(b) $y = x(1 - x);y = 0;-1leq xleq1$

Explanation:

Step1: Expand the function

$y = x(1 - x)=x - x^{2}$

Step2: Use the definite - integral formula for area

The area $A$ between the curve $y = f(x)$ and the $x$ - axis from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)|dx$. First, find where $y=x - x^{2}=x(1 - x)$ is non - negative and non - positive on the interval $[-1,1]$. Set $y = 0$, we get $x = 0$ and $x = 1$. We can split the integral based on the sign of the function. On the interval $[-1,0]$, $y=x - x^{2}\leq0$, and on the interval $[0,1]$, $y=x - x^{2}\geq0$. $A=\int_{-1}^{0}-(x - x^{2})dx+\int_{0}^{1}(x - x^{2})dx$

Step3: Integrate term - by - term

For $\int-(x - x^{2})dx=-\left(\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)+C=-\frac{x^{2}}{2}+\frac{x^{3}}{3}+C$. Evaluating $\int_{-1}^{0}-\left(x - x^{2}\right)dx=\left[-\frac{x^{2}}{2}+\frac{x^{3}}{3}\right]{-1}^{0}=0-\left(-\frac{(-1)^{2}}{2}+\frac{(-1)^{3}}{3}\right)=\frac{1}{2}+\frac{1}{3}=\frac{3 + 2}{6}=\frac{5}{6}$. For $\int(x - x^{2})dx=\frac{x^{2}}{2}-\frac{x^{3}}{3}+C$. Evaluating $\int{0}^{1}(x - x^{2})dx=\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{2}-\frac{1}{3}=\frac{3 - 2}{6}=\frac{1}{6}$.

Step4: Sum the two integral results

$A=\frac{5}{6}+\frac{1}{6}=1$

Answer:

$1$