Problem 44 - Single Variable Calculus

Question

In Exercises 43 and 44, use integration to find the area of the triangle with the given vertices. $$ (-2,4),(0,-2),(6,2) $$

Step-by-Step Solution

Answer

The area of the triangle with vertices \((-2,4)\), \((0,-2)\), and \((6,2)\) can be determined by integrating each line segment of the triangle and combining the results. To find the area, perform the following calculations: 1. Determine the equations of the lines: - Line 1: \(y = -3x - 2\) - Line 2: \(y = \frac{2}{3}x - 2\) - Line 3: \(y = -\frac{1}{4}x + \frac{9}{2}\) 2. Determine the intercepts: - Between lines 1 and 3: \(x = -\frac{26}{11}\) - Between lines 2 and 3: \(x = \frac{78}{11}\) 3. Integrate each line segment: - Area under line 1: \(\int_{-\frac{26}{11}}^0 (-3x - 2)dx\) - Area under line 2: \(\int_0^{\frac{78}{11}} (\frac{2}{3}x - 2)dx\) - Area under line 3: \(\int_{-\frac{26}{11}}^{\frac{78}{11}} (-\frac{1}{4}x + \frac{9}{2})dx\) 4. Calculate the total area: Total area \(= 21\) square units

Step 1: Determine the equations of the lines

First, we need to find the equations of the lines that form the sides of the triangle. To do this, we can use the slope-intercept form of a line, which is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. Slope between points \((-2, 4)\) and \((0, -2)\): \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 4}{0 - (-2)} = -3\) Plugging point \((-2, 4)\) into the equation to find the y-intercept: \(4 = -3(-2) + b\) \(b = -2\) Equation of line 1: \(y = -3x - 2\) Slope between points \((0, -2)\) and \((6, 2)\): \(m = \frac{2 - (-2)}{6 - 0} = \frac{4}{6} = \frac{2}{3}\) Plugging point \((0, -2)\) into the equation to find the y-intercept: \(-2 = 0 + b\) \(b = -2\) Equation of line 2: \(y = \frac{2}{3}x - 2\) Slope between points \((-2, 4)\) and \((6, 2)\): \(m = \frac{2 - 4}{6 - (-2)} = \frac{-2}{8} = -\frac{1}{4}\) Plugging point \((-2, 4)\) into the equation to find the y-intercept: \(4 = (-\frac{1}{4})(-2) + b\) \(b = \frac{9}{2}\) Equation of line 3: \(y = -\frac{1}{4}x + \frac{9}{2}\)

Step 2: Determine the intercepts

Now, we need to find the x-intercepts of the lines where they intersect each other. To do this, we need to solve the equations simultaneously. Between line 1 and line 3: \(-3x - 2 = -\frac{1}{4}x + \frac{9}{2}\) Multiplying both sides by 4 to eliminate fractions: \(-12x - 8 = -x + 18\) Solving for x: \(-11x = 26\) \(x = -\frac{26}{11}\) Between line 2 and line 3: \(\frac{2}{3}x - 2 = -\frac{1}{4}x + \frac{9}{2}\) Multiplying both sides by 12 to eliminate fractions: \(8x - 24 = -3x + 54\) Solving for x: \(11x = 78\) \(x = \frac{78}{11}\)

Step 3: Integrate each line segment

Now, we need to integrate each of the line segments that form the sides of the triangle. To do this, we evaluate the definite integral of the absolute value of each equation on the determined range of integration. Area under line 1: \(\int_{-\frac{26}{11}}^0 (-3x - 2)dx\) Area under line 2: \(\int_0^{\frac{78}{11}} (\frac{2}{3}x - 2)dx\) Area under line 3: \(\int_{-\frac{26}{11}}^{\frac{78}{11}} (-\frac{1}{4}x + \frac{9}{2})dx\)

Step 4: Calculate the total area

Finally, we calculate the total area of the triangle by adding the integrals obtained in the previous step. Total area: \(\int_{-\frac{26}{11}}^0 (-3x - 2)dx + \int_0^{\frac{78}{11}} (\frac{2}{3}x - 2)dx - \int_{-\frac{26}{11}}^{\frac{78}{11}} (-\frac{1}{4}x + \frac{9}{2})dx\) This gives us the final answer: Total area \(= 21\) square units

Key Concepts

Area of Triangle

When we talk about the area of a triangle, we're discussing the amount of space contained within the triangle's boundaries. Typically, the area can be easily calculated using the formula:

• For a triangle with a base and height: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
• Using vertices coordinates: In cases where vertices are known, integration or other methods may be employed.

In this exercise, the vertices are given in a coordinate system, which means we can use the integration method to determine the area under or between lines that create the triangle. By evaluating the definite integrals of these lines, we find the total area that these lines enclose.

Slope-Intercept Form

The slope-intercept form of a line's equation is a straightforward formula given as \( y = mx + b \). This representation helps us identify:

• \( m \) as the slope of the line, which indicates the line's steepness and direction.
• \( b \) as the y-intercept, the point where the line crosses the y-axis.

This form is integral to defining a line between any two points on the Cartesian plane, such as finding equations of the triangle's sides. The slope is calculated from two given points \((x_1, y_1)\) and \((x_2, y_2)\) using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Once computed, substituting a point back into the slope-intercept equation allows us to solve for \( b \). Thus, we effectively model each line forming the triangle.

Definite Integral

A definite integral is a concept from calculus that helps calculate the area between a curve and the x-axis, between two points.

• The notation \( \int_{a}^{b} f(x) \, dx \) signifies finding the integral of \( f(x) \), the function, from \( a \) to \( b \), the bounds.
• The result represents the net area between the curve \( y = f(x) \) and the x-axis, considering areas above the axis as positive and areas below as negative.

In the context of this problem, definite integrals are used to compute the triangular region's area through the equations of its sides. By integrating parts of each line segment forming the triangle, we precisely determine the total enclosed area.

Simultaneous Equations

Solving simultaneous equations involves finding a set of values for variables that satisfy multiple equations at the same time. In this exercise, it’s crucial for determining the points of intersection between the lines forming the triangle. The process involves:

• Equating two line equations (derived from the slope-intercept form) to find their intersection.
• Substituting one variable from one equation into another to solve for outcomes completely.

By identifying the correct solution to these simultaneous equations, we validate the shared points or intersections, ensuring the lines truly converge and accurately define the construction of the triangle.