Let R be the region bounded by the lines y = x + 1, y = 3 − x and the…

The correct answer is Option (3) → (B), (C) and (D) only

The line y = x + 1 meets the x-axis at x = −1 and the line y = 3 − x meets it at x = 3. The lines meet where

    \(x+1 = 3-x,\ \ x = 1,\ \ y = 2\)

so R is the triangle with vertices (−1, 0), (3, 0) and (1, 2).

Integrating in x, the top of R is y = x + 1 up to x = 1 and y = 3 − x after it:

    \( \displaystyle \int_{-1}^{1}(x+1)\,dx = \left[\frac{x^{2}}{2}+x\right]_{-1}^{1} = \frac{3}{2}-\left(-\frac{1}{2}\right) = 2 \)

    \( \displaystyle \int_{1}^{3}(3-x)\,dx = \left[3x-\frac{x^{2}}{2}\right]_{1}^{3} = \frac{9}{2}-\frac{5}{2} = 2 \)

Integrating in y, for \(0 \le y \le 2\) the region runs from x = y − 1 to x = 3 − y, a width of \(4-2y\):

    \( \displaystyle \int_{0}^{2}(4-2y)\,dy = \left[4y-y^{2}\right]_{0}^{2} = 4 \)

So the lines meet at (1, 2), and both the integral in (B) and the integral in (C) give the area of R, which is 4.

Checking the statements

(A) The two lines meet at the point (2, 1) — Incorrect. They meet at (1, 2); (2, 1) is not on y = x + 1.

(B) The area of R is \(\displaystyle\int_{-1}^{1}(x+1)\,dx+\int_{1}^{3}(3-x)\,dx\) — Correct. The two parts are 2 and 2, total 4.

(C) The area of R is \(\displaystyle\int_{0}^{2}(4-2y)\,dy\) — Correct. The width at height y is \((3-y)-(y-1) = 4-2y\), and the integral is 4.

(D) The area of R is 4 sq. units — Correct. Also \(\frac{1}{2}\times4\times2 = 4\) for the triangle.