Straight Lines and Enclosed Area Calculation

To determine the area enclosed between two straight lines, we must first establish their individual equations. Both lines are specified to pass through the origin, which is the point \((x_0, y_0) = (0, 0)\).

Lines Equations Determination

Line 1: This line is defined by passing through the origin \((0, 0)\) and the point \((x_1, y_1) = (1, 3)\).

• The formula for the slope \(m\) of a straight line passing through two points \((x_a, y_a)\) and \((x_b, y_b)\) is given by \(m = \frac{y_b - y_a}{x_b - x_a}\).
• Applying this formula for Line 1, the slope \(m_1\) is calculated as:
• \(m_1 = \frac{3 - 0}{1 - 0} = \frac{3}{1} = 3\)
• Since the line passes through the origin, its equation takes the simplified form \(y = mx\), where \(m\) is the slope.
• Consequently, the equation for Line 1 is: \(y_1 = 3x\)

Line 2: This second straight line also passes through the origin \((0, 0)\) and goes through the point \((x_2, y_2) = (1, 2)\).

• Similarly, for Line 2, the slope \(m_2\) is determined using the same slope formula:
• \(m_2 = \frac{2 - 0}{1 - 0} = \frac{2}{1} = 2\)
• As Line 2 also passes through the origin, its equation is also of the form \(y = mx\).
• Therefore, the equation for Line 2 is: \(y_2 = 2x\)

Area Enclosed Calculation

Our objective is to calculate the area enclosed between these two straight lines, \(y_1 = 3x\) and \(y_2 = 2x\), specifically over the interval \([0, 1]\) on the x-axis.

Within the interval \([0, 1]\), for any non-negative value of \(x\), it is evident that \(3x \ge 2x\). This indicates that the graph of \(y_1\) is always above or coincident with the graph of \(y_2\) in this interval.

The area \(A\) between two curves, \(f(x)\) and \(g(x)\), from \(x = a\) to \(x = b\), where \(f(x) \ge g(x)\) over the interval \([a, b]\), is found by evaluating the definite integral:

\(A = \int_{a}^{b} [f(x) - g(x)] dx\)

In this problem, \(f(x) = y_1 = 3x\), \(g(x) = y_2 = 2x\), and the limits of integration are from \(a = 0\) to \(b = 1\).

Substituting these values into the integral formula, we get:

\(A = \int_{0}^{1} (3x - 2x) dx\)

Simplifying the integrand:

\(A = \int_{0}^{1} x dx\)

Now, we proceed to evaluate this definite integral:

\(A = \left[ \frac{x^{1+1}}{1+1} \right]_{0}^{1}\)

\(A = \left[ \frac{x^2}{2} \right]_{0}^{1}\)

Applying the upper and lower limits of integration:

\(A = \frac{(1)^2}{2} - \frac{(0)^2}{2}\)

\(A = \frac{1}{2} - 0\)

\(A = 0.5\)

Final Result and Conclusion

The calculated area enclosed between the two straight lines within the specified interval of \([0, 1]\) on the x-axis is \(0.5\) square units.