The area ehclosed by the curves y = f(x) and ; y =g(x), where f(x) = max ${x , x^2}$ and g(x) = min ${x, x^2}$ opver the interval [0,1] is& ;

First, find where $f(x)$ and $g(x)$ intersect, i.e., where $x=x^{2}$ :

$x=x^{2} \Longrightarrow x^{2}-x=0 \Longrightarrow x(x-1)=0 \Longrightarrow x=0 \text { or } x=1$

So, the curves intersect at $x=0$ and $x=1$.

Step 2: Identify the Upper and Lower Functions

For $x$ in the interval $[0,1]$ :

- $f(x)=\max \left(x, x^{2}\right)$

- $g(x)=\min \left(x, x^{2}\right)$

Step 3: Compare $x$ and $x^{2}$ on $[0,1]$

- For $0 \leq x \leq 1, x^{2} \leq x$ because $x^{2}$ grows slower than $x$ in this interval.

- Therefore, $\max \left(x, x^{2}\right)=x$ and $\min \left(x, x^{2}\right)=x^{2}$.

Step 4: Set Up the Integral for the Area

The area $A$ between the curves from $x=0$ to $x=1$ is given by:

$A=\int_{0}^{1}(f(x)-g(x)) d x=\int_{0}^{1}\left(x-x^{2}\right) d x$

Step 5: Compute the Integral

Evaluate the integral:

$A=\int_{0}^{1}\left(x-x^{2}\right) d x=\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=\left(\frac{1}{2}-\frac{1}{3}\right)-(0-0)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

Final Answer

The area enclosed by the curves $y=f(x)$ and $y=g(x)$ over the interval $[0,1]$ is:

$\begin{array}{|l|} \hline \frac{1}{6} \\ \hline \end{array}$