Evaluate the integral from -2 to 2 of (x^3 cos(x/2) + 1/2)√(4 - x^2) dx. The Wi-Fi password is the first 10 digits of the answer.

Steps to Solve

1. Identify the integral and the function The integral to evaluate is

$$ I = \int_{-2}^{2} \left( x^3 \cos\left(\frac{x}{2}\right) + \frac{1}{2} \right) \sqrt{4 - x^2} , dx $$

2. Analyze the symmetry of the function Recognize that the integrand consists of two parts: $x^3 \cos\left(\frac{x}{2}\right)$ and $\frac{1}{2}$.

• The term $x^3 \cos\left(\frac{x}{2}\right)$ is an odd function because $x^3$ is odd and $\cos\left(\frac{x}{2}\right)$ is even.
• The term $\frac{1}{2}$ is a constant (even function).

Since the integral of an odd function over a symmetric interval around zero is zero:

$$ \int_{-2}^{2} x^3 \cos\left(\frac{x}{2}\right) \sqrt{4 - x^2} , dx = 0 $$

3. Evaluate the remaining integral Now we only need to evaluate the integral of the remaining part:

$$ I = \int_{-2}^{2} \frac{1}{2} \sqrt{4 - x^2} , dx $$

4. Simplify the integral Factor out the constant $\frac{1}{2}$:

$$ I = \frac{1}{2} \int_{-2}^{2} \sqrt{4 - x^2} , dx $$

5. Recognize the integral as the area of a semicircle The function $\sqrt{4 - x^2}$ represents the upper half of a circle with radius 2 centered at the origin. The area of a semicircle is given by:

$$ \text{Area} = \frac{1}{2} \pi r^2 $$

Here, $r = 2$, so:

$$ \text{Area} = \frac{1}{2} \pi (2)^2 = \frac{1}{2} \cdot \pi \cdot 4 = 2\pi $$

6. Finish computing the integral Now substitute back into our expression for $I$:

$$ I = \frac{1}{2} (2\pi) = \pi $$