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$$ \begin{array}{|l|c|c|c|c|c|} \hline \mbox{ t (minutes)} & 0 & 4 & 9 & 15 & 20 \\ \mbox{ W(t) (degrees Fahrenheight)} & 55.0 & 57.1 & 61.8 & 67.9 & 71.0 \\ \hline \end{array} $$

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

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@yunone, how did you make the array work? – yiyi Nov 16 '12 at 6:43
I just removed the \begin{center} and \end{center} parts. I think it might have been throwing off MathJax. I surrounded it with $$ to force it to center. – yunone Nov 16 '12 at 6:50
up vote 3 down vote accepted

It must be that you are allowed to use a hand calculator which has numerical integration. When I put the integral into maple, a fairly good antiderivative finder, it didn't find one in closed form. So then I found the numerical approximation to the integral, and that was 2.043144699.. which it seems the answer rounded off to 2.043155, then rounded again in the final answer given, to three decimals.

Or are you just wondering why the formula $$f(b)=f(a)+\int_a^b f'(x) dx$$was used? If that's your question then look up the "fundamental theorem of calculus" which says exactly the above, in one of its forms.

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But it asks for the value at the point $t = 25$. That is what is confusing me, the formula given by college board and you, show the total area underneath the curve, but it just asks for the value at one point? Shouldn't that be zero, because dt is very very small? – yiyi Nov 16 '12 at 6:47
When an integral is done, it is finding area under the curve by a limit of rectangles. True, the width of the rectangles goes to zero, but the number of them is going to infinity, and in the limit the total area tends to the area under the curve, which is not zero. – coffeemath Nov 16 '12 at 7:22
Just confused because it asks for the value at $W\left(25\right)$, so that exact pt its zero, but the area from [0,25], I understand is not zero. I would understand if the question asked how much did the temp change from $t = 0$ to $t = 25$. – yiyi Nov 17 '12 at 11:48
The integral is from $t=20$ to $t=25$. They tell you the temp at $t=20$ is 71.0, and the integral gives the change in temp from $t=20$ to $t=25$. So 71.0 + [change] = temp at $t=25$. – coffeemath Nov 17 '12 at 15:26

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