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. $16 \text{ sq units}$

Answer 1

Answer: (D)

$16 \text{ sq units}$

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 $x + 2y = 12$:

$y = \frac{12 - x}{2}$

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.