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

5

The correct answer is Option (2) → 5

Line: $2y+x=8 \Rightarrow y=4-\frac{x}{2}$.

Area between $x=2$ and $x=4$ above $x$-axis:

$\displaystyle A=\int_{2}^{4}\left(4-\frac{x}{2}\right)\,dx$

$\displaystyle =\left[4x-\frac{x^2}{4}\right]_{2}^{4}$

$\displaystyle =\big(16-\frac{16}{4}\big)-\big(8-\frac{4}{4}\big)=12-7=5$

Area = 5 square units