let the region r be the area enclosed by the function f(x)=√x - 1 and g(x)=1/2x - 1. write an integral in…

Explanation:

Step1: Find intersection points

Set $\sqrt{x}-1=\frac{1}{2}x - 1$. Then $\sqrt{x}=\frac{1}{2}x$. Let $t = \sqrt{x}(t\geq0)$, so $t=\frac{1}{2}t^{2}$. Rearranging gives $\frac{1}{2}t^{2}-t = 0$, factoring out $t$ we get $t(\frac{1}{2}t - 1)=0$. So $t = 0$ or $t = 2$. When $t = 0$, $x=0$; when $t = 2$, $x = 4$. So the intersection - points are $x_1 = 0$ and $x_2=4$. For $x\in[0,4]$, $\sqrt{x}-1\geq\frac{1}{2}x - 1$. The area in terms of $x$ is $A_x=\int_{0}^{4}[(\sqrt{x}-1)-(\frac{1}{2}x - 1)]dx=\int_{0}^{4}(\sqrt{x}-\frac{1}{2}x)dx$.

Step2: Rewrite functions in terms of y

From $y=\sqrt{x}-1$, we can solve for $x$: $y + 1=\sqrt{x}$, so $x=(y + 1)^2$. From $y=\frac{1}{2}x-1$, we get $x = 2y + 2$. Find the $y$ - values of the intersection points. When $x = 0$, for $y=\sqrt{x}-1$, $y=-1$; for $y=\frac{1}{2}x - 1$, $y=-1$. When $x = 4$, for $y=\sqrt{x}-1$, $y = 1$; for $y=\frac{1}{2}x - 1$, $y = 1$. For $y\in[-1,1]$, $2y + 2\geq(y + 1)^2$. The area in terms of $y$ is $A_y=\int_{-1}^{1}[(2y + 2)-(y + 1)^2]dy$.

Answer:

$x_1 = 0$, $x_2 = 4$, $\int_{0}^{4}(\sqrt{x}-\frac{1}{2}x)dx$ $y_1=-1$, $y_2 = 1$, $\int_{-1}^{1}[(2y + 2)-(y + 1)^2]dy$