question\nfind the area bounded by y = 2 - x, y = √x and the x - axis.\nsubmit your answer in fractional…

Explanation:

Step1: Find intersection point

Set $2 - x=\sqrt{x}$. Let $t = \sqrt{x}(t\geq0)$, then $2 - t^{2}=t$, or $t^{2}+t - 2=0$. Factoring gives $(t + 2)(t - 1)=0$. Since $t\geq0$, $t = 1$, so $x = 1$.

Step2: Set up integral for area

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

Step3: Integrate $\int_{0}^{1}\sqrt{x}dx$

Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, for $y=\sqrt{x}=x^{\frac{1}{2}}$, $\int_{0}^{1}x^{\frac{1}{2}}dx=\left[\frac{2}{3}x^{\frac{3}{2}}\right]_{0}^{1}=\frac{2}{3}(1)^{\frac{3}{2}}-\frac{2}{3}(0)^{\frac{3}{2}}=\frac{2}{3}$.

Step4: Integrate $\int_{1}^{2}(2 - x)dx$

$\int_{1}^{2}(2 - x)dx=\int_{1}^{2}2dx-\int_{1}^{2}xdx$. $\int_{1}^{2}2dx=2x\big|{1}^{2}=2(2)-2(1)=2$, $\int{1}^{2}xdx=\left[\frac{1}{2}x^{2}\right]{1}^{2}=\frac{1}{2}(2)^{2}-\frac{1}{2}(1)^{2}=2-\frac{1}{2}=\frac{3}{2}$. So $\int{1}^{2}(2 - x)dx=2-\frac{3}{2}=\frac{1}{2}$.

Step5: Calculate total area

$A=\frac{2}{3}+\frac{1}{2}=\frac{4 + 3}{6}=\frac{7}{6}$.

Answer:

$\frac{7}{6}$