`int_(0)^(2) |x^(2)+2x-3|dx` is equal to

Options

A. 4
B. 6
C. 3
D. 2

Explanation

To solve the integral \( \int_{0}^{2} |x^{2} + 2x - 3| \, dx \), we will follow these steps:

### Step 1: Factor the quadratic expression
First, we need to factor the expression inside the absolute value:
\[ x^{2} + 2x - 3 = (x + 3)(x - 1) \]
This means the roots of the equation \( x^{2} + 2x - 3 = 0 \) are \( x = -3 \) and \( x = 1 \).

### Step 2: Determine the sign of the expression
Next, we need to determine where the expression \( x^{2} + 2x - 3 \) is positive or negative. We can do this by testing intervals based on the roots:
- For \( x < -3 \): Choose \( x = -4 \) → \( (-4 + 3)(-4 - 1) = (-1)(-5) > 0 \)
- For \( -3 < x < 1 \): Choose \( x = 0 \) → \( (0 + 3)(0 - 1) = (3)(-1) < 0 \)
- For \( x > 1 \): Choose \( x = 2 \) → \( (2 + 3)(2 - 1) = (5)(1) > 0 \)

Thus, we have:
- \( x^{2} + 2x - 3 > 0 \) for \( x < -3 \) and \( x > 1 \)
- \( x^{2} + 2x - 3 < 0 \) for \( -3 < x < 1 \)

### Step 3: Break the integral into intervals
Since we are integrating from \( 0 \) to \( 2 \), we note that the expression is negative between \( 0 \) and \( 1 \) and positive from \( 1 \) to \( 2 \). Therefore, we can break the integral as follows:
\[ \int_{0}^{2} |x^{2} + 2x - 3| \, dx = \int_{0}^{1} -(x^{2} + 2x - 3) \, dx + \int_{1}^{2} (x^{2} + 2x - 3) \, dx \]

### Step 4: Evaluate the first integral
Now we evaluate the first integral:
\[ \int_{0}^{1} -(x^{2} + 2x - 3) \, dx = \int_{0}^{1} (-x^{2} - 2x + 3) \, dx \]
Calculating this:
\[ = \left[-\frac{x^{3}}{3} - x^{2} + 3x\right]_{0}^{1} = \left[-\frac{1}{3} - 1 + 3\right] - [0] = -\frac{1}{3} - 1 + 3 = -\frac{1}{3} - \frac{3}{3} + \frac{9}{3} = \frac{5}{3} \]

### Step 5: Evaluate the second integral
Now we evaluate the second integral:
\[ \int_{1}^{2} (x^{2} + 2x - 3) \, dx \]
Calculating this:
\[ = \left[\frac{x^{3}}{3} + x^{2} - 3x\right]_{1}^{2} = \left[\frac{8}{3} + 4 - 6\right] - \left[\frac{1}{3} + 1 - 3\right] \]
Calculating the values:
\[ = \left[\frac{8}{3} + 4 - 6\right] = \left[\frac{8}{3} - \frac{6}{3} + \frac{12}{3}\right] = \frac{14}{3} \]
And for the lower limit:
\[ = \left[\frac{1}{3} + 1 - 3\right] = \left[\frac{1}{3} + \frac{3}{3} - \frac{9}{3}\right] = -\frac{5}{3} \]
Thus, the second integral becomes:
\[ \frac{14}{3} - (-\frac{5}{3}) = \frac{14}{3} + \frac{5}{3} = \frac{19}{3} \]

### Step 6: Combine the results
Now we combine the results of both integrals:
\[ \int_{0}^{2} |x^{2} + 2x - 3| \, dx = \frac{5}{3} + \frac{19}{3} = \frac{24}{3} = 8 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{2} |x^{2} + 2x - 3| \, dx = 8 \]