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

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Question

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

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$2$

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(A) The equation of the line is $2x - y + 1 = 0$, which can be rewritten as $x = \frac{y - 1}{2}$.The region is bounded by $y = -1$, $y = 3$, $x = 0$ (y-axis), and $x = \frac{y - 1}{2}$.The area $A$ is given by the integral $\int_{-1}^{3} |x| \, dy = \int_{-1}^{3} |\frac{y - 1}{2}| \, dy$.We split the integral at $y = 1$ where the expression changes sign:$A = \int_{-1}^{1} -(\frac{y - 1}{2}) \, dy + \int_{1}^{3} (\frac{y - 1}{2}) \, dy$.$A = -\frac{1}{2} [\frac{y^2}{2} - y]_{-1}^{1} + \frac{1}{2} [\frac{y^2}{2} - y]_{1}^{3}$.$A = -\frac{1}{2} [(\frac{1}{2} - 1) - (\frac{1}{2} + 1)] + \frac{1}{2} [(\frac{9}{2} - 3) - (\frac{1}{2} - 1)]$.$A = -\frac{1}{2} [-\frac{1}{2} - \frac{3}{2}] + \frac{1}{2} [\frac{3}{2} - (-\frac{1}{2})] = -\frac{1}{2}(-2) + \frac{1}{2}(2) = 1 + 1 = 2$.

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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