Answer
$\frac{1}{20}$
Work Step by Step
Given:
$r(t) = (t^3)i + (t^2)j$
we get:
$x = t^3$
$y = t^2$
Derive $r(t)$:
$r'(t) = (3t^2, 2t)$
The line integral of $f$ is calculated as follows:
$\int_c F dr = \int F(r(t))\times r'(t) dt$
$ = \int_0^1 (((t^3)(t^2)^2, -(t^3)^2) \times (3t^2, 2t)) dt$
$ = \int_0^1 ((t^7, -t^6) \times (3t^2, 2t)) dt$
$ = \int_0^1 (3t^9 - 2t^7) dt$
$ = \frac{3}{10}t^{10} - \frac{2}{8}t^8 $
$ = (\frac{3}{10}(1)^{10} - \frac{2}{8}(1)^8) - (\frac{3}{10}(0)^{10} - \frac{2}{8}(0)^8)$
$ =(\frac{3}{10} - \frac{2}{8}) - (0 - 0)$
$ = \frac{1}{20}$
