The problem states:

If $\int_{-1}^2 g(x) \, dx = 4$ and $\int_{-1}^2 3g(x) \, dx = 6$, evaluate $\int_2^{-1} g(x) \, dx$.

Solution:

1. Property of Constants in Integration: Given $\int_{-1}^2 3g(x) \, dx = 6$, we can factor out the constant 3: $3 \int_{-1}^2 g(x) \, dx = 6$ Divide by 3: $\int_{-1}^2 g(x) \, dx = 2$

2. Relation Between Limits of Integration: By definition of definite integrals, reversing the limits changes the sign: $\int_2^{-1} g(x) \, dx = -\int_{-1}^2 g(x) \, dx$

3. Substitute the Value: From Step 1, $\int_{-1}^2 g(x) \, dx = 2$. Therefore: $\int_2^{-1} g(x) \, dx = -2$

Final Answer:

$\boxed{-2}$

Let me know if you have further questions or would like details!

Here are some related questions to expand understanding:

1. What is the interpretation of a definite integral's sign in terms of the area under the curve?
2. How does multiplying the integrand by a constant affect the definite integral?
3. Can you demonstrate reversing the limits for any definite integral example?
4. How does symmetry in the function $g(x)$ affect the evaluation of definite integrals?
5. What happens if the bounds of integration are the same?

Tip: Always double-check your limits of integration when working with definite integrals, as reversing them changes the sign!