The expression

$\int_{8}^{11} w'(t) \, dt$

represents the change in the child's weight (in pounds) between the ages of 8 and 11.

Explanation:

1. The integral of a rate of change, $w'(t)$, over a specific interval $[a, b]$ gives the total change in the quantity $w(t)$ over that interval.
2. Here, $w'(t)$ is the rate of growth in pounds per year, and the integral from $t=8$ to $t=11$ sums up all the changes in the child's weight over this period.
3. Therefore, the result of this integral gives the total change in the child's weight between ages 8 and 11.

Correct Answer:

The change in the child's weight (in pounds) between the ages of 8 and 11.

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Would you like further details or have any questions? Here are 5 related questions for you to explore:

1. What is the physical interpretation of a definite integral in terms of real-world problems?
2. How would the interpretation change if $w'(t)$ were a negative value over part of the interval?
3. What if the integral limits were reversed, such as $\int_{11}^{8} w'(t) \, dt$?
4. How would you calculate $w(11)$ if you were given $w(8)$ and $\int_{8}^{11} w'(t) \, dt$?
5. Can you explain the connection between the Fundamental Theorem of Calculus and this problem?

Tip: Always interpret the units of the rate and the limits of integration to understand the physical meaning of an integral.