To evaluate the integral \( \int_{-4}^{3} |x^2 - 4| \, dx \), we will follow these steps:
### Step 1: Identify where the expression inside the absolute value is zero
We start by finding the points where \( x^2 - 4 = 0 \):
\[
x^2 = 4 \implies x = \pm 2
\]
This means the expression \( |x^2 - 4| \) changes at \( x = -2 \) and \( x = 2 \).
### Step 2: Break the integral into intervals
We will break the integral into three parts based on the points we found:
1. From \( -4 \) to \( -2 \)
2. From \( -2 \) to \( 2 \)
3. From \( 2 \) to \( 3 \)
Thus, we can write:
\[
\int_{-4}^{3} |x^2 - 4| \, dx = \int_{-4}^{-2} |x^2 - 4| \, dx + \int_{-2}^{2} |x^2 - 4| \, dx + \int_{2}^{3} |x^2 - 4| \, dx
\]
### Step 3: Determine the sign of \( x^2 - 4 \) in each interval
- For \( x \in [-4, -2] \): \( x^2 - 4 \) is positive, so \( |x^2 - 4| = x^2 - 4 \).
- For \( x \in [-2, 2] \): \( x^2 - 4 \) is negative, so \( |x^2 - 4| = -(x^2 - 4) = 4 - x^2 \).
- For \( x \in [2, 3] \): \( x^2 - 4 \) is positive, so \( |x^2 - 4| = x^2 - 4 \).
### Step 4: Rewrite the integral with the correct expressions
Now we can rewrite the integral:
\[
\int_{-4}^{3} |x^2 - 4| \, dx = \int_{-4}^{-2} (x^2 - 4) \, dx + \int_{-2}^{2} (4 - x^2) \, dx + \int_{2}^{3} (x^2 - 4) \, dx
\]
### Step 5: Evaluate each integral
1. **First Integral**:
\[
\int_{-4}^{-2} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{-4}^{-2}
\]
Evaluating at the limits:
\[
= \left( \frac{(-2)^3}{3} - 4(-2) \right) - \left( \frac{(-4)^3}{3} - 4(-4) \right)
= \left( -\frac{8}{3} + 8 \right) - \left( -\frac{64}{3} + 16 \right)
= \left( -\frac{8}{3} + \frac{24}{3} \right) - \left( -\frac{64}{3} + \frac{48}{3} \right)
= \frac{16}{3} + \frac{16}{3} = \frac{32}{3}
\]
2. **Second Integral**:
\[
\int_{-2}^{2} (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2}
\]
Evaluating at the limits:
\[
= \left( 4(2) - \frac{(2)^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right)
= \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right)
= \left( \frac{24}{3} - \frac{8}{3} \right) + \left( 8 - \frac{8}{3} \right)
= \frac{16}{3} + \frac{24}{3} = \frac{40}{3}
\]
3. **Third Integral**:
\[
\int_{2}^{3} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{2}^{3}
\]
Evaluating at the limits:
\[
= \left( \frac{(3)^3}{3} - 4(3) \right) - \left( \frac{(2)^3}{3} - 4(2) \right)
= \left( 9 - 12 \right) - \left( \frac{8}{3} - 8 \right)
= -3 + \left( 8 - \frac{8}{3} \right)
= -3 + \frac{24}{3} - \frac{8}{3} = -3 + \frac{16}{3} = -\frac{9}{3} + \frac{16}{3} = \frac{7}{3}
\]
### Step 6: Combine all parts
Now we combine all the results:
\[
\int_{-4}^{3} |x^2 - 4| \, dx = \frac{32}{3} + \frac{40}{3} + \frac{7}{3} = \frac{79}{3}
\]
### Final Answer
Thus, the final result is:
\[
\int_{-4}^{3} |x^2 - 4| \, dx = \frac{79}{3}
\]
