Let R be the region enclosed between the parabola y = 4x-x^2 and the…

The correct answer is Option (2) → (A), (B) and (D) only

The parabola and the line meet where

    \(4x-x^{2} = x\)
    \(x(3-x) = 0\)
    \(x = 0\) or \(x = 3\)

The points are (0, 0) and (3, 3). Between them the parabola is above the line (at x = 1, the parabola gives 3 and the line gives 1), so the height of R is \((4x-x^{2})-x = 3x-x^{2}\):

    \( \displaystyle \text{Area of R} = \int_{0}^{3}(3x-x^{2})\,dx \)

    \( \displaystyle = \left[\frac{3x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{3} = \frac{27}{2}-9 = \frac{9}{2} \)

The parabola meets the x-axis where \(x(4-x) = 0\), at x = 0 and x = 4:

    \( \displaystyle \int_{0}^{4}(4x-x^{2})\,dx = \left[2x^{2}-\frac{x^{3}}{3}\right]_{0}^{4} = 32-\frac{64}{3} = \frac{32}{3} \)

So the curves meet at (0, 0) and (3, 3), the area of R is \(\displaystyle\frac{9}{2}\), and the parabola cuts off \(\displaystyle\frac{32}{3}\) with the x-axis.

Checking the statements

(A) The parabola and the line meet at the points (0, 0) and (3, 3) — Correct. \(4x-x^{2} = x\) gives x = 0 and x = 3.

(B) The area of R is \(\displaystyle\frac{9}{2}\) sq. units — Correct. \(\displaystyle\frac{27}{2}-9 = \frac{9}{2}\).

(C) The area of the region bounded by the parabola and the x-axis is \(\displaystyle\frac{16}{3}\) sq. units — Incorrect. The area from x = 0 to x = 4 is \(\displaystyle\frac{32}{3}\).

(D) The area of R is \(\displaystyle\int_{0}^{3}(3x-x^{2})\,dx\) — Correct. The height of R is parabola minus line, \(3x-x^{2}\).