Let R be the region bounded by the line 2x + y = 8 and the coordinate…

The correct answer is Option (1) → (A) and (B) only

Putting x = 0 and then y = 0 in 2x + y = 8 gives the points where the line meets the axes:

    \(x = 0 \Rightarrow y = 8\)
    \(y = 0 \Rightarrow x = 4\)

So R is the triangle with vertices (0, 0), (4, 0) and (0, 8). Its area is

    \( \displaystyle \frac{1}{2}\times 4\times 8 = 16 \)

In terms of x, the upper boundary is y = 8 − 2x, for \(0 \le x \le 4\):

    \( \displaystyle \int_{0}^{4}(8-2x)\,dx = \left[8x-x^{2}\right]_{0}^{4} = 32-16 = 16 \)

In terms of y, the right boundary is \(x = \frac{8-y}{2}\), for \(0 \le y \le 8\). The integrand 8 − 2y is not this boundary:

    \( \displaystyle \int_{0}^{8}(8-2y)\,dy = \left[8y-y^{2}\right]_{0}^{8} = 64-64 = 0 \)

So the line meets the y-axis at (0, 8), the area of R is 16, the x-integral gives 16 and the given y-integral gives 0.

Checking the statements

(A) The line meets the y-axis at the point (0, 8) — Correct. x = 0 gives y = 8.

(B) The area of R is \(\displaystyle\int_{0}^{4}(8-2x)\,dx\) — Correct. The integrand is the line y = 8 − 2x, and the integral is 16.

(C) The area of R is \(\displaystyle\int_{0}^{8}(8-2y)\,dy\) — Incorrect. In terms of y the line is \(x = \frac{8-y}{2}\); the given integral is 0, not 16.

(D) The area of R is 32 sq. units — Incorrect. The area is \(\frac{1}{2}\times 4\times 8 = 16\).