To evaluate the double integral

$\int_{0}^{2} \int \sqrt{4-x^2} \, dx \, dx,$

let's interpret and solve it step by step.

Step 1: Understanding the Integral

This integral appears to involve the region bounded by $x$ from 0 to 2, and within this, we are integrating the function $\sqrt{4 - x^2}$. This is the upper half of a circle with radius $2$, centered at the origin.

However, the outer integration is redundant because $x$ is already bounded. This simplifies the problem to a single integral:

$\int_0^2 \sqrt{4 - x^2} \, dx.$

Step 2: Solve the Single Integral

The integral $\int \sqrt{4 - x^2} \, dx$ is a standard form and can be solved using trigonometric substitution.

Substitution:

Let: $x = 2\sin\theta, \quad dx = 2\cos\theta \, d\theta, \quad \text{where } \theta \in [0, \pi/2].$ Then: $\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4(1 - \sin^2\theta)} = \sqrt{4\cos^2\theta} = 2\cos\theta.$

Substituting into the integral: $\int_0^2 \sqrt{4 - x^2} \, dx = \int_0^{\pi/2} 2\cos\theta \cdot 2\cos\theta \, d\theta = \int_0^{\pi/2} 4\cos^2\theta \, d\theta.$

Simplify $\cos^2\theta$ Using a Trigonometric Identity:

$\cos^2\theta = \frac{1 + \cos(2\theta)}{2}.$ Thus: $\int_0^{\pi/2} 4\cos^2\theta \, d\theta = \int_0^{\pi/2} 4 \cdot \frac{1 + \cos(2\theta)}{2} \, d\theta = \int_0^{\pi/2} (2 + 2\cos(2\theta)) \, d\theta.$

Separate the Terms:

$\int_0^{\pi/2} (2 + 2\cos(2\theta)) \, d\theta = \int_0^{\pi/2} 2 \, d\theta + \int_0^{\pi/2} 2\cos(2\theta) \, d\theta.$

1. First Term: $\int_0^{\pi/2} 2 \, d\theta = 2 \cdot \left[\theta\right]_0^{\pi/2} = 2 \cdot \frac{\pi}{2} = \pi.$

2. Second Term: [ \int_0^{\pi/2} 2\cos(2\theta) , d\theta = 2 \cdot \left[\frac{\sin(2\theta)}{2}\right]_0^{\pi/2} = \left[\sin(2\theta)\right]_0^{\pi/2}. ] At $\theta = \pi/2$, $\sin(2\theta) = \sin(\pi) = 0$. At $\theta = 0$, $\sin(2\theta) = \sin(0) = 0$. Therefore: $\int_0^{\pi/2} 2\cos(2\theta) \, d\theta = 0.$

Combine Results:

$\int_0^2 \sqrt{4 - x^2} \, dx = \pi + 0 = \pi.$

Final Answer:

$\int_0^2 \sqrt{4 - x^2} \, dx = \pi.$

Let me know if you would like additional details or explanations! 😊

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Here are 5 related questions for further exploration:

1. How do trigonometric substitutions help in solving integrals of radical functions?
2. Can you compute the area of the entire circle using a similar method?
3. What is the geometric interpretation of the integral in polar coordinates?
4. How would you compute $\int_{-2}^{2} \sqrt{4 - x^2} \, dx$?
5. How can numerical methods approximate such integrals?

Tip: Always visualize the problem geometrically when dealing with integrals involving square roots of quadratic expressions. It often relates to areas of circles or ellipses!