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The given integral is:

\[
\int_{-2}^{1} |x| \, dx
\]

To solve this, we need to handle the absolute value function. The absolute value function, \(|x|\), behaves differently depending on whether \(x\) is positive or negative. Specifically:

\[
|x| = 
\begin{cases} 
-x & \text{if } x < 0 \\
x  & \text{if } x \geq 0
\end{cases}
\]

So, we will break the integral into two parts:

1. From \(-2\) to \(0\), where \(|x| = -x\).
2. From \(0\) to \(1\), where \(|x| = x\).

The integral becomes:

\[
\int_{-2}^{1} |x| \, dx = \int_{-2}^{0} -x \, dx + \int_{0}^{1} x \, dx
\]

Now, let's compute these integrals separately.

1. \(\int_{-2}^{0} -x \, dx\)
   \[
   = \left[ -\frac{x^2}{2} \right]_{-2}^{0}
   = -\frac{(0)^2}{2} + \frac{(-2)^2}{2} = 0 + \frac{4}{2} = 2
   \]

2. \(\int_{0}^{1} x \, dx\)
   \[
   = \left[ \frac{x^2}{2} \right]_{0}^{1}
   = \frac{(1)^2}{2} - \frac{(0)^2}{2} = \frac{1}{2} - 0 = \frac{1}{2}
   \]

Now, adding these results together:

\[
2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}
\]

So, the value of the integral is:

\[
\int_{-2}^{1} |x| \, dx = \frac{5}{2}
\]

Would you like any further details or clarification?

Here are some related questions you could explore:
1. How does the absolute value function affect integration?
2. How do we handle improper integrals involving absolute values?
3. What is the geometric interpretation of this integral?
4. How do you compute integrals involving piecewise functions?
5. What happens if the limits of integration include singular points?

**Tip:** When integrating functions with absolute values, breaking the integral into intervals where the expression is continuous is key!

Evaluate the integral ∫ from -2 to 1 of |x| dx

Learn how to evaluate the definite integral ∫ from -2 to 1 of |x| dx by breaking it into two parts and applying the Fundamental Theorem of Calculus. This problem involves key calculus concepts, including handling absolute value functions and piecewise integrals. Ideal for Grades 11-12 or early college students.

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