Answer
$\dfrac{83}{3}$
Work Step by Step
$I=\int_{C_1}x^2 dx+y^2 dy+\int_{C_2}x^2 dx+y^2 dy=(8) \times \int_{0}^{\pi/2} \sin^2 (2t) \cos (t)-\cos^2 (t) \times \sin (t) dt+\int_{0}^{1} (4t)^2 (4 dt) +(2+t)^2 dt$
Suppose $m=\sin t$ and $n=\cos t $ and $dm=\cos t dt; dn=-\sin t dt$
Thus, $I=8 \times \int_{0}^{\pi/2} m^2 dm+8 \times \int n^2 dn+\int_0^1 (64t^2+4+4t+t^2) dt$
or, $=\dfrac{8}{3} \times [ \sin^3(t)]_0^{\pi/2}+\dfrac{8}{3} \times (\cos^3 t)_0^{\pi/2}+[64 \times (\dfrac{t^3}{3})+4t+4 \times (\dfrac{t^2}{2})+(\dfrac{t^3}{3})] _0^1$
Hence, $I= \dfrac{8}{3}(1-0+0-1)+\dfrac{83}{3}=\dfrac{83}{3}$
