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The given integral is:

\[
\int_{-2\sqrt{5}}^{2\sqrt{5}} \frac{6x}{\sqrt{x^2 + 16}} \, dx
\]

This is a definite integral of an odd function over a symmetric interval. Let's break this down:

### 1. The Function Inside the Integral
The function being integrated is:

\[
f(x) = \frac{6x}{\sqrt{x^2 + 16}}
\]

Here, \( f(x) \) is an **odd function** because it satisfies the condition:

\[
f(-x) = -f(x)
\]

This can be checked as follows:
\[
f(-x) = \frac{6(-x)}{\sqrt{(-x)^2 + 16}} = \frac{-6x}{\sqrt{x^2 + 16}} = -f(x)
\]

### 2. Integral of Odd Function Over a Symmetric Interval
When integrating an odd function over a symmetric interval, such as \([-a, a]\), the result is always **0**. This is because the positive and negative parts of the function cancel each other out.

In this case, since the integral is over the symmetric interval \([-2\sqrt{5}, 2\sqrt{5}]\), and the integrand is an odd function, the result is:

\[
\int_{-2\sqrt{5}}^{2\sqrt{5}} \frac{6x}{\sqrt{x^2 + 16}} \, dx = 0
\]

Hence, the answer is \( 0 \).

Would you like more details or have any questions?

### Related Questions:
1. What is the general definition of an odd function?
2. How does symmetry affect definite integrals?
3. Can an integral of an even function over a symmetric interval be zero?
4. What would change if the integrand was an even function?
5. How can you check if a function is odd or even?

### Tip:
Always check the symmetry of the integrand when dealing with definite integrals over symmetric intervals—this can help simplify the evaluation!

Why is the answer 0 for the integral \int_{-2\sqrt{5}}^{2\sqrt{5}} \frac{6x}{\sqrt{x^2 + 16}} \, dx?

Discover why the integral of the odd function 6x/√(x^2 + 16) from -2√5 to 2√5 equals 0, by using symmetry in calculus and properties of odd functions.

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