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
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5
From \(2y + x = 8\),
\( \displaystyle y = 4 - \frac{x}{2} \)
which is positive for \(2 \le x \le 4\), so the line lies above the x-axis there:
\( \displaystyle \text{Area} = \int_{2}^{4}\left(4 - \frac{x}{2}\right)dx \)
\( \displaystyle = \left[4x - \frac{x^2}{4}\right]_{2}^{4} \)
\(= (16 - 4) - (8 - 1)\)
\(= 12 - 7 = 5\) sq. units
Shortcut: The region is a trapezium with parallel sides y(2) = 3 and y(4) = 2 and width 2, so its area is \(\frac{1}{2}(3 + 2)\times 2 = 5\).