evaluate the integral -2 to 2 if f(x) = 2 for all x between -2 and 0 and f(x) equals 4-x^2 for all x between 0 and 2

To evaluate the integral from \(-2\) to \(2\) for the piecewise function \( f(x) \), we'll break the integral into two parts based on the definition of \(f(x)\). The function is given as:

\[
f(x) = 
\begin{cases} 
2 & \text{for } -2 \leq x \leq 0 \\
4 - x^2 & \text{for } 0 \leq x \leq 2
\end{cases}
\]

The total integral can be written as the sum of two integrals:

\[
\int_{-2}^{2} f(x) \, dx = \int_{-2}^{0} 2 \, dx + \int_{0}^{2} (4 - x^2) \, dx
\]

### Step 1: Evaluate \( \int_{-2}^{0} 2 \, dx \)
Since \( f(x) = 2 \) over the interval \([-2, 0]\), this is a simple integral of a constant function:

\[
\int_{-2}^{0} 2 \, dx = 2[x]_{-2}^{0} = 2(0 - (-2)) = 2 \times 2 = 4
\]

### Step 2: Evaluate \( \int_{0}^{2} (4 - x^2) \, dx \)
Now, we'll compute the integral of \( 4 - x^2 \) over \([0, 2]\):

\[
\int_{0}^{2} (4 - x^2) \, dx = \int_{0}^{2} 4 \, dx - \int_{0}^{2} x^2 \, dx
\]

The first part:

\[
\int_{0}^{2} 4 \, dx = 4[x]_{0}^{2} = 4(2 - 0) = 8
\]

The second part:

\[
\int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}
\]

So:

\[
\int_{0}^{2} (4 - x^2) \, dx = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}
\]

### Step 3: Add the results
Now, sum the two integrals:

\[
\int_{-2}^{2} f(x) \, dx = 4 + \frac{16}{3} = \frac{12}{3} + \frac{16}{3} = \frac{28}{3}
\]

### Final Answer:
\[
\boxed{\frac{28}{3}}
\]

Let me know if you need further clarification or details!

Here are 5 related questions to deepen your understanding:
1. How would the answer change if the limits of integration were from \(-3\) to \(3\)?
2. Can you express the definite integral graphically using areas?
3. What happens if the function \( f(x) \) were defined as \( f(x) = x^2 \) for all \( x \)?
4. How do you handle piecewise functions in integrals if they have more than two pieces?
5. Can you apply this method if the upper bound is a variable, say \( x \), instead of 2?

**Tip:** When dealing with piecewise functions, always break the integral into parts corresponding to the intervals where the function is defined separately.

Learn how to evaluate the integral of the piecewise function f(x), where f(x) = 2 for -2 ≤ x ≤ 0 and f(x) = 4 - x^2 for 0 ≤ x ≤ 2. This problem involves integral calculus and requires computing integrals of constant and polynomial functions. Suitable for college-level mathematics students.

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