To solve the problem using Part 1 of the Fundamental Theorem of Calculus, we need to recognize that if $f(x) = \int_{a(x)}^{b(x)} g(t) \, dt$, then the derivative $f'(x)$ can be computed as:

$f'(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$

For the given function $f(x) = \int_{x}^{2} \left(\frac{1}{3} t^3 - 1\right)^9 dt$, observe that:

• $g(t) = \left(\frac{1}{3} t^3 - 1\right)^9$
• $a(x) = x$
• $b(x) = 2$

Since $a(x) = x$ and $b(x) = 2$, we can apply the Fundamental Theorem of Calculus as follows:

1. Evaluate $g(t)$ at the upper and lower limits of integration.
2. Differentiate the limits with respect to $x$.

Given:

• $g(t) = \left(\frac{1}{3} t^3 - 1\right)^9$
• $b(x) = 2$ (so $b'(x) = 0$)
• $a(x) = x$ (so $a'(x) = 1$)

Thus:

$f'(x) = g(2) \cdot 0 - g(x) \cdot 1$

Since $g(2) \cdot 0$ is zero, the derivative simplifies to:

$f'(x) = -\left(\frac{1}{3} x^3 - 1\right)^9$

So, the derivative of $f(x)$ is:

$f'(x) = -\left(\frac{1}{3} x^3 - 1\right)^9$

Would you like more details or have any further questions?

Here are 5 related questions:

1. What is Part 2 of the Fundamental Theorem of Calculus, and how does it differ from Part 1?
2. How would you handle the derivative if the limits of integration were both functions of $x$?
3. What are the conditions under which the Fundamental Theorem of Calculus applies?
4. How do you interpret the physical meaning of the Fundamental Theorem of Calculus in applied mathematics?
5. How would the solution change if the integrand had an additional function multiplied by it?

Tip: When using the Fundamental Theorem of Calculus, always check if the limits of integration are functions of $x$, as this will determine how you differentiate the expression.