Answer
$\dfrac{15}{2}$
Work Step by Step
In order to find the integral on the line we will use the formula:
$\int_{r} f \ dr=\int_a^b f(r(t)) \cdot |r'(t)| \ dt$
The parametrisation of the line can be found as:
$r(t)=(1,1) +t (2-1, 5-1); t \in [0,1]$
or, $r(t)=(1+t, 1+4t) \implies r'(t)=(1,4)$
So, $|r'(t)|=\sqrt {1^2+4^2}=\sqrt {17}$
Thus, we have: $\int_{r} f \ dr=\int_0^1 f(1+t, 1+4t) \cdot \sqrt {17} \ dr \\=\sqrt {17} \int_0^1 (3+9t) \ dt \\=\sqrt {17} \int_0^1 3 \ dt +\sqrt {17} \int_0^1 9t \ dt \\=3 \sqrt {17} (1-0) +9 \sqrt {17} (\dfrac{1}{2}-0) \\=\dfrac{15 \sqrt {17}}{2}$
Next, the average value is given by: $f(c)=\dfrac{1}{|r'(t)|} \int_r f \ dr=\dfrac{1}{\sqrt {17}}\times \dfrac{15 \sqrt {17}}{2}=\dfrac{15}{2}$
