Let's go through each problem one by one, providing solutions and a brief explanation of each.

Problem 25:

Given function: $y = x\sqrt{4 - x^2}$

This represents the shaded area under a curve from $x = -2$ to $x = 2$. To find the area, we need to compute the definite integral:

$\text{Area} = \int_{-2}^{2} x \sqrt{4 - x^2} \, dx$

This can be solved using trigonometric substitution (such as $x = 2\sin{\theta}$), and after solving, you get:

$\text{Area} = \frac{16}{3}$

Problem 26:

Given function: $y = (1 - \cos{x})\sin{x}$

Here, the area is between $x = 0$ and $x = \pi$. So we compute:

$\text{Area} = \int_0^\pi (1 - \cos{x}) \sin{x} \, dx$

Using integration by parts or substitution, this results in:

$\text{Area} = 2$

Problem 27:

Given function: $y = 3(\sin{x})\sqrt{1 + \cos{x}}$

Here, the limits of integration are from $x = -\pi$ to $x = 0$. So we compute:

$\text{Area} = \int_{-\pi}^{0} 3 (\sin{x})\sqrt{1 + \cos{x}} \, dx$

This is another trigonometric integral that requires substitution, and after solving:

$\text{Area} = 12$

Problem 28:

Given function: $y = \frac{\pi}{2} \cos{x} \sin(\pi + \pi \sin{x})$

The limits of integration are from $x = -\pi$ to $x = 0$. To find the area, we compute the integral:

$\text{Area} = \int_{-\pi}^{0} \frac{\pi}{2} \cos{x} \sin(\pi + \pi \sin{x}) \, dx$

This integral is more complicated and involves periodic functions, but it simplifies to:

$\text{Area} = 0$

The area cancels out due to symmetry, so no net area exists in this case.

Would you like more details on any specific part of the calculations or methods used?

Related Questions:

1. How do you solve definite integrals using trigonometric substitution?
2. Can you explain integration by parts with an example?
3. What are the key steps to solving an integral involving periodic functions?
4. How can symmetry affect the total area in a graph problem?
5. How do we handle negative values of $y$ when finding the total area under a curve?

Tip: When solving integrals with trigonometric functions, substitution is often the key to simplifying the expression and making the integral solvable.