by (3x? + y²) dx+ (2.xy) dy 7. C: 3D1 from (3,0) to (0,4) %3D - = 1 6.

We are given the following integral:
${\int }_C(3x^2+y^2) dx + \left(2xy\right) dy$ over the path $C: \frac{x^2}{9}+\frac{y^2}{16} = 1$ from (3, 0) to (0, 4).
The path C can be drawn as

Now, we form the parameterization of the path C.
Let x = 3 cos t, y = 4 sin t $\Rightarrow dx = -3 \sin t dt; dy = 4 \cos t dt.$ From the figure we can see that the path C is a quarter of the circle. The value of t ranges from 0 to $\frac{\pi }{2}$. 
Therefore, 
$$\begin{gathered} I = {\int }_C\left(3x^2+y^2\right) dx + \left(2xy\right) dy \\ = {\int }_0^{\frac{\pi }{2}} \left(3(9 {\cos }^{2}t) + (16 {\sin }^{2}t)\right)(-3 \sin t) + \left(2\left(3 \cos t\right)\right(4 \sin t\left)\right)\left(4 \cos t\right) dt \\ = {\int }_0^{\frac{\pi }{2}}\left(27 {\cos }^{2}t + 16 {\sin }^{2}t\right)(- \sin t)+\hspace{0.17em}(96 {\cos }^{2}t \sin t) dt \\ ={\int }_0^{\frac{\pi }{2}}\left(11 {\cos }^{2}t + 1\right)(-3 \sin t)+\hspace{0.17em}(96 {\cos }^{2}t \sin t) dt \\ ={\int }_0^{\frac{\pi }{2}}-11 {\cos }^{2}t \sin t -3 \sin t+\hspace{0.17em}96 {\cos }^{2}t \sin t dt \\ ={\int }_0^{\frac{\pi }{2}}63 {\cos }^{2}t \sin t - 3 \sin t dt \end{gathered}$$