The area (in sq.units) of the region bounded by the line $2y + x = 8$, the x-axis and the lines $x = 2$ and $x = 4$ is

The region is bounded by the line $2y + x = 8$, the x-axis, and the vertical lines $x = 2$ and $x = 4$.

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Solving for $y$:

$2y + x = 8$

$2y = 8 - x$

$y = \frac{8-x}{2}$

$y = 4 - \frac{x}{2}$

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Finding the heights at the boundaries:

At $x = 2$:

$y = 4 - \frac{2}{2}$

$y = 4 - 1$

$y = 3$

At $x = 4$:

$y = 4 - \frac{4}{2}$

$y = 4 - 2$

$y = 2$

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The region forms a trapezoid with:

• Left side height $= 3$ units
• Right side height $= 2$ units
• Width $= 4 - 2 = 2$ units

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Using the trapezoid area formula:

$\text{Area} = \frac{1}{2}(b_1 + b_2) \times h$

$\text{Area} = \frac{1}{2}(3 + 2) \times 2$

$= \frac{1}{2} \times 5 \times 2$

$= 5$

Therefore, the area of the bounded region is $5$ square units.