Answer:
The value of the line integral is [tex]\frac{157}{2}[/tex].
Step-by-step explanation:
Given a path C, with parametrization r(t) for [tex]t_0\leq t \leq t_1[/tex], we have that
[tex]\int_{C}F dr = \int_{t_0}^{t_1} F(r(t)) \cdot r'(t) dt [/tex] where [tex] \cdot [/tex] is the dot product between two vectors.
In our case, we have [tex] r(t) = (6t,11t^2), 0\leq t\leq 1[/tex]. Then [tex]r'(t) = (6,22t)[/tex]. In this case we have that F(x,y) = (x,y). Then,[tex] F(r(t)) = (6t, 11t^2)[/tex]
So
[tex]\int_{C}F dr = \int_{0}^{1} (6t,11t^2)\cdot(6,22t) dt = \int_{0}^{1} 36t+11\cdot 22 t^3= \left.(36\frac{t^2}{2}+11\cdot 22 \frac{t^4}{4})\right|_{0}^{1} = \frac{36}{2}+\frac{11\cdot 22}{4}= \frac{157}{2}[/tex]
