in problems 33 - 46, find the area of the of the given equations. 33. y = 4 - x², y = x²

Explanation:

Step1: Find intersection points

Set $4 - x^{2}=x^{2}$, then $2x^{2}=4$, $x^{2} = 2$, so $x=\pm\sqrt{2}$.

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 $g(x)=x^{2}$, $a =-\sqrt{2}$, $b=\sqrt{2}$. So $A=\int_{-\sqrt{2}}^{\sqrt{2}}[(4 - x^{2})-x^{2}]dx=\int_{-\sqrt{2}}^{\sqrt{2}}(4 - 2x^{2})dx$.

Step3: Evaluate the integral

We know that $\int(4 - 2x^{2})dx=4x-\frac{2}{3}x^{3}+C$. Then $\left[4x-\frac{2}{3}x^{3}\right]_{-\sqrt{2}}^{\sqrt{2}}=(4\sqrt{2}-\frac{2}{3}(\sqrt{2})^{3})-( - 4\sqrt{2}-\frac{2}{3}(-\sqrt{2})^{3})=(4\sqrt{2}-\frac{4\sqrt{2}}{3})-(-4\sqrt{2}+\frac{4\sqrt{2}}{3})=\frac{8\sqrt{2}}{3}+\frac{8\sqrt{2}}{3}=\frac{16\sqrt{2}}{3}$.

Answer:

$\frac{16\sqrt{2}}{3}$