The image shows a graph of a piecewise linear function $y = f(x)$, and the task is to calculate definite integrals using geometry.

Explanation of Approach:

To calculate the integrals geometrically, we will find the areas of the geometric shapes (triangles and rectangles) formed between the graph and the x-axis over specific intervals. Areas above the x-axis contribute positively to the integral, and areas below the x-axis contribute negatively.

a. $\int_0^1 f(x) \, dx$:

• The region between $x = 0$ and $x = 1$ is a triangle with height 4 and base 1.
• Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 4 = 2$.

So, $\int_0^1 f(x) \, dx = 2$.

b. $\int_1^2 f(x) \, dx$:

• The region between $x = 1$ and $x = 2$ is another triangle with height 2 and base 1.
• Area of triangle = $\frac{1}{2} \times 1 \times 2 = 1$.

So, $\int_1^2 f(x) \, dx = 1$.

c. $\int_0^2 f(x) \, dx$:

• This integral covers the area from $x = 0$ to $x = 2$. From part a and b, we know that: $\int_0^2 f(x) \, dx = \int_0^1 f(x) \, dx + \int_1^2 f(x) \, dx = 2 + 1 = 3$.

So, $\int_0^2 f(x) \, dx = 3$.

d. $\int_2^5 f(x) \, dx$:

• From $x = 2$ to $x = 3$, there is a rectangle with height $-1$ (below the x-axis) and base 1. Its area is $-1 \times 1 = -1$.
• From $x = 3$ to $x = 4$, there is a triangle with base 1 and height $-2$. Its area is $\frac{1}{2} \times 1 \times (-2) = -1$.
• From $x = 4$ to $x = 5$, there is a triangle with base 1 and height $3$. Its area is $\frac{1}{2} \times 1 \times 3 = 1.5$.

So, the total area is: $-1 + (-1) + 1.5 = -0.5$.

Thus, $\int_2^5 f(x) \, dx = -0.5$.

e. $\int_0^5 f(x) \, dx$:

• From $x = 0$ to $x = 2$, the area is 3 (from part c).
• From $x = 2$ to $x = 5$, the area is $-0.5$ (from part d).

So, the total area is: $3 + (-0.5) = 2.5$.

Thus, $\int_0^5 f(x) \, dx = 2.5$.

Summary of Results:

• a. $\int_0^1 f(x) \, dx = 2$
• b. $\int_1^2 f(x) \, dx = 1$
• c. $\int_0^2 f(x) \, dx = 3$
• d. $\int_2^5 f(x) \, dx = -0.5$
• e. $\int_0^5 f(x) \, dx = 2.5$

Would you like further details or have any other questions?

Here are five related questions to consider:

1. How does the sign of the area change when the graph is below the x-axis?
2. How do definite integrals relate to calculating areas under curves in general?
3. Can you explain why the area of a triangle is used for some segments in these integrals?
4. How would the result change if the graph had different shapes (e.g., curves instead of straight lines)?
5. What is the significance of the definite integral when the result is negative?

Tip: Always check if the graph is above or below the x-axis when computing definite integrals geometrically, as this affects the sign of the area.