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To solve for \( G'(x) \) where \( G(x) = \int_{0}^{x} (2t^2 + \sqrt{t}) \, dt \), we can apply the **Fundamental Theorem of Calculus**. This theorem states that if we have a function \( G(x) \) defined as:

\[
G(x) = \int_{a}^{x} f(t) \, dt
\]

then its derivative \( G'(x) \) is simply \( f(x) \), assuming \( f(t) \) is continuous. 

### Solution:

1. Given:
   \[
   G(x) = \int_{0}^{x} (2t^2 + \sqrt{t}) \, dt
   \]

2. By the Fundamental Theorem of Calculus, we can directly differentiate \( G(x) \) with respect to \( x \):
   \[
   G'(x) = 2x^2 + \sqrt{x}
   \]

### Answer:
\[
G'(x) = 2x^2 + \sqrt{x}
\]

Would you like further details on this process, or have any questions?

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Here are 5 related questions to further your understanding:

1. How would you evaluate \( G(x) \) itself if you were given a specific value of \( x \)?
2. What if the lower bound of the integral was \( x \) instead of \( 0 \)? How would that change \( G'(x) \)?
3. How would you find \( G'(x) \) if \( G(x) \) had a different upper limit, say \( 2x \)?
4. How does the Fundamental Theorem of Calculus relate to the Mean Value Theorem for integrals?
5. What would happen if the integrand \( f(t) \) was discontinuous at some point in the interval?

**Tip:** Remember, the Fundamental Theorem of Calculus is very useful for differentiating integral functions where the limits are variable.

Find G'(x) if G(x) = ∫[0 to x] (2t^2 + √t) dt.

Learn how to find G'(x) for the function G(x) = ∫[0 to x] (2t^2 + √t) dt using the Fundamental Theorem of Calculus. This problem involves calculus concepts such as definite integrals and differentiation, suitable for high school students in Grades 11-12.

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