The area of the region bounded by the curves $y = x^2 + 2$, $y = x$, $x = 0$ and $x = 2$ is

The region is bounded by:

• Parabola: $y = x^2 + 2$
• Line: $y = x$
• Vertical boundaries: $x = 0$ and $x = 2$

To determine which curve is on top, test at $x = 1$:

For $y = x^2 + 2$: $y = (1)^2 + 2 = 3$

For $y = x$: $y = 1$

Since $3 > 1$, the parabola $y = x^2 + 2$ is above the line $y = x$ in this interval.

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The area between curves is given by:

$\text{Area} = \int_0^2 [(x^2 + 2) - x] \, dx$

$= \int_0^2 (x^2 - x + 2) \, dx$

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Integrating term by term:

$= \left[\frac{x^3}{3} - \frac{x^2}{2} + 2x\right]_0^2$

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At $x = 2$:

$\frac{(2)^3}{3} - \frac{(2)^2}{2} + 2(2)$

$= \frac{8}{3} - \frac{4}{2} + 4$

$= \frac{8}{3} - 2 + 4$

At $x = 0$:

$\frac{0}{3} - \frac{0}{2} + 0 = 0$

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$\text{Area} = \frac{8}{3} - 2 + 4$

$= \frac{8}{3} - \frac{6}{3} + \frac{12}{3}$

$= \frac{8 - 6 + 12}{3}$

$= \frac{14}{3}$

Therefore, the area of the region is $\frac{14}{3}$ square units.