Putting Santul through his paces

Santul completes two 20 mile training runs with different pace profiles:

1) A constant 9 minute mile pace (which is impressive both physically and for its consistency); and

2) His pace starts at 10 minutes per mile and then linearly decreases to 8 minutes per mile over the course of the race. So for instance, when Santul is halfway done with the race (in time, not distance), his pace is 9 minutes per mile.

The question is: which run did he complete faster and what were the two times?

Santul's 9 minute mile run will complete in $9 \,\text{min} \, / \, \text{mile}\, \cdot 20 \,\text{miles} = 180 \,\text{minutes}.$

Let's now look at the second run. Let

$$p(t) = a + bt$$ be the linear function of Santul's pace at time $t$. From the initial data we know that $$p(0) = a = 10$$ and that if $T$ is that time that Santul finishes his second run, then $$p(T) = 10 + b T = 8.$$ So $a = 10$ and $b = -\frac{2}{T}.$

We can convert Santul's pace into distance by integrating as follows $d(s) = \int_0^t \,\frac{ds}{p(s)},$ so given the 20 mile course was finished in $T$, we have

$$20 = \int_0^T \,\frac{ds}{10 - \frac{2t}{T}} = -\frac{T}{2} \ln (\frac{8}{10}).$$ Solving for $T$ we get $$T = \frac{40}{\ln 1.25} = 179.256804709...$$ which is just a shade under the constant pace time.