To solve the integral $\int_{0}^{21} f(x) \, dx$, we need to compute the total signed area under the curve $f(x)$ from $x = 0$ to $x = 21$. Here's the breakdown based on the graph provided:

Observations:

1. Regions: The curve can be broken into multiple regions:
   • A triangular region from $x = 0$ to $x = 2$.
   • A quarter-circle (positive area) from $x = 2$ to $x = 7$.
   • A quarter-circle (negative area) from $x = 7$ to $x = 12$.
   • A triangular region from $x = 12$ to $x = 21$.

Calculations:

1. Area of the first triangle ($x = 0$ to $x = 2$):

• The base is 2, and the height is 7.
• Area of a triangle: $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 7 = 7$.

2. Area of the quarter-circle ($x = 2$ to $x = 7$):

• Radius: 7.
• Area of a quarter-circle: $\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (7^2) = \frac{49\pi}{4}$.

3. Area of the quarter-circle ($x = 7$ to $x = 12$):

• Radius: 7.
• Since this quarter-circle is below the x-axis, its area is negative.
• Area: $-\frac{1}{4} \pi r^2 = -\frac{49\pi}{4}$.

4. Area of the second triangle ($x = 12$ to $x = 21$):

• The base is $21 - 12 = 9$, and the height is 7.
• Area of a triangle: $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9 \times 7 = 31.5$.

Total Area:

Now, sum up the areas: $\text{Total Area} = 7 + \frac{49\pi}{4} - \frac{49\pi}{4} + 31.5 = 7 + 31.5 = 38.5.$

Thus, the value of the integral is: $\int_{0}^{21} f(x) \, dx = 38.5.$

Would you like a detailed visualization or further explanations?

Related Questions:

1. What does the signed area under the curve represent in a practical context?
2. How would the integral change if the graph were mirrored about the x-axis?
3. Can this result be approximated for irregular regions using numerical integration methods?
4. How would we compute $\int_{a}^{b} f(x) \, dx$ if only parts of the graph were visible?
5. What are other applications of definite integrals in geometry and physics?

Tip:

For problems involving geometric areas under curves, break the regions into simple shapes like triangles, rectangles, and known segments of circles. This simplifies calculations!