Finding Area Bounded by Lines and X-axis

The problem asks us to find the area of a specific region in the coordinate plane. This region is enclosed by four boundaries: a linear equation, two vertical lines, and the x-axis.

Identifying the Boundaries of the Region

The given boundaries are:

• Line 1: \(x + 2y = 12\)
• Line 2: \(x = 2\)
• Line 3: \(x = 6\)
• Line 4: The x-axis, which is the line \(y = 0\).

To find the area of the region bounded by a curve \(y = f(x)\), the lines \(x=a\), \(x=b\), and the x-axis, we typically evaluate the definite integral \( \int_{a}^{b} f(x) \, dx \). First, we need to express the equation of Line 1 in the form \(y = f(x)\).

Expressing the Line in \(y = f(x)\) Form

We rearrange the equation \(x + 2y = 12\) to solve for \(y\):

\(2y = 12 - x\)

\(y = \frac{12 - x}{2}\)

\(y = 6 - \frac{x}{2}\)

So, the function defining the upper boundary of the region is \(f(x) = 6 - \frac{x}{2}\).

Setting Up the Definite Integral for Area

The region is bounded by the vertical lines \(x = 2\) and \(x = 6\). These will be our limits of integration. The lower boundary is the x-axis (\(y=0\)). We need to ensure that the function \(f(x) = 6 - \frac{x}{2}\) is non-negative over the interval \(x \in [2, 6]\).

When \(x=2\), \(y = 6 - \frac{2}{2} = 6 - 1 = 5\). When \(x=6\), \(y = 6 - \frac{6}{2} = 6 - 3 = 3\).

Since the function values are positive for \(x=2\) and \(x=6\), and the function is linear (monotonically decreasing), it remains positive throughout the interval \([2, 6]\). Therefore, the area of the region is given by the definite integral:

\( \text{Area} = \int_{2}^{6} \left(6 - \frac{x}{2}\right) \, dx \)

Evaluating the Definite Integral

We now evaluate the integral step-by-step:

First, find the antiderivative of \(6 - \frac{x}{2}\):

\( \int \left(6 - \frac{x}{2}\right) \, dx = \int 6 \, dx - \int \frac{x}{2} \, dx \)

\( = 6x - \frac{1}{2} \int x \, dx \)

\( = 6x - \frac{1}{2} \left(\frac{x^{1+1}}{1+1}\right) + C \)

\( = 6x - \frac{1}{2} \left(\frac{x^2}{2}\right) + C \)

\( = 6x - \frac{x^2}{4} + C \)

Now, evaluate the definite integral using the limits 2 and 6:

\( \text{Area} = \left[6x - \frac{x^2}{4}\right]_{2}^{6} \)

\( = \left(6(6) - \frac{6^2}{4}\right) - \left(6(2) - \frac{2^2}{4}\right) \)

\( = \left(36 - \frac{36}{4}\right) - \left(12 - \frac{4}{4}\right) \)

\( = (36 - 9) - (12 - 1) \)

\( = 27 - 11 \)

\( = 16 \)

The area of the region bounded by the lines \(x + 2y = 12\), \(x = 2\), \(x = 6\), and the x-axis is 16 square units.

Conclusion

By setting up and evaluating the definite integral of the function \(y = 6 - \frac{x}{2}\) from \(x=2\) to \(x=6\), we found the area of the specified region.

Revision Table: Area Calculation

Additional Information: Area Between Curves

If the region were bounded by two curves, say \(y = f(x)\) and \(y = g(x)\) from \(x=a\) to \(x=b\), where \(f(x) \ge g(x)\) over \([a,b]\), the area would be given by \( \int_{a}^{b} [f(x) - g(x)] \, dx \). In our problem, the lower boundary is the x-axis, which is the curve \(y=0\). So, we were essentially finding the area between \(y = 6 - \frac{x}{2}\) and \(y = 0\) over the interval \([2, 6]\). The formula \( \int_{a}^{b} f(x) \, dx \) is a special case of the area between curves formula when \(g(x) = 0\).

It's also possible to calculate areas by integrating with respect to \(y\). If a region is bounded by curves \(x = g(y)\), \(x = h(y)\), and horizontal lines \(y=c\), \(y=d\), where \(g(y) \ge h(y)\) over \([c,d]\), the area is \( \int_{c}^{d} [g(y) - h(y)] \, dy \). However, for this specific problem, integrating with respect to \(x\) was more straightforward given the form of the equations and vertical boundaries.