Evaluate: `int_-2^2 f(x) dx` where
`f(x)={:{(2x-1,-2 le x le1),(3x-2,1lex le2):}`

Answer

To evaluate the integral \( \int_{-2}^{2} f(x) \, dx \) where

\[ f(x) = \begin{cases} 2x - 1 & \text{for } -2 \leq x \leq 1 \\ 3x - 2 & \text{for } 1 < x \leq 2 \end{cases} \]

we will split the integral into two parts based on the definition of \( f(x) \).

### Step 1: Split the Integral
We can break the integral into two parts:
\[ \int_{-2}^{2} f(x) \, dx = \int_{-2}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx \]

### Step 2: Evaluate the First Integral
For the first integral, where \( f(x) = 2x - 1 \):
\[ \int_{-2}^{1} (2x - 1) \, dx \]

Calculating this integral:
\[ = \left[ x^2 - x \right]_{-2}^{1} \]

Now, we evaluate the limits:
\[ = \left(1^2 - 1\right) - \left((-2)^2 - (-2)\right) \]
\[ = (1 - 1) - (4 + 2) = 0 - 6 = -6 \]

### Step 3: Evaluate the Second Integral
For the second integral, where \( f(x) = 3x - 2 \):
\[ \int_{1}^{2} (3x - 2) \, dx \]

Calculating this integral:
\[ = \left[ \frac{3}{2}x^2 - 2x \right]_{1}^{2} \]

Now, we evaluate the limits:
\[ = \left( \frac{3}{2}(2^2) - 2(2) \right) - \left( \frac{3}{2}(1^2) - 2(1) \right) \]
\[ = \left( \frac{3}{2}(4) - 4 \right) - \left( \frac{3}{2}(1) - 2 \right) \]
\[ = \left( 6 - 4 \right) - \left( \frac{3}{2} - 2 \right) \]
\[ = 2 - \left( \frac{3}{2} - \frac{4}{2} \right) = 2 - \left( -\frac{1}{2} \right) = 2 + \frac{1}{2} = \frac{5}{2} \]

### Step 4: Combine the Results
Now, we combine the results of both integrals:
\[ \int_{-2}^{2} f(x) \, dx = -6 + \frac{5}{2} \]
To combine these, we convert -6 into a fraction:
\[ -6 = -\frac{12}{2} \]
Thus,
\[ \int_{-2}^{2} f(x) \, dx = -\frac{12}{2} + \frac{5}{2} = -\frac{7}{2} \]

### Final Answer
The value of the integral is:
\[ \int_{-2}^{2} f(x) \, dx = -\frac{7}{2} \]