Question:

The region is bounded by the curves $y = x^2 + 2$ and $y = x$, and the vertical lines $x = 0$ and $x = 3$. First, determine which curve is above the other in the interval $[0, 3]$. Consider the difference between the two functions: $$(x^2 + 2) - x = x^2 - x + 2$$ The discriminant of the quadratic $x^2 - x + 2$ is $\Delta = (-1)^2 - 4(1)(2) = 1 - 8 = -7$. Since the discriminant is negative and the leading coefficient (1) is positive, the quadratic $x^2 - x + 2$ is always positive for all real $x$. Therefore, $x^2 + 2 > x$ for all $x$, which means the curve $y = x^2 + 2$ is always above the line $y = x$ in the interval $[0, 3]$. The area $A$ of the region is given by the definite integral: $$A = \int_0^3 ((x^2 + 2) - x) \, dx$$ $$A = \int_0^3 (x^2 - x + 2) \, dx$$ Now, evaluate the integral: $$A = \left[ \frac{x^3}{3} - \frac{x^2}{2} + 2x \right]_0^3$$ Substitute the limits of integration: $$A = \left( \frac{3^3}{3} - \frac{3^2}{2} + 2(3) \right) - \left( \frac{0^3}{3} - \frac{0^2}{2} + 2(0) \right)$$ $$A = \left( \frac{27}{3} - \frac{9}{2} + 6 \right) - (0)$$ $$A = \left( 9 - \frac{9}{2} + 6 \right)$$ $$A = \left( 15 - \frac{9}{2} \right)$$ $$A = \left( \frac{30}{2} - \frac{9}{2} \right)$$ $$A = \frac{21}{2}$$