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

let the region r be the area enclosed by the function f(x) = √x + 2 and g(x) = 1/2x + 2. write an integral in terms of x and also an integral in terms of y that would represent the area of the region r. if necessary, round limit values to the nearest thousandth. answer attempt 1 out of 3 x1 = x2 = ∫x1^x2 dx y1 = y2 = ∫y1^y2 dy

let the region r be the area enclosed by the function f(x) = √x + 2 and g(x) = 1/2x + 2. write an integral in terms of x and also an integral in terms of y that would represent the area of the region r. if necessary, round limit values to the nearest thousandth. answer attempt 1 out of 3 x1 = x2 = ∫x1^x2 dx y1 = y2 = ∫y1^y2 dy

Answer

Explanation:

Step1: Find intersection - points

Set $\sqrt{x}+2=\frac{1}{2}x + 2$. Subtract 2 from both sides: $\sqrt{x}=\frac{1}{2}x$. Let $t = \sqrt{x}(t\geq0)$, then $t=\frac{1}{2}t^{2}$. Rearrange to $\frac{1}{2}t^{2}-t = 0$, factor out $t$: $t(\frac{1}{2}t - 1)=0$. So $t = 0$ or $t = 2$. When $t = 0,x = 0$; when $t = 2,x = 4$. So $x_1=0,x_2 = 4$.

Step2: Integral in terms of $x$

The upper - function is $y_1=\sqrt{x}+2$ and the lower - function is $y_2=\frac{1}{2}x + 2$. The area in terms of $x$ is $\int_{0}^{4}[(\sqrt{x}+2)-(\frac{1}{2}x + 2)]dx=\int_{0}^{4}(\sqrt{x}-\frac{1}{2}x)dx$.

Step3: Rewrite functions in terms of $y$

From $y=\sqrt{x}+2$, we get $x=(y - 2)^2$ ($y\geq2$). From $y=\frac{1}{2}x + 2$, we get $x = 2y-4$.

Step4: Find $y$ - limits

The intersection points in terms of $y$: when $x = 0,y = 2$; when $x = 4,y = 4$. So $y_1=2,y_2 = 4$.

Step5: Integral in terms of $y$

The right - function is $x_1=2y - 4$ and the left - function is $x_2=(y - 2)^2$. The area in terms of $y$ is $\int_{2}^{4}[(2y-4)-(y - 2)^2]dy$.

Answer:

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