The image shows a graph of a piecewise function $f(x)$, with a semi-circular portion from $-a$ to $a$, and two straight segments from $a$ to $3a$. We need to evaluate the integral $\int_{-a}^{3a} f(x) \, dx$ for $a = 12$, and round the final answer to 4 decimal places.

Step 1: Break down the integral

The integral can be split into the following parts based on the graph:

$\int_{-a}^{3a} f(x) \, dx = \int_{-a}^{a} f(x) \, dx + \int_{a}^{2a} f(x) \, dx + \int_{2a}^{3a} f(x) \, dx$

Step 2: Evaluate the semi-circle portion from $-a$ to $a$

The graph shows that $f(x)$ is a semi-circle on the interval $[-a, a]$. The area under this semi-circle is given by the area of a circle of radius $a$, which is $\pi a^2$, but since we only have a semi-circle, the area is:

$\text{Area of semi-circle} = \frac{1}{2} \pi a^2$

Given that the semi-circle is above the x-axis, this area is positive.

Step 3: Evaluate the straight line portions from $a$ to $3a$

• From $a$ to $2a$, the function is a straight line at $-a$, so the integral is:

$\int_{a}^{2a} f(x) \, dx = (-a) \times (2a - a) = -a^2$

• From $2a$ to $3a$, the function is zero, so this integral contributes nothing:

$\int_{2a}^{3a} f(x) \, dx = 0$

Step 4: Combine the results

Now, we combine the results from each interval:

$\int_{-a}^{3a} f(x) \, dx = \frac{1}{2} \pi a^2 + (-a^2) + 0$

Simplifying this expression:

$\int_{-a}^{3a} f(x) \, dx = \frac{1}{2} \pi a^2 - a^2$

Step 5: Plug in $a = 12$

Substituting $a = 12$ into the equation:

$\int_{-12}^{36} f(x) \, dx = \frac{1}{2} \pi (12)^2 - (12)^2$ $= \frac{1}{2} \pi \times 144 - 144$ $= 72\pi - 144$

Step 6: Calculate the numerical value

Using $\pi \approx 3.1416$:

$72 \times 3.1416 = 226.1942$ $\int_{-12}^{36} f(x) \, dx \approx 226.1942 - 144 = 82.1942$

Thus, the value of the integral is approximately:

$\boxed{82.1942}$

Do you want further details or have any questions?

Here are 5 related questions:

1. How do you calculate the area of a full circle?
2. What is the geometric interpretation of an integral?
3. Why is the integral of the straight-line portion negative?
4. How does the integral change if $a$ takes on a different value?
5. What does it mean when a function has an integral equal to zero?

Tip: When splitting an integral, always pay attention to where the function crosses the x-axis, as this can change the sign of the integral.