find the area of each of the region bounded by the following curves. write your solutions on the space…

Explanation:

Step1: Find intersection points

Set $4 - x^{2}=x$. Rearrange to $x^{2}+x - 4=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=1$, $c=-4$, we have $x=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-4)}}{2\times1}=\frac{-1\pm\sqrt{17}}{2}$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, $f(x)=4 - x^{2}$ and $g(x)=x$, and $a=\frac{-1-\sqrt{17}}{2}$, $b=\frac{-1 + \sqrt{17}}{2}$. So $A=\int_{\frac{-1-\sqrt{17}}{2}}^{\frac{-1+\sqrt{17}}{2}}((4 - x^{2})-x)dx$.

Step3: Integrate

$\int((4 - x^{2})-x)dx=4x-\frac{1}{3}x^{3}-\frac{1}{2}x^{2}+C$.

Step4: Evaluate definite - integral

$A=\left[4x-\frac{1}{3}x^{3}-\frac{1}{2}x^{2}\right]_{\frac{-1-\sqrt{17}}{2}}^{\frac{-1+\sqrt{17}}{2}}$ [ \begin{align*} &=\left(4\times\frac{-1 + \sqrt{17}}{2}-\frac{1}{3}\left(\frac{-1+\sqrt{17}}{2}\right)^{3}-\frac{1}{2}\left(\frac{-1+\sqrt{17}}{2}\right)^{2}\right)-\left(4\times\frac{-1-\sqrt{17}}{2}-\frac{1}{3}\left(\frac{-1-\sqrt{17}}{2}\right)^{3}-\frac{1}{2}\left(\frac{-1-\sqrt{17}}{2}\right)^{2}\right)\ &=\frac{17\sqrt{17}}{6} \end{align*} ]

Answer:

$\frac{17\sqrt{17}}{6}$