Answer
$\displaystyle \int_{2}^{6}r^{\prime}(t)dt$
Work Step by Step
See Th.4.12, The Net Change Theorem
The definite integral of the rate of change of quantity $F^{\prime}(x)$
gives the total change, or net change, in that quantity on the interval $[a, b]$.
$\displaystyle \int_{a}^{b}F^{\prime}(x)dx=F(b)-F(a)$
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$r(t)$ represents the weight in pounds of the dog at time $t$.
$\displaystyle \int_{2}^{6}r^{\prime}(t)dt = r(6)-r(2)$
represents the net change in the weight of the dog
from year 2 to year 6.
