The area of the region bounded by the lines 2 | vedclass

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(A) The given line is $2x - y + 1 = 0$, which can be rewritten as $x = \frac{y - 1}{2}$.The region is bounded by $y = -1$, $y = 3$, and the y-axis $(x

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$2$

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(A) The given line is $2x - y + 1 = 0$, which can be rewritten as $x = \frac{y - 1}{2}$.The region is bounded by $y = -1$, $y = 3$, and the y-axis $(x = 0)$.The area $A$ is given by the integral $\int_{-1}^{3} |x| \, dy$.Since $x = \frac{y - 1}{2}$, we have $A = \int_{-1}^{3} |\frac{y - 1}{2}| \, dy$.At $y = 1$, $x = 0$. For $y \in [-1, 1]$, $x \le 0$, so $|x| = -\frac{y - 1}{2} = \frac{1 - y}{2}$.For $y \in [1, 3]$, $x \ge 0$, so $|x| = \frac{y - 1}{2}$.$A = \int_{-1}^{1} \frac{1 - y}{2} \, dy + \int_{1}^{3} \frac{y - 1}{2} \, dy$.$A = \frac{1}{2} [y - \frac{y^2}{2}]_{-1}^{1} + \frac{1}{2} [\frac{y^2}{2} - y]_{1}^{3}$.$A = \frac{1}{2} [(1 - 0.5) - (-1 - 0.5)] + \frac{1}{2} [(4.5 - 3) - (0.5 - 1)]$.$A = \frac{1}{2} [0.5 + 1.5] + \frac{1}{2} [1.5 + 0.5] = \frac{1}{2}(2) + \frac{1}{2}(2) = 1 + 1 = 2$ square units.

The area of the region bounded by the lines $2x - y + 1 = 0$, $y = -1$, $y = 3$ and the y-axis is

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