To solve this, we will use the linearity of integrals:

$\int_{-83}^{-21} [4f(x) + 6g(x) - h(x)] \, dx = 4 \int_{-83}^{-21} f(x) \, dx + 6 \int_{-83}^{-21} g(x) \, dx - \int_{-83}^{-21} h(x) \, dx$

From the problem: $\int_{-83}^{-21} f(x) \, dx = 11, \quad \int_{-83}^{-21} g(x) \, dx = 27, \quad \int_{-83}^{-21} h(x) \, dx = 29$

Substitute these values into the equation:

$4 \int_{-83}^{-21} f(x) \, dx = 4(11) = 44$ $6 \int_{-83}^{-21} g(x) \, dx = 6(27) = 162$ $-\int_{-83}^{-21} h(x) \, dx = -29$

Add these results together:

$44 + 162 - 29 = 177$

Thus, the value of the integral is:

$\boxed{177}$

Would you like a breakdown of any specific step or have any further questions?

Related Questions

1. How does linearity simplify integral calculations like this one?
2. What happens if the limits of integration change to $[-21, -83]$?
3. Can this process be extended to functions multiplied by other constants?
4. How would this problem change if one of the integral values was unknown?
5. What are some real-world applications of evaluating integrals like this?

Tip

Always check if properties like linearity or symmetry of integrals can simplify the problem before attempting complex calculations!