is $\frac{14}{3}$ square units.

2. Second problem with rational function

   For this integral, we need to remember that the antiderivative of $\frac{1}{x^2 + 1}$ is $\arctan(x)$.

   $$A = \int_{0}^{1} \frac{1}{x^2 + 1} \, dx$$

   $$A = [\arctan(x)]_{0}^{1}$$

   $$A = \arctan(1) - \arctan(0) = \frac{\pi}{4}$$

   The area of that region is $\frac{\pi}{4}$ square units.

3. Third problem with parabola

   First, let's find where the curve intersects the $x$-axis:

   $$2x - x^2 = 0 \Rightarrow x(2 - x) = 0$$

   So the intersection points are at $x = 0$ and $x = 2$. Since this function is positive between these two points, we can directly integrate:

   $$A = \int_{0}^{2} (2x - x^2) \, dx$$

   $$A = \left[x^2 - \frac{x^3}{3}\right]_{0}^{2}$$

   $$A = 4 - \frac{8}{3} = \frac{4}{3}$$

   The area of that region is $\frac{4}{3}$ square units.