Let's go through each question one by one:

Question 1

We are given: $\int_2^5 f(x) \, dx = 8 \quad \text{and} \quad \int_2^5 g(x) \, dx = 3$ and need to compute: $\int_5^2 \left(2f(x) - 3g(x)\right) \, dx$

To solve this, we can use properties of definite integrals:

1. Property of Constants: If $a$ is a constant, then $\int_a b \, dx = a \int f(x) \, dx$.

2. Reversing Limits: $\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$.

Using these properties, we get: $\int_5^2 \left(2f(x) - 3g(x)\right) \, dx = -\int_2^5 \left(2f(x) - 3g(x)\right) \, dx$ Now, break it down into two parts: $= -\left(2 \int_2^5 f(x) \, dx - 3 \int_2^5 g(x) \, dx\right)$ Substitute the given values: $= -\left(2 \cdot 8 - 3 \cdot 3\right) = -\left(16 - 9\right) = -7$

So, the answer for Question 1 is: $\int_5^2 \left(2f(x) - 3g(x)\right) \, dx = -7$

Question 2

We are asked to evaluate: $\frac{1}{\pi} \int_0^4 \sqrt{16 - x^2} \, dx$

The integral $\int_0^4 \sqrt{16 - x^2} \, dx$ represents the area of a quarter circle with radius 4 (since $\sqrt{16 - x^2}$ describes a circle with radius $4$).

The area of a full circle with radius $r = 4$ is: $\pi \cdot r^2 = \pi \cdot 4^2 = 16\pi$

Since the integral only covers one-quarter of the circle, the area represented by the integral is: $\frac{1}{4} \cdot 16\pi = 4\pi$

Therefore: $\frac{1}{\pi} \cdot 4\pi = 4$

So, the answer for Question 2 is: $\frac{1}{\pi} \int_0^4 \sqrt{16 - x^2} \, dx = 4$

Would you like further explanations or details on these solutions?

Here are five related questions that expand on this information:

1. How would the solution change in Question 1 if the limits of integration were from $5$ to $2$ instead of $2$ to $5$?
2. Can the integral in Question 2 be solved using trigonometric substitution instead of geometric interpretation?
3. What are other properties of definite integrals that might simplify computations like in Question 1?
4. How can definite integrals be applied to calculate areas under curves that aren't simple geometric shapes?
5. Why does reversing the limits of integration change the sign of the result?

Tip: When working with integrals that represent geometric shapes, remember to leverage symmetry and area formulas for quicker solutions.