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?
 A: 
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:
$$
\begin{aligned}
 \int_{-1}^3\left[1+y-\frac{1}{2}\left(y^2-1\right)\right] d y 
= & {\left[\frac{(1+y)^2}{2}-\frac{y^3}{6}+\frac{y}{2}\right]_{-1}^3 } \\
= & 8-\frac{9}{2}+\frac{3}{2}-\frac{1}{6}+\frac{1}{2} \\
= & \frac{16}{3}
\end{aligned}
$$
A: $$\left[\int_{x=-1/2}^{x=0} 2\sqrt{2t+1} ~dt\right] ~+~
\left[\int_{x=0}^{x=4} \sqrt{2t+1} - (t-1) ~dt\right] \tag1 $$
$$= \left[\frac{2}{3} (2t+1)^{3/2} ~\bigg|_{t=-1/2}^{t=0}
 ~~\right]
+ 
\left[\frac{1}{3} (2t+1)^{3/2} - \frac{(t-1)^2}{2} ~\bigg|_{t=0}^{t=4} ~~\right]
$$
$$= \left[\frac{2}{3}(1 - 0)\right]
~+~ \left[ ~\left(9 - \frac{9}{2}\right) - \left(\frac{1}{3} - \frac{1}{2}\right) ~\right]$$
$$= \frac{2}{3} + \frac{9}{2} + \frac{1}{6} = \frac{32}{6} = \frac{16}{3}.$$
(1) above takes some explaining.
I integrated vertically [i.e. using the $x$ coordinate to determine the start/stop of the region(s)], rather than horizontally, letting $x$ vary from $(-1/2)$ to $(4)$ and computing the height at each value of $x$.
For $(-1/2) \leq x \leq 0$, the region is bounded above and below by $+\sqrt{2x+1}$ and $-\sqrt{2x+1}$, respectively.
For $0 \leq x \leq 4$, the upper bound continues to be $+\sqrt{2x+1},$ and the lower bound is $(x-1).$

My approach isn't really more elegant than your approach.  Integrating vertically seems more natural to me.  When integrating vertically, I don't see how the region can reasonably be expressed in only one integral.  That is, you have to accommodate that between $x=-1/2$ and $x=0$, the lower bound of the region is different than when $x$ is between $0$ and $4$.
In hindsight, I think that the answer of Lai is superior to mine.  That is, suppose that you integrate horizontally rather than vertically, using the $y$ coordinate to determine the start/stop of the region.
Then the rightmost boundary is the same throughout the region and the leftmost boundary is also the same throughout the region.
Therefore, the area can then be expressed with only one integral.
