Sketch the graph of :
`f (x) = {{:(|x-2|+2"," x le 2),( " "x ^(2) -2 "," x gt 2):}.`
Evaluate `int _(0) ^(4)f (x) dx.` What does the value of this integral represent on the graph ?

Answer

To solve the problem, we need to follow these steps:

### Step 1: Define the function and its pieces
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} |x - 2| + 2 & \text{if } x \leq 2 \\ x^2 - 2 & \text{if } x > 2 \end{cases} \]

### Step 2: Rewrite the absolute value
For \( x \leq 2 \):
\[ |x - 2| = 2 - x \quad \text{(since \( x \) is less than or equal to 2)} \]
Thus,
\[ f(x) = (2 - x) + 2 = 4 - x \]

For \( x > 2 \):
\[ f(x) = x^2 - 2 \]

### Step 3: Sketch the graph
1. **For \( x \leq 2 \)**: The line \( f(x) = 4 - x \) intersects the y-axis at \( (0, 4) \) and the x-axis at \( (4, 0) \). The point at \( x = 2 \) is \( (2, 2) \).
2. **For \( x > 2 \)**: The parabola \( f(x) = x^2 - 2 \) starts at \( (2, 2) \) and opens upwards. At \( x = 4 \), \( f(4) = 16 - 2 = 14 \).

### Step 4: Identify the area to be calculated
We need to evaluate the integral:
\[ \int_0^4 f(x) \, dx \]
This can be split into two parts:
\[ \int_0^2 f(x) \, dx + \int_2^4 f(x) \, dx \]

### Step 5: Calculate the first integral
For \( x \in [0, 2] \):
\[ \int_0^2 (4 - x) \, dx \]
Calculating this:
\[ = \left[ 4x - \frac{x^2}{2} \right]_0^2 = \left[ 4(2) - \frac{(2)^2}{2} \right] - \left[ 4(0) - \frac{(0)^2}{2} \right] \]
\[ = (8 - 2) - 0 = 6 \]

### Step 6: Calculate the second integral
For \( x \in [2, 4] \):
\[ \int_2^4 (x^2 - 2) \, dx \]
Calculating this:
\[ = \left[ \frac{x^3}{3} - 2x \right]_2^4 \]
Calculating at the limits:
\[ = \left[ \frac{(4)^3}{3} - 2(4) \right] - \left[ \frac{(2)^3}{3} - 2(2) \right] \]
\[ = \left[ \frac{64}{3} - 8 \right] - \left[ \frac{8}{3} - 4 \right] \]
\[ = \left[ \frac{64}{3} - \frac{24}{3} \right] - \left[ \frac{8}{3} - \frac{12}{3} \right] \]
\[ = \left[ \frac{40}{3} \right] - \left[ -\frac{4}{3} \right] = \frac{40}{3} + \frac{4}{3} = \frac{44}{3} \]

### Step 7: Combine the results
Now, we combine both integrals:
\[ \int_0^4 f(x) \, dx = 6 + \frac{44}{3} = \frac{18}{3} + \frac{44}{3} = \frac{62}{3} \]

### Final Result
Thus, the value of the integral is:
\[ \int_0^4 f(x) \, dx = \frac{62}{3} \]

### Interpretation of the Integral
The value of this integral represents the area under the curve of \( f(x) \) from \( x = 0 \) to \( x = 4 \).