Question

Use integration to find the area of the triangular region having the given vertices. $$ \begin{aligned} &(0,0),(4,0),(4,4) \\ &(0,0),(4,0),(6,4) \end{aligned} $$

Short answer

The area of the first triangle is 8 square units and the area of the second triangle is 12 square units.

Step-by-step solution

Setup the integral

The first point of both triangles is at the origin (0,0), meaning there's no need for dealing with negative space. Since the triangles are right triangles, the integration is straightforward. For the first integral, the function representing the area below the line connecting (0,0) and (4,4) is y=x. For the second triangle, the function representing the area below the line connecting (0,0) and (6,4) is y=\(\frac{2}{3}x\). The area for each triangle can be calculated with integral from 0 to 4 for the first triangle and from 0 to 6 for the second triangle.

Solve the integrals

Now each integral is calculated separately. For the first triangle, the integral is: \[\int_{0}^{4} x dx \] The solution of this integral is \( \frac{1}{2}*4^2 - \frac{1}{2}*0^2 = 8\). Similar for the second triangle, the integral is: \[\int_{0}^{6} \frac{2}{3}x dx \] The solution of this integral is \( \frac{1}{3}*2*6^2 - \frac{1}{3}*0^2 = 12\).

Finalize

The areas for both triangles have been calculated with the respective integrals. The first triangle's area is 8 square units and the second triangle's area is 12 square units.

Key concepts

Definite Integrals

In calculus, a definite integral is fundamental for computing areas, volumes, and for solving numerous real-world problems. Think of a definite integral as a tool that accumulates the values of a function over a specific interval. Mathematically, it looks like this: \[ \int_{a}^{b} f(x) dx \] where \( f(x) \) is the function to be integrated, and \( a \) and \( b \) are the limits of integration representing the interval over which you want to calculate the area.

When applying this to our problem with triangles, the function represents the height of the triangle at any point along the base. The limits of the integral correspond to the endpoints of the base. By calculating the definite integral, we can find the total area under the curve of the function, giving us the area of the triangular region defined by its vertices.

Area of Triangles

The area of a triangle can be found using several methods, depending on the information provided. A simple formula for the area is \(\frac{1}{2} \times \text{base} \times \text{height}\), which works perfectly for right-angled triangles. In our case, we use integration as the method since the triangle's sides are expressed by linear functions.

For Right-Angled Triangles

These types of triangles have one internal angle equal to 90 degrees, which makes the height coincide with one of the sides if the triangle is positioned accordingly. The base can be along the x-axis, and the height can be expressed as a function in terms of x. By integrating this function over the interval representing the base's length, you find the area encapsulated by the base, the function, and the x-axis.

Calculus Applications

Calculus, especially integration, has a vast range of applications in various fields beyond finding the area under a curve. It includes understanding changes in quantities, optimizing functions, solving differential equations, and describing physical phenomena. In physics, calculus is used to calculate the motion of objects, while in engineering, it is fundamental for designing structures and understanding materials' behavior.

Calculus can also model biological systems in medicine, aiding drug dosage optimization, predicting the spread of diseases, and understanding the dynamics of populations in ecology. In every application, calculus often provides critical insights that are impossible to obtain via mere algebra or geometry.