$ \begin{array}{|l|c|c|c|c|c|} \hline \mbox{ t (minutes)} & 0 & 4 & 9 & 15 & 20 \\ \mbox{ W(t) (degrees Fahrenheight)} & 55.0 & 57.1 & 61.8 & 67.9 & 71.0 \\ \hline \end{array} $
The Temperature of water in tub at time $t$ is modeled by a strictly increasing, twice-differentiable function $W$, where $W\left(t\right)$ is measured in degrees Fahrenheit and $t$ is measured in minutes. At time $t = 0$, the temperature of the water is 55$\circ$F. The water is heated for 30 minutes, beginning at time $t =0$. Values of $W\left(t\right)$ at selected times $t$ for the first 20 minutes are given in the table above.
(d) For $20 \leq t \leq 25$, the function $W$ that models the water tempature has first derivative given by $W^{\prime}\left(t\right) = 0.4\sqrt{t}\cos\left(0.06t\right)$. Based on the model, what is the temperature of the water at time $t = 25$?
The answer given by the College board
(d) $W\left(25\right) = 71.0 + \int_{20}^{25} W^{\prime}\left(t\right)\: \textrm{d}t = 71.0 + 2.043155 = 73.043$
What I don't understand is the answer for question d.
the question can be found here and sorry the array didn't work