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?

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.