Area between two equations I want to calculate the area between $y^2=2x+1$ and $x-y-1=0$ . I did this:
$$x_1 = \frac {y^2-1}{2},   x_2 = y+1$$
And the intersection points are, $y=-1$ and $y=3$ . So the area is:
$$\int_{-1}^3|x_1 - x_2| dy$$
$$\implies\int_{-1}^3\left|\frac{y^2-1}{2} - y-1\right| dy = \frac{16}{3}$$
And this is the correct answer. I tried to calculate this, using $dx$. For the area above the x-axis:
$$\int_{-\frac{1}{2}}^4 \sqrt{2x+1}\text{ }dx-\int_1^4(x-1)\text{ }dx = \frac {9}{2}$$
And for the area below the x-axis:
$$\int_{-\frac{1}{2}}^4\left|-\sqrt{2x+1}\right|\text{ }dx-\int_0^4\left|x-1+\sqrt{2x+1}\right|\text{ }dx + \int_1^4(x-1)\text{ }dx=\frac{5}{6}$$
And if we add these numbers we get to $\frac {16}{3}$ . I want to know is there a better way to calculate this area using $dx$? One integration for area above the x-axis and one for another, maybe?


Vertical Strips
$$
\begin{aligned}
\textrm{  Area between the curve }& =\int_{-\frac{1}{2}}^0  [\sqrt{2 x+1}-(-\sqrt{2 x+1})] d x+\int_0^4[\sqrt{2 x+1}-(x-1)] d x \\
& =\left[\frac{2}{3}(2 x+1)^{\frac{3}{2}}\right]_{-\frac{1}{2}}^0+\left[\frac{1}{3}(2 x+1)^{\frac{3}{2}}-\frac{(x-1)^2}{2}\right]_0^4 \\
& =\frac{2}{3}+9-\frac{9}{2}-\frac{1}{3}+\frac{1}{2} \\
& =\frac{16}{3}
\end{aligned}
$$
Horizontal Strips
Since the straight line is always on the right of the parabola for $x\in [-1,4]$, therefore we can find conveniently(without absolute sign) the enclosed area by only one integral as below:
$$
\begin{aligned}
 \int_{-1}^3\left[1+y-\frac{1}{2}\left(y^2-1\right)\right] d y 
= & {\left[\frac{(1+y)^2}{2}-\frac{y^3}{6}+\frac{y}{2}\right]_{-1}^3 } \\
= & 8-\frac{9}{2}+\frac{3}{2}-\frac{1}{6}+\frac{1}{2} \\
= & \frac{16}{3}
\end{aligned}
$$
