1. **Problem Statement:** We need to find the area of the region enclosed by the given curves. Since the curves are not explicitly provided, let's assume the problem involves two functions $y=f(x)$ and $y=g(x)$ that intersect and enclose a region. 2. **Formula for Area Between Curves:** The area $A$ enclosed between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by: $$ A = \int_a^b |f(x) - g(x)| \, dx $$ If $f(x) \geq g(x)$ on $[a,b]$, then: $$ A = \int_a^b (f(x) - g(x)) \, dx $$ 3. **Steps to Find the Area:** - Find the points of intersection $a$ and $b$ by solving $f(x) = g(x)$. - Determine which function is on top (greater) between $a$ and $b$. - Set up the integral of the difference of the functions over $[a,b]$. - Evaluate the integral to find the area. 4. **Example:** Suppose the curves are $y = x^2$ and $y = 4 - x^2$. - Find intersection points: $$ x^2 = 4 - x^2 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \pm \sqrt{2} $$ - On $[-\sqrt{2}, \sqrt{2}]$, $4 - x^2 \geq x^2$. - Area: $$ A = \int_{-\sqrt{2}}^{\sqrt{2}} \left(4 - x^2 - x^2\right) dx = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) dx $$ - Evaluate the integral: $$ \int (4 - 2x^2) dx = 4x - \frac{2x^3}{3} $$ - Compute definite integral: $$ A = \left[4x - \frac{2x^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}} = \left(4\sqrt{2} - \frac{2(\sqrt{2})^3}{3}\right) - \left(-4\sqrt{2} + \frac{2(-\sqrt{2})^3}{3}\right) $$ - Simplify: $$ (4\sqrt{2} - \frac{2(2\sqrt{2})}{3}) - (-4\sqrt{2} + \frac{2(-2\sqrt{2})}{3}) = (4\sqrt{2} - \frac{4\sqrt{2}}{3}) + (4\sqrt{2} - \frac{-4\sqrt{2}}{3}) $$ $$ = \left(4\sqrt{2} - \frac{4\sqrt{2}}{3}\right) + \left(4\sqrt{2} + \frac{4\sqrt{2}}{3}\right) = 8\sqrt{2} $$ **Final answer:** $$ \boxed{8\sqrt{2}} $$ This is the area enclosed by the curves $y=x^2$ and $y=4-x^2$. If you provide the specific curves, I can compute the exact area for your problem.