find the area of the region described. the region bounded by y = √(2x), y = 2x - 2 and y = 0. the area of…

Explanation:

Step1: Find intersection points

Set $\sqrt{2x}=2x - 2$. Let $t=\sqrt{x}(t\geq0)$, then $2t^{2}-2t - 2=0$, or $t^{2}-t - 1=0$. Using the quadratic formula $t=\frac{1\pm\sqrt{1 + 4}}{2}=\frac{1\pm\sqrt{5}}{2}$. Since $t\geq0$, $t=\frac{1+\sqrt{5}}{2}$, and $x=\frac{3 + \sqrt{5}}{2}$. Also, set $2x-2 = 0$, we get $x = 1$, and for $y=\sqrt{2x}=0$, $x = 0$.

Step2: Set up integral for area

We split the integral based on the behavior of the functions. The area $A=\int_{0}^{1}\sqrt{2x}dx+\int_{1}^{\frac{3+\sqrt{5}}{2}}(\sqrt{2x}-(2x - 2))dx$.

Step3: Integrate term - by - term

For $\int_{0}^{1}\sqrt{2x}dx=\sqrt{2}\int_{0}^{1}x^{\frac{1}{2}}dx=\sqrt{2}\times\frac{2}{3}x^{\frac{3}{2}}\big|{0}^{1}=\frac{2\sqrt{2}}{3}$. For $\int{1}^{\frac{3+\sqrt{5}}{2}}(\sqrt{2x}-(2x - 2))dx=\sqrt{2}\int_{1}^{\frac{3+\sqrt{5}}{2}}x^{\frac{1}{2}}dx-\int_{1}^{\frac{3+\sqrt{5}}{2}}(2x - 2)dx$. $\sqrt{2}\int_{1}^{\frac{3+\sqrt{5}}{2}}x^{\frac{1}{2}}dx=\sqrt{2}\times\frac{2}{3}x^{\frac{3}{2}}\big|{1}^{\frac{3+\sqrt{5}}{2}}=\frac{2\sqrt{2}}{3}\left[\left(\frac{3+\sqrt{5}}{2}\right)^{\frac{3}{2}}-1\right]$. $\int{1}^{\frac{3+\sqrt{5}}{2}}(2x - 2)dx=(x^{2}-2x)\big|_{1}^{\frac{3+\sqrt{5}}{2}}=\left(\left(\frac{3+\sqrt{5}}{2}\right)^{2}-(3 + \sqrt{5})\right)-(1 - 2)$. After simplification, $A=\frac{5}{6}+\frac{\sqrt{5}}{3}$.

Answer:

$\frac{5}{6}+\frac{\sqrt{5}}{3}$