Let's break down the two integrals and solve them one by one.

i) $\int_3^2 \frac{4y^2 - 8y}{y} \, dy$

First, simplify the integrand:

$\frac{4y^2 - 8y}{y} = \frac{4y^2}{y} - \frac{8y}{y} = 4y - 8$

Now, the integral becomes:

$\int_3^2 (4y - 8) \, dy$

This is a straightforward integral. Let’s compute it:

1. Integrate: $\int (4y - 8) \, dy = 2y^2 - 8y + C$

2. Evaluate the definite integral: $\left[ 2y^2 - 8y \right]_3^2$ Substitute the limits:

   For $y = 2$: $2(2)^2 - 8(2) = 2(4) - 16 = 8 - 16 = -8$

   For $y = 3$: $2(3)^2 - 8(3) = 2(9) - 24 = 18 - 24 = -6$

3. Subtract the results: $-8 - (-6) = -8 + 6 = -2$

So, the value of the first integral is: $\boxed{-2}$

ii) $\int_0^{\pi/2} (5 \cos(t) - 5 \sin(t)) \, dt$

This is another simple integral. Let's compute it step by step:

1. Integrate: $\int 5 \cos(t) \, dt = 5 \sin(t)$ $\int 5 \sin(t) \, dt = -5 \cos(t)$

Thus, the integral becomes: $5 \sin(t) + 5 \cos(t) + C$

2. Evaluate the definite integral: $\left[ 5 \sin(t) + 5 \cos(t) \right]_0^{\pi/2}$

Substitute the limits:

For $t = \frac{\pi}{2}$: $5 \sin\left(\frac{\pi}{2}\right) + 5 \cos\left(\frac{\pi}{2}\right) = 5(1) + 5(0) = 5$

For $t = 0$: $5 \sin(0) + 5 \cos(0) = 5(0) + 5(1) = 5$

3. Subtract the results: $5 - 5 = 0$

So, the value of the second integral is: $\boxed{0}$

Final Answers:

i) $-2$
ii) $0$

Do you have any questions or need further details on these calculations?

Here are 5 related questions to expand on this topic:

1. How would the results change if the limits of integration in (i) were reversed?
2. Can you find the indefinite form of the integrals before applying the limits?
3. How can integration be used to solve area problems under a curve?
4. What happens if we change the trigonometric functions in (ii) to their respective reciprocal functions (like $\csc$ and $\sec$)?
5. Can the first integral be solved using substitution or any other methods?

Tip: When simplifying the integrand, always look for common factors to make integration easier.