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The question asks for the area enclosed by the curve \( y = x^2 - x - 2 \) and the x-axis.

To find the area between the curve and the x-axis, we first need to determine the points where the curve intersects the x-axis. These points are found by setting \( y = 0 \) and solving for \( x \):

\[
x^2 - x - 2 = 0
\]

We can factor this quadratic equation as follows:

\[
(x - 2)(x + 1) = 0
\]

So, the roots are \( x = 2 \) and \( x = -1 \). These are the x-values where the curve intersects the x-axis.

The area \( A \) between the curve and the x-axis from \( x = -1 \) to \( x = 2 \) is given by the integral of the function between these two points:

\[
A = \int_{-1}^{2} (x^2 - x - 2) \, dx
\]

Let's calculate this integral.

\[
A = \int_{-1}^{2} (x^2 - x - 2) \, dx
\]

First, integrate each term:

\[
\int x^2 \, dx = \frac{x^3}{3}, \quad \int x \, dx = \frac{x^2}{2}, \quad \int 2 \, dx = 2x
\]

So, we have:

\[
A = \left[\frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-1}^{2}
\]

Now, substitute the limits:

\[
A = \left(\frac{2^3}{3} - \frac{2^2}{2} - 2(2)\right) - \left(\frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1)\right)
\]

Simplifying each term:

For \( x = 2 \):

\[
\frac{8}{3} - 2 - 4 = \frac{8}{3} - 6 = \frac{8 - 18}{3} = -\frac{10}{3}
\]

For \( x = -1 \):

\[
-\frac{1}{3} - \frac{1}{2} + 2 = -\frac{1}{3} - \frac{1}{2} + \frac{4}{2} = -\frac{1}{3} - \frac{1}{2} + 2
\]

\[
= -\frac{1}{3} - \frac{3}{6} + \frac{6}{3} = \frac{-2 - 3 + 12}{6} = \frac{7}{6}
\]

So:

\[
A = \left(-\frac{10}{3}\right) - \left(\frac{7}{6}\right) = -\frac{20}{6} - \frac{7}{6} = -\frac{27}{6} = -\frac{9}{2}
\]

Since area cannot be negative, we take the absolute value:

\[
A = \frac{9}{2}
\]

So the correct answer is 1) \( \frac{9}{2} \).

Would you like further details or have any questions? Here are some related questions:

1. How do you factor a quadratic equation like \( ax^2 + bx + c = 0 \)?
2. How do you determine the intersection points of a curve and the x-axis?
3. What is the significance of the absolute value when calculating areas under a curve?
4. How would you solve a definite integral with more complex functions?
5. What methods are used to approximate the area under a curve when an integral cannot be solved analytically?

**Tip:** When solving for areas between curves and axes, always consider the correct limits and check the sign of your final result to ensure the area is positive.

Find the area enclosed by the curve y = x^2 - x - 2 and the x-axis.

Learn how to calculate the area enclosed by the quadratic curve y = x^2 - x - 2 and the x-axis. The problem covers important concepts such as solving quadratic equations and using definite integrals to find areas under curves. Suitable for high school students.

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