Question

The area of the region bounded by the lines $x + 2y = 12$, $x = 2$, $x = 6$, and the $x$-axis is:

• A. $34 \text{ sq units}$
• B. $20 \text{ sq units}$
• C. $24 \text{ sq units}$
• D. $16 \text{ sq units}$

Hint:

When calculating the area under a curve, always express the equation in terms of \( y \) or use the given function directly. For linear functions, the integral simplifies significantly, and evaluating the bounds at each step helps ensure accuracy. Remember to apply the fundamental theorem of calculus when substituting limits into the integral expression. This method is very effective for finding areas between curves and the x-axis.

Answer 1

The Correct Option is D

To find the area of the region bounded by \( x + 2y = 12 \), \( x = 2 \), \( x = 6 \), and the x-axis, we start by expressing \( y \) in terms of \( x \) from the equation \( x + 2y = 12 \):

\[ y = \frac{12 - x}{2} \]

The area between \( x = 2 \) and \( x = 6 \) under the line \( y = \frac{12 - x}{2} \) is given by:

\[ \text{Area} = \int_{2}^{6} \frac{12 - x}{2} \, dx \]

Evaluating this integral:

\[= \int_{2}^{6} \frac{12 - x}{2} \, dx = \frac{1}{2} \int_{2}^{6} (12 - x) \, dx\]

\[= \frac{1}{2} \left[ 12x - \frac{x^2}{2} \right]_{2}^{6}\]

\[= \frac{1}{2} \left[ \left( 12 \times 6 - \frac{6^2}{2} \right) - \left( 12 \times 2 - \frac{2^2}{2} \right) \right]\]

\[= \frac{1}{2} \left[ (72 - 18) - (24 - 2) \right]\]

\[= \frac{1}{2} [54 - 22] = \frac{1}{2} \times 32 = 16\]

Therefore, the area of the region is 16 sq units.

Answer 2

To find the area of the region bounded by \( x + 2y = 12 \), \( x = 2 \), \( x = 6 \), and the x-axis, we start by expressing \( y \) in terms of \( x \) from the equation \( x + 2y = 12 \):

\[ y = \frac{12 - x}{2} \]

Step 1: Set up the integral to find the area between \( x = 2 \) and \( x = 6 \) under the line \( y = \frac{12 - x}{2} \):

The area between \( x = 2 \) and \( x = 6 \) under the curve is given by the integral:

\[ \text{Area} = \int_{2}^{6} \frac{12 - x}{2} \, dx \]

Step 2: Simplify the integral expression:

\[ \text{Area} = \frac{1}{2} \int_{2}^{6} (12 - x) \, dx \]

Step 3: Evaluate the integral:

\[ = \frac{1}{2} \left[ 12x - \frac{x^2}{2} \right]_{2}^{6} \]

Step 4: Substitute the limits of integration:

\[ = \frac{1}{2} \left[ \left( 12 \times 6 - \frac{6^2}{2} \right) - \left( 12 \times 2 - \frac{2^2}{2} \right) \right] \]

Step 5: Perform the arithmetic calculations:

\[ = \frac{1}{2} \left[ (72 - 18) - (24 - 2) \right] \] \[ = \frac{1}{2} [54 - 22] = \frac{1}{2} \times 32 = 16 \]

Conclusion: Therefore, the area of the region is 16 square units.