Problem 55 - Calculus of a Single Variable

Question

Finding the Area of a Figure In Exercises \(55-58,\) use integration to find the area of the figure having the given vertices. $$ (2,-3),(4,6),(6,1) $$

Step-by-Step Solution

Answer

Given that the vertices of the triangle are (2,-3), (4,6), and (6,1), the area of the triangle can be found by integrating the difference of the equations of the lines that make up the sides of the triangle, over the interval of x-values for each side.

Step 1: Find the equations of the lines

To start, find the equations of the lines that pass through the points (2,-3), (4,6), and (6,1). The equation of a line can be obtained by using the two point form which is \(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\) where \((x_1,y_1)\) and \((x_2,y_2)\) are two points on the line. Apply this formula to find equations of each of the three lines forming the triangle.

Step 2: Express y as a function of x for each line

After obtaining the equations of the lines, make y the subject for each equation. This gives you three new equations, each expressing \(y\) as a function of \(x\) for a different side of the triangle. Call them \(y1(x)\), \(y2(x)\), and \(y3(x)\).

Step 3: Set up the integral

The area of the triangle can be found by integrating the difference between the \(y\) values of the functions that define the upper and lower limits of the triangle. Write the integral in the form \(\int_a^b [y1(x) - y2(x)]\, dx\) or \(\int_a^b [y2(x) - y1(x)]\, dx\), depending on whether \(y1(x) > y2(x)\) or \(y2(x) > y1(x)\) in the interval [a, b]. Do the same for the other side of the triangle.

Step 4: Calculate the area

Finally, evaluate the definite integral to find the area of the triangle. The area will be the sum of the absolute values of the two integrals.

Key Concepts

Equations of Lines

Lines are fundamental in geometry and analytical math. Each line in a plane can be represented by a unique equation. These equations help determine various properties like slope, intersections, and more.
To find an equation of a line, you need at least two distinct points on that line. Using these points, you craft a relationship between the x-values and y-values.

• Understanding the behavior of lines is crucial for solving geometry problems.
• Equations simplify complex geometric relationships.

Two Point Form

The two point form is a straightforward method to derive the equation of a line. It is particularly useful when you have two points \((x_1,y_1)\) and \((x_2,y_2)\).
The formula is given by:\[y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\]This formula captures the essence of linearity by using the slope, which is the change in y divided by the change in x.

• Slope (\(m\)) is a measure of steepness.
• Helps in transforming point coordinates into line equations.
• Vital for achieving accurate graph plots.

Definite Integral

In calculus, the definite integral provides a way to calculate the area under a curve between two points, say \(a\) and \(b\). Essentially, it sums up infinitely small rectangles beneath the curve.
Definite integral provides the total accumulation of a quantity, such as area. The notation is:\[\int_a^b f(x) \, dx\]This integral finds the exact area bounded by the graph of \(f(x)\) from \(x = a\) to \(x = b\).

• Represents the signed area under the curve.
• Useful in geometric, physical, and statistical applications.
• Key element for calculating total change in a function over an interval.

Triangle Vertices

Vertices are the corners or points of intersection in geometric shapes. In triangles, the three vertices uniquely define the shape. In this exercise, we have vertices at \(\((2,-3),(4,6),(6,1)\)\).
The vertices help establish the perimeter and area of the triangle. Knowing these points allows us to understand the triangle’s geometric properties.

• Vertices provide critical information for forming line equations.
• Determine spatial relationships in geometry.
• Essential for geometric constructions and proofs.