sketch the region enclosed by the given curves. decide whether to integrate with respect to x or y. then…

Explanation:

Step1: Find intersection points

Set (2x^{2}=x^{2}+3). Subtract (x^{2}) from both sides: (2x^{2}-x^{2}=x^{2}+3 - x^{2}), so (x^{2}=3), then (x =-\sqrt{3}) and (x=\sqrt{3}). Since (x^{2}+3\geq2x^{2}) for (x\in[-\sqrt{3},\sqrt{3}]), and we integrate with respect to (x) (because the functions are given as (y) in terms of (x) and the limits are easier to find for (x)).

Step2: Set up the integral

The area (A=\int_{a}^{b}(upper - lower)dx). Here (a =-\sqrt{3}), (b=\sqrt{3}), (upper=x^{2}+3), (lower = 2x^{2}). So (A=\int_{-\sqrt{3}}^{\sqrt{3}}[(x^{2}+3)-2x^{2}]dx=\int_{-\sqrt{3}}^{\sqrt{3}}(3 - x^{2})dx).

Step3: Evaluate the integral

Use the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)). (\int(3 - x^{2})dx=3x-\frac{x^{3}}{3}+C). Then (A=\left[3x-\frac{x^{3}}{3}\right]_{-\sqrt{3}}^{\sqrt{3}}) [ \begin{align*} &=\left(3\sqrt{3}-\frac{(\sqrt{3})^{3}}{3}\right)-\left(3(-\sqrt{3})-\frac{(-\sqrt{3})^{3}}{3}\right)\ &=\left(3\sqrt{3}-\sqrt{3}\right)-\left(-3\sqrt{3}+\sqrt{3}\right)\ &=2\sqrt{3}-(-2\sqrt{3})\ &=4\sqrt{3} \end{align*} ]

Answer:

(4\sqrt{3})