Questions: Find the area of the shaded region. (i) square units Enter a fraction, integer, or exact decimal. Do not approximate.

Solution Steps

To find the intersection points of the two curves, set $y = x - 2$ and $y = \sqrt{x}$ equal to each other: $x - 2 = \sqrt{x}$ Square both sides to get rid of the square root: $(x - 2)^2 = x$ $x^2 - 4x + 4 = x$ $x^2 - 5x + 4 = 0$ $(x - 1)(x - 4) = 0$ $x = 1$ or $x = 4$ When $x = 1$, $y = 1 - 2 = -1$, but since $y = \sqrt{x}$, $y$ cannot be negative. The graph visually confirms this, and the graphs do not intersect at x=1. When $x = 4$, $y = 4 - 2 = 2$ and $y = \sqrt{4} = 2$. Thus, the intersection point is $(4, 2)$.

The area of the shaded region can be found by integrating the difference of the two functions with respect to x from the intersection points. The lower limit of integration is where $y = \sqrt{x}$ intersects the x-axis which is at $x=0$. The upper limit is the $x$ value at the point of intersection ($x=4$). Area $= \int_{0}^{4} ( \sqrt{x} - (x - 2)) \,dx = \int_0^4 (\sqrt x - x + 2)\, dx$.

Area $= \int_0^4 (x^{1/2} - x + 2)\, dx = [\frac{2}{3}x^{3/2} - \frac{1}{2}x^2 + 2x]_0^4 $ $= (\frac{2}{3}(4^{3/2}) - \frac{1}{2}(4^2) + 2(4)) - 0$ $= \frac{2}{3}(8) - 8 + 8 = \frac{16}{3}$

Final Answer: The area of the shaded region is $\frac{16}{3}$ square units.