current attempt in progress\nfind the area between y = 2x and y = 2√x.\nround your answer to four decimal…

Explanation:

Step1: Find intersection points

Set $2x = 2\sqrt{x}$. Let $t=\sqrt{x}(t\geq0)$, then $2t^{2}-2t = 0$, $2t(t - 1)=0$. So $t = 0$ or $t = 1$. When $t = 0,x = 0$; when $t = 1,x = 1$.

Step2: Set up integral for area

The area $A=\int_{0}^{1}(2\sqrt{x}-2x)dx$.

Step3: Integrate term - by - term

$\int(2\sqrt{x}-2x)dx=2\int x^{\frac{1}{2}}dx-2\int xdx$. Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $2\times\frac{2}{3}x^{\frac{3}{2}}-2\times\frac{1}{2}x^{2}=\frac{4}{3}x^{\frac{3}{2}}-x^{2}+C$.

Step4: Evaluate definite integral

$A=\left[\frac{4}{3}x^{\frac{3}{2}}-x^{2}\right]_{0}^{1}=\frac{4}{3}\times1^{\frac{3}{2}}-1^{2}-\left(\frac{4}{3}\times0^{\frac{3}{2}}-0^{2}\right)=\frac{4}{3}-1=\frac{1}{3}\approx0.3333$.

Answer:

$0.3333$