The value of the integral ` int _(0)^(2) | x^(2)-1| dx` is

Options

A. 0
B. 2
C. `-(1)/(3)`
D. `-2`

Explanation

To solve the integral \( \int_{0}^{2} |x^2 - 1| \, dx \), we need to analyze the expression inside the absolute value, \( |x^2 - 1| \).

### Step 1: Determine where \( x^2 - 1 \) changes sign
The expression \( x^2 - 1 \) is zero when \( x^2 = 1 \), which gives us \( x = 1 \) (since we are considering the interval from 0 to 2).

- For \( x < 1 \) (i.e., in the interval [0, 1]), \( x^2 - 1 < 0 \).
- For \( x > 1 \) (i.e., in the interval [1, 2]), \( x^2 - 1 > 0 \).

### Step 2: Rewrite the integral using the properties of absolute value
We can split the integral into two parts based on the point where the expression changes sign:

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

### Step 3: Simplify the integrals
Now we can rewrite the integrals:

1. For the first integral:
\[ \int_{0}^{1} -(x^2 - 1) \, dx = \int_{0}^{1} (1 - x^2) \, dx \]

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

### Step 4: Calculate the first integral
Now we calculate \( \int_{0}^{1} (1 - x^2) \, dx \):

\[ \int (1 - x^2) \, dx = x - \frac{x^3}{3} + C \]

Evaluating from 0 to 1:

\[ \left[ x - \frac{x^3}{3} \right]_{0}^{1} = \left( 1 - \frac{1^3}{3} \right) - \left( 0 - 0 \right) = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 5: Calculate the second integral
Now we calculate \( \int_{1}^{2} (x^2 - 1) \, dx \):

\[ \int (x^2 - 1) \, dx = \frac{x^3}{3} - x + C \]

Evaluating from 1 to 2:

\[ \left[ \frac{x^3}{3} - x \right]_{1}^{2} = \left( \frac{2^3}{3} - 2 \right) - \left( \frac{1^3}{3} - 1 \right) = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 1 \right) \]

Calculating this gives:

\[ \frac{8}{3} - \frac{6}{3} = \frac{2}{3} \quad \text{and} \quad \frac{1}{3} - 1 = -\frac{2}{3} \]

Thus,

\[ \left( \frac{2}{3} \right) - \left( -\frac{2}{3} \right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

### Step 6: Combine the results
Now we combine both integrals:

\[ \int_{0}^{2} |x^2 - 1| \, dx = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2 \]

### Final Answer
The value of the integral \( \int_{0}^{2} |x^2 - 1| \, dx \) is \( \boxed{2} \).