Evaluate : `int_(-6)^(3)|x+3|dx`

Answer

To evaluate the integral \(\int_{-6}^{3} |x + 3| \, dx\), we need to consider the modulus function, which requires us to break the integral into parts based on the point where the expression inside the modulus changes sign.

### Step-by-Step Solution:

1. **Identify the point where the expression inside the modulus is zero:**
\[ x + 3 = 0 \implies x = -3 \]
This means we will split the integral at \(x = -3\).

2. **Set up the integral with the appropriate limits:**
We can express the integral as:
\[ \int_{-6}^{3} |x + 3| \, dx = \int_{-6}^{-3} |x + 3| \, dx + \int_{-3}^{3} |x + 3| \, dx \]

3. **Determine the sign of \(x + 3\) in each interval:**
- For \(x < -3\) (i.e., from \(-6\) to \(-3\)), \(x + 3 < 0\), so \(|x + 3| = -(x + 3)\).
- For \(x \geq -3\) (i.e., from \(-3\) to \(3\)), \(x + 3 \geq 0\), so \(|x + 3| = x + 3\).

4. **Rewrite the integral based on the sign of the expression:**
\[ \int_{-6}^{3} |x + 3| \, dx = \int_{-6}^{-3} -(x + 3) \, dx + \int_{-3}^{3} (x + 3) \, dx \]

5. **Evaluate the first integral:**
\[ \int_{-6}^{-3} -(x + 3) \, dx = -\int_{-6}^{-3} (x + 3) \, dx \]
Now, compute:
\[ = -\left[ \frac{x^2}{2} + 3x \right]_{-6}^{-3} \]
Calculate the limits:
\[ = -\left[ \left( \frac{(-3)^2}{2} + 3(-3) \right) - \left( \frac{(-6)^2}{2} + 3(-6) \right) \right] \]
\[ = -\left[ \left( \frac{9}{2} - 9 \right) - \left( 18 - 18 \right) \right] \]
\[ = -\left[ \frac{9}{2} - 9 \right] = -\left[ \frac{9}{2} - \frac{18}{2} \right] = -\left[ -\frac{9}{2} \right] = \frac{9}{2} \]

6. **Evaluate the second integral:**
\[ \int_{-3}^{3} (x + 3) \, dx = \left[ \frac{x^2}{2} + 3x \right]_{-3}^{3} \]
Calculate the limits:
\[ = \left[ \left( \frac{3^2}{2} + 3(3) \right) - \left( \frac{(-3)^2}{2} + 3(-3) \right) \right] \]
\[ = \left[ \left( \frac{9}{2} + 9 \right) - \left( \frac{9}{2} - 9 \right) \right] \]
\[ = \left[ \frac{9}{2} + \frac{18}{2} - \left( \frac{9}{2} - \frac{18}{2} \right) \right] \]
\[ = \left[ \frac{27}{2} - \frac{-9}{2} \right] = \frac{27}{2} + \frac{9}{2} = \frac{36}{2} = 18 \]

7. **Combine the results:**
\[ \int_{-6}^{3} |x + 3| \, dx = \frac{9}{2} + 18 = \frac{9}{2} + \frac{36}{2} = \frac{45}{2} \]

### Final Answer:
\[ \int_{-6}^{3} |x + 3| \, dx = \frac{45}{2} \]