The area (in sq. units) of the region bounded by the lines y = x, y = 3x and x = 2 is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
Both lines pass through the origin, and for \(0 \le x \le 2\) the line y = 3x lies above the line y = x. The region is the strip between them, so its height at x is \(3x-x\):
\( \displaystyle A = \int_{0}^{2}(3x-x)\,dx \)
\( \displaystyle = \int_{0}^{2}2x\,dx \)
\( \displaystyle = \left[x^{2}\right]_{0}^{2} = 4 \)