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To find the area of the region below the function \( f(x) = 2x + \sin(2x) \), from \( x = 0 \) to \( x = \frac{1}{2} \pi \), we need to compute the definite integral of the function between these limits.

The area under the curve is given by:

\[
\text{Area} = \int_{0}^{\frac{\pi}{2}} \left( 2x + \sin(2x) \right) \, dx
\]

Now, let's compute this integral.

1. The integral of \( 2x \) is \( x^2 \).
2. The integral of \( \sin(2x) \) requires substitution. Let \( u = 2x \), hence \( du = 2dx \). This results in:

\[
\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)
\]

Thus, the complete integral is:

\[
\int_0^{\frac{\pi}{2}} (2x + \sin(2x)) \, dx = \left[ x^2 - \frac{1}{2} \cos(2x) \right]_0^{\frac{\pi}{2}}
\]

Now I will compute this expression.The area under the curve of the function \( f(x) = 2x + \sin(2x) \) from \( x = 0 \) to \( x = \frac{\pi}{2} \) is:

\[
\text{Area} = 1 + \frac{\pi^2}{4}
\]

Would you like further details on how to interpret this result, or do you have any specific questions?

Here are 5 related questions to explore further:
1. How do you evaluate the area under a curve if the function is piecewise?
2. What is the geometric interpretation of definite integrals?
3. How can you use numerical methods to approximate integrals when exact solutions are difficult to obtain?
4. How do integrals involving trigonometric functions differ from polynomials in their solutions?
5. How do the limits of integration affect the area calculation?

**Tip:** Always double-check your limits of integration when calculating areas, especially for periodic functions like sine or cosine.

Find the area of the region below the function f(x) = 2x + sin(2x), from x = 0 to x = π/2.

This problem involves finding the area under the curve of the function f(x) = 2x + sin(2x) using definite integrals from x = 0 to x = π/2. It includes concepts from calculus and trigonometric integration, making it suitable for college or AP Calculus students.

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