From the graph and the provided integrals, let's analyze and calculate the definite integrals step by step:

1. Integral $\int_{-2}^2 f(x) \, dx$:

• The graph shows symmetry with respect to the y-axis, and the area above the x-axis from $x = -2$ to $x = 0$ is positive.
• The area below the x-axis from $x = 0$ to $x = 2$ is negative, and both areas have equal magnitude.

Thus: $\int_{-2}^2 f(x) \, dx = 0$

2. Integral $\int_{3}^2 f(x) \, dx$:

• This represents the area from $x = 2$ to $x = 3$, which is negative because we're moving in the reverse direction.
• From the graph, the triangle under the x-axis from $x = 2$ to $x = 3$ has:
  • Base = $3 - 2 = 1$,
  • Height = $-2$.
  • Area = $\frac{1}{2} \cdot 1 \cdot 2 = 1$.

Thus: $\int_{3}^2 f(x) \, dx = -1$

3. Integral $\int_{3}^6 f(x) \, dx$:

• From $x = 3$ to $x = 6$, the graph forms a right triangle above the x-axis:
  • Base = $6 - 3 = 3$,
  • Height = $4$.
  • Area = $\frac{1}{2} \cdot 3 \cdot 4 = 6$.

Thus: $\int_{3}^6 f(x) \, dx = 6$

4. Sum of $\int_{-2}^0 f(x) \, dx$ and $\int_{0}^2 f(x) \, dx$:

• From symmetry and earlier calculations:
  • $\int_{-2}^0 f(x) \, dx = A$,
  • $\int_{0}^2 f(x) \, dx = -A$,
    where $A$ is the positive area above the x-axis from $x = -2$ to $0$.
• Since $A + (-A) = 0$, the sum is: $\int_{-2}^0 f(x) \, dx + \int_{0}^2 f(x) \, dx = 0$

If you'd like to verify or have detailed breakdowns for any part, let me know! 😊

5 Related Questions:

1. How is the area of geometric shapes (e.g., triangles) used to calculate definite integrals?
2. Why do definite integrals of symmetric graphs over symmetric intervals sometimes result in 0?
3. How does reversing the limits of integration affect the sign of the integral?
4. Can you explain how to split integrals when evaluating piecewise functions?
5. How do definite integrals relate to the Fundamental Theorem of Calculus?

Tip:

Always visualize the graph and symmetry before calculating integrals—it can simplify the work significantly!