1. **State the problem:** Find the area of the region bounded by the curves $y = x^2 + 3$, $y = x$, and the vertical lines $x = -1$ and $x = 1$. 2. **Identify the curves and limits:** The region is bounded between $x = -1$ and $x = 1$. The upper curve is $y = x^2 + 3$ and the lower curve is $y = x$. 3. **Formula for area between curves:** The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is given by: $$A = \int_a^b |f(x) - g(x)| \, dx$$ Since $x^2 + 3 > x$ for all $x$ in $[-1,1]$, we have: $$A = \int_{-1}^1 \left( (x^2 + 3) - x \right) dx = \int_{-1}^1 (x^2 - x + 3) \, dx$$ 4. **Calculate the integral:** $$\int_{-1}^1 (x^2 - x + 3) \, dx = \int_{-1}^1 x^2 \, dx - \int_{-1}^1 x \, dx + \int_{-1}^1 3 \, dx$$ Calculate each separately: - $$\int_{-1}^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^1 = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}$$ - $$\int_{-1}^1 x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^1 = \frac{1^2}{2} - \frac{(-1)^2}{2} = \frac{1}{2} - \frac{1}{2} = 0$$ - $$\int_{-1}^1 3 \, dx = 3x \Big|_{-1}^1 = 3(1) - 3(-1) = 3 + 3 = 6$$ 5. **Sum the results:** $$A = \frac{2}{3} - 0 + 6 = \frac{2}{3} + 6 = \frac{2}{3} + \frac{18}{3} = \frac{20}{3}$$ 6. **Final answer:** The area of the region bounded by the curves is: $$\boxed{\frac{20}{3}}$$