To solve this, we will use Part 1 of the Fundamental Theorem of Calculus. This theorem states:

If $g(s) = \int_a^s f(t) \, dt$, then $g'(s) = f(s)$.

Here, the integral defines $g(s)$, and the derivative $g'(s)$ is obtained by substituting the upper limit of the integral (in this case $s$) into the integrand $f(t)$.

Problem:

We are given:

$g(s) = \int_7^s (t - t^2)^3 \, dt$

Step 1: Identify $f(t)$:

The integrand $f(t)$ is:

$f(t) = (t - t^2)^3$

Step 2: Apply the Fundamental Theorem:

The derivative of $g(s)$ is:

$g'(s) = f(s) = (s - s^2)^3$

Final Answer:

$g'(s) = (s - s^2)^3$

Would you like a more detailed explanation of the theorem or the steps?

Here are some related questions:

1. How does the Fundamental Theorem of Calculus Part 2 differ from Part 1?
2. Can you apply this theorem to cases where the limits of integration are functions?
3. What happens if the upper limit is $s^2$ instead of $s$?
4. How can you interpret this result graphically?
5. How would you compute the derivative if the integral had a variable lower limit?

Tip: Always remember that when the upper limit is variable, simply substitute it into the integrand to find the derivative.