We are given that:

$\int_a^b f(x) \, dx = a + 2b$

We want to find:

$\int_a^b \big(f(x) + 5\big) \, dx$

Solution:

Using the property of integrals, the integral of a sum is the sum of the integrals. Thus:

$\int_a^b \big(f(x) + 5\big) \, dx = \int_a^b f(x) \, dx + \int_a^b 5 \, dx$

Step 1: Substitute the given value of $\int_a^b f(x) \, dx$:

$\int_a^b f(x) \, dx = a + 2b$

Step 2: Calculate $\int_a^b 5 \, dx$:

The integral of a constant $5$ over $[a, b]$ is given by: $\int_a^b 5 \, dx = 5 \cdot (b - a)$

Step 3: Add the results:

$\int_a^b \big(f(x) + 5\big) \, dx = \big(a + 2b\big) + \big(5(b - a)\big)$

Simplify: $\int_a^b \big(f(x) + 5\big) \, dx = a + 2b + 5b - 5a$

Combine terms: $\int_a^b \big(f(x) + 5\big) \, dx = -4a + 7b$

Final Answer:

$\boxed{-4a + 7b}$

Do you want further details or have any questions?

Here are 5 related questions:

1. How does the property of integrals help in breaking down sums?
2. What happens if the added constant was a function instead?
3. How would this change if the limits of integration were swapped?
4. Can you generalize this approach for any constant $c$ instead of 5?
5. How does this relate to definite vs indefinite integrals?

Tip: Always simplify integral expressions step-by-step to avoid errors in complex combinations.