Evaluate: `int_0^2 |x^2+2x-3| dx`

Answer

To evaluate the integral \( \int_0^2 |x^2 + 2x - 3| \, dx \), we first need to determine where the expression inside the absolute value is positive or negative.

### Step 1: Factor the quadratic expression
The expression \( x^2 + 2x - 3 \) can be factored. We look for two numbers that multiply to \(-3\) (the constant term) and add to \(2\) (the coefficient of \(x\)). The factors are \(3\) and \(-1\).

Thus, we can write:
\[ x^2 + 2x - 3 = (x - 1)(x + 3) \]

### Step 2: Find the roots
Setting the expression equal to zero gives us the roots:
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

### Step 3: Determine the sign of the expression in the interval [0, 2]
We need to evaluate the sign of \( |x^2 + 2x - 3| \) in the interval from \(0\) to \(2\).

- For \(x < -3\), the expression is positive.
- For \(-3 < x < 1\), the expression is negative.
- For \(x > 1\), the expression is positive.

In the interval \([0, 2]\):
- At \(x = 0\): \(0^2 + 2(0) - 3 = -3\) (negative)
- At \(x = 1\): \(1^2 + 2(1) - 3 = 0\) (zero)
- At \(x = 2\): \(2^2 + 2(2) - 3 = 5\) (positive)

Thus, we have:
- From \(0\) to \(1\), \(x^2 + 2x - 3 < 0\)
- From \(1\) to \(2\), \(x^2 + 2x - 3 > 0\)

### Step 4: Rewrite the integral
We can split the integral at the point where the expression changes sign:
\[ \int_0^2 |x^2 + 2x - 3| \, dx = \int_0^1 -(x^2 + 2x - 3) \, dx + \int_1^2 (x^2 + 2x - 3) \, dx \]

### Step 5: Evaluate the integrals
1. **First Integral**:
\[ \int_0^1 -(x^2 + 2x - 3) \, dx = \int_0^1 (-x^2 - 2x + 3) \, dx \]
Calculating this:
\[ = \left[-\frac{x^3}{3} - x^2 + 3x\right]_0^1 = \left[-\frac{1}{3} - 1 + 3\right] - [0] = -\frac{1}{3} - 1 + 3 = -\frac{1}{3} + 2 = \frac{5}{3} \]

2. **Second Integral**:
\[ \int_1^2 (x^2 + 2x - 3) \, dx \]
Calculating this:
\[ = \left[\frac{x^3}{3} + x^2 - 3x\right]_1^2 \]
Calculating at the limits:
\[ = \left[\frac{2^3}{3} + 2^2 - 3(2)\right] - \left[\frac{1^3}{3} + 1^2 - 3(1)\right] \]
\[ = \left[\frac{8}{3} + 4 - 6\right] - \left[\frac{1}{3} + 1 - 3\right] \]
\[ = \left[\frac{8}{3} - 2\right] - \left[\frac{1}{3} - 2\right] \]
\[ = \left[\frac{8}{3} - \frac{6}{3}\right] - \left[\frac{1}{3} - \frac{6}{3}\right] \]
\[ = \left[\frac{2}{3}\right] - \left[-\frac{5}{3}\right] = \frac{2}{3} + \frac{5}{3} = \frac{7}{3} \]

### Step 6: Combine the results
Now, we combine the results of the two integrals:
\[ \int_0^2 |x^2 + 2x - 3| \, dx = \frac{5}{3} + \frac{7}{3} = \frac{12}{3} = 4 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^2 |x^2 + 2x - 3| \, dx = 4 \]