Mathematics - Area Under Curves Question with Solution | TestHub

The given line is $2x-y+1=0\Rightarrow x=\frac{y-1}{2}$

The required area is bounded by the curve $x=\frac{y-1}{2}$, the y-axis ($x=0$), and the lines $y=-1$ and $y=3$.

Area $={\int }_{-1}^3|x|dy={\int }_{-1}^3\left|\frac{y-1}{2}\right|dy$

Since $\frac{y-1}{2}\le 0$ for $y\in [-1,1]$ and $\frac{y-1}{2}\ge 0$ for $y\in [1,3]$:

Area $={\int }_{-1}^1\frac{1-y}{2}dy+{\int }_1^3\frac{y-1}{2}dy$

Area $=\frac{1}{2}{\left[y-\frac{y^2}{2}\right]}_{-1}^1+\frac{1}{2}{\left[\frac{y^2}{2}-y\right]}_1^3$

Area $=\frac{1}{2}\left(\left(1-\frac{1}{2}\right)-\left(-1-\frac{1}{2}\right)\right)+\frac{1}{2}\left(\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)\right)$

Area $=\frac{1}{2}\left(\frac{1}{2}+\frac{3}{2}\right)+\frac{1}{2}\left(\frac{3}{2}+\frac{1}{2}\right)$

Area $=\frac{1}{2}(2)+\frac{1}{2}(2)=1+1=2$

Answer: $2$