Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 32

Evaluate $\displaystyle \int^4_0 |\sqrt{x + 2} - x| dx$ and interpret it as the area of a region. Sketch the region.

The integral can be interpreted as the area between the curves $y = \sqrt{x + 2}$ and $y = x$ in the region from $x = 0$ to $x = 4$








Notice that the orientation of the area differs at the point of intersection. To evaluate it, we can divide the area in to two sub region. $A_1$ be the area to the left of point of intersection. While $A_2$ be the area to the right of the point of intersection. So by using vertical strips


$\begin{equation} \begin{aligned} A_1 =& \int^2_0 [\sqrt{x + 2} - x] dx \\ \\ A_1 =& \left[ \frac{(x + 2)^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} - \frac{x^2}{2} \right] ^2_0 \\ \\ A_1 =& 1.4477 \text{ square units} \end{aligned} \end{equation}$


Then,


$\begin{equation} \begin{aligned} A_2 =& \int^4 _2 [x - \sqrt{x + 2}] dx \\ \\ A_2 =& \left[ \frac{x^2}{2} - \frac{(x + 2)^{\frac{3}{32}}}{\displaystyle \frac{3}{2}} \right] \\ \\ A_2 =& 1.5354 \text{ square units} \end{aligned} \end{equation}$


Therefore, the total area is $A_1 + A_2 = 2.9831$ square units