To evaluate each integral by interpreting the areas under the curve of $f(x)$, I'll break down the areas by using the shapes formed under each segment of the curve:

1. For $\int_0^{16} f(x) \, dx$:

   • The segment from $x = 0$ to $x = 16$ forms a triangle with a base of $16$ and a height of $16$.
   • The area of this triangle is: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 16 \times 16 = 128$
   • So, $\int_0^{16} f(x) \, dx = 128$.

2. For $\int_0^{40} f(x) \, dx$:

   • This integral includes the areas from $x = 0$ to $x = 40$.
   • From $x = 16$ to $x = 32$, we have a rectangle with a width of $16$ and height $16$: $\text{Area} = 16 \times 16 = 256$
   • From $x = 32$ to $x = 40$, there is a triangle with a base of $8$ and height of $16$: $\text{Area} = \frac{1}{2} \times 8 \times 16 = 64$
   • Adding up the areas from $0$ to $40$: $128 + 256 + 64 = 448$
   • So, $\int_0^{40} f(x) \, dx = 448$.

3. For $\int_{40}^{56} f(x) \, dx$:

   • This interval goes from $x = 40$ to $x = 56$, where the function is negative.
   • From $x = 40$ to $x = 48$, we have a triangle with a base of $8$ and height of $16$: $\text{Area} = \frac{1}{2} \times 8 \times 16 = 64$
   • From $x = 48$ to $x = 56$, we have another triangle with the same dimensions: $\text{Area} = \frac{1}{2} \times 8 \times 16 = 64$
   • Since both areas are below the x-axis, they are negative: $\int_{40}^{56} f(x) \, dx = -64 - 64 = -128$

4. For $\int_0^{72} f(x) \, dx$:

   • This covers the entire interval from $x = 0$ to $x = 72$.
   • We already calculated the areas from $0$ to $40$ (448) and from $40$ to $56$ (-128).
   • From $x = 56$ to $x = 72$, there is a triangle with a base of $16$ and a height of $8$: $\text{Area} = \frac{1}{2} \times 16 \times 8 = 64$
   • Summing all these areas: $448 - 128 + 64 = 384$
   • So, $\int_0^{72} f(x) \, dx = 384$.

Here are the evaluated integrals:

• (a) $\int_0^{16} f(x) \, dx = 128$
• (b) $\int_0^{40} f(x) \, dx = 448$
• (c) $\int_{40}^{56} f(x) \, dx = -128$
• (d) $\int_0^{72} f(x) \, dx = 384$

Would you like further explanations on any part of these solutions?

Here are some related questions for further exploration:

1. How does interpreting integrals as areas help in understanding definite integrals?
2. Can you explain why areas below the x-axis contribute negatively to the integral?
3. What would change if the curve of $f(x)$ was shifted vertically by a constant?
4. How would you approximate the integral if the shapes were irregular?
5. Can you calculate the integral for other intervals on the graph?

Tip: When interpreting integrals in terms of areas, remember that any segment below the x-axis contributes negatively to the total integral value.