Answer
$\int_C \textbf F .\textbf {dr} = \frac{1}{2}$
Work Step by Step
$\textbf r(t) = \cos t \textbf i + \sin t \textbf j $
it follows that, $x(t) = \cos t , y(t) = \sin t $
$\textbf F(x,y) = 3x \textbf i +4y\textbf j \\
\therefore \textbf F(x(t),y(t)) = 3\cos t \textbf i + 4\sin t \textbf j\\
\textbf r' = -\sin t \textbf i +\cos t \textbf j\\
\int_C \textbf F \cdot d\textbf r = \int_{t=0}^{\pi/2}\textbf F \cdot \textbf r' dt\\
= \int_0^{\pi/2}(3\cos t \textbf i + 4\sin t \textbf j)\cdot (-\sin t \textbf i +\cos t \textbf j) dt\\
= \int_0^{\pi/2} (-3\cos t . \sin tr + 4 \sin t . \cos t) dt\\
=[\frac{\sin^2t}{2}]_0^{\pi/2} = \frac{1}{2}$
