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

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(A) Step $1$: Express $x$ in terms of $y$ from the equation $2x - y + 1 = 0$. We get $2x = y - 1$, so $x = \frac{y - 1}{2}$.Step $2$: The area $A$ bou

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(A) Step $1$: Express $x$ in terms of $y$ from the equation $2x - y + 1 = 0$. We get $2x = y - 1$, so $x = \frac{y - 1}{2}$.Step $2$: The area $A$ bounded by the curve $x = f(y)$, the $y$-axis, and the lines $y = c$ and $y = d$ is given by $A = \int_{c}^{d} |x| \, dy$.Step $3$: Substitute the limits $y = -1$ to $y = 3$: $A = \int_{-1}^{3} |\frac{y - 1}{2}| \, dy$.Step $4$: The expression $\frac{y - 1}{2}$ changes sign at $y = 1$. Thus, $A = \int_{-1}^{1} -(\frac{y - 1}{2}) \, dy + \int_{1}^{3} (\frac{y - 1}{2}) \, dy$.Step $5$: Calculate the integrals: $\frac{1}{2} [-( \frac{y^2}{2} - y )]_{-1}^{1} + \frac{1}{2} [ \frac{y^2}{2} - y ]_{1}^{3} = \frac{1}{2} [-( (\frac{1}{2} - 1) - (\frac{1}{2} + 1) )] + \frac{1}{2} [ (\frac{9}{2} - 3) - (\frac{1}{2} - 1) ] = \frac{1}{2} [ -(-\frac{1}{2} - \frac{3}{2}) ] + \frac{1}{2} [ \frac{3}{2} - (-\frac{1}{2}) ] = \frac{1}{2} [2] + \frac{1}{2} [2] = 1 + 1 = 2$ 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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