Evaluate : `int _(-2)^(3) |1 - x^(2) | dx `

Answer

To evaluate the integral \( \int_{-2}^{3} |1 - x^2| \, dx \), we will first determine where the expression inside the absolute value, \( 1 - x^2 \), is positive or negative. This will help us to rewrite the integral without the absolute value.

### Step 1: Find the points where \( 1 - x^2 = 0 \)

Set the expression inside the absolute value to zero:
\[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 \]
Thus, the critical points are \( x = -1 \) and \( x = 1 \).

### Step 2: Determine the sign of \( 1 - x^2 \) in the intervals

We will check the sign of \( 1 - x^2 \) in the intervals \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \):

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ 1 - (-2)^2 = 1 - 4 = -3 \quad (\text{negative}) \]

- For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\[ 1 - 0^2 = 1 - 0 = 1 \quad (\text{positive}) \]

- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ 1 - 2^2 = 1 - 4 = -3 \quad (\text{negative}) \]

### Step 3: Rewrite the integral based on the sign of \( 1 - x^2 \)

Now we can rewrite the integral as follows:
\[ \int_{-2}^{3} |1 - x^2| \, dx = \int_{-2}^{-1} -(1 - x^2) \, dx + \int_{-1}^{1} (1 - x^2) \, dx + \int_{1}^{3} -(1 - x^2) \, dx \]

### Step 4: Simplify the integrals

Now we can simplify each integral:
1. For \( \int_{-2}^{-1} -(1 - x^2) \, dx \):
\[ = \int_{-2}^{-1} (x^2 - 1) \, dx \]

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

3. For \( \int_{1}^{3} -(1 - x^2) \, dx \):
\[ = \int_{1}^{3} (x^2 - 1) \, dx \]

### Step 5: Evaluate each integral

1. Evaluate \( \int_{-2}^{-1} (x^2 - 1) \, dx \):
\[ = \left[ \frac{x^3}{3} - x \right]_{-2}^{-1} = \left( \frac{(-1)^3}{3} - (-1) \right) - \left( \frac{(-2)^3}{3} - (-2) \right) \]
\[ = \left( -\frac{1}{3} + 1 \right) - \left( -\frac{8}{3} + 2 \right) = \left( \frac{2}{3} \right) - \left( -\frac{8}{3} + \frac{6}{3} \right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

2. Evaluate \( \int_{-1}^{1} (1 - x^2) \, dx \):
\[ = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = \left( 1 - \frac{1}{3} \right) - \left( -1 + \frac{1}{3} \right) = \left( \frac{2}{3} \right) - \left( -\frac{2}{3} \right) = \frac{4}{3} \]

3. Evaluate \( \int_{1}^{3} (x^2 - 1) \, dx \):
\[ = \left[ \frac{x^3}{3} - x \right]_{1}^{3} = \left( \frac{27}{3} - 3 \right) - \left( \frac{1}{3} - 1 \right) = (9 - 3) - \left( \frac{1}{3} - 1 \right) = 6 - \left( -\frac{2}{3} \right) = 6 + \frac{2}{3} = \frac{18}{3} + \frac{2}{3} = \frac{20}{3} \]

### Step 6: Combine the results

Now combine all the results:
\[ \int_{-2}^{3} |1 - x^2| \, dx = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-2}^{3} |1 - x^2| \, dx = \frac{28}{3} \]