The four corners of the square C are $(0,0), (0,2), (2,2) , (0,2) $ and the given integral is $$\int_C y^4dx+5x^2dy$$ Parameterizing the square is where I am stuck.
I have $$\left<2t,0 \right> for \space 0 \leq t \leq 1$$
$$\left<2, 2(t-1) \right> for \space 1\leq t\leq 2$$ $$\left<6-2t, 2 \right> for \space 2\leq t\leq 3$$ $$\left<0,0 \right> for\space 3\leq t\leq 4$$
There are two ways to go about this -
The four corners of the square are $A(0,0),B(2,0),C(2,2),D(0,2)$ (assuming there was a typo on the second one. Then, we have a path with counterclockwise orientation.
If you are going by the second method,
Most of your parametrization looked ok but you do not necessarily have to choose increasing values for $t$. These are all individual line segments.
For $AB,$ we have $(0,0) + (2-0, 0-0)t = (2t,0) \,$ for $0 \leq t \leq 1$
For $BC,$ we have $(2,0) + (2-2, 2-0)t = (2,2t) \,$ for $0 \leq t \leq 1$
Similarly for $CD,$ we have $(2-2t,2) \,$ for $0 \leq t \leq 1$
For $DA,$ we have $(0,2-2t) \,$ for $0 \leq t \leq 1$
Now I will take one of these line integrals as an example and you can do the rest and add,
Over path $BC (2, 2t)$,
$\frac{dx}{dt} = 0, \frac{dy}{dt} = 2$
Line integral $I_{BC} = \displaystyle \int_0^1 ((2t)^4 \frac{dx}{dt} + 5(2)^2 \frac{dy}{dt}) dt = \int_0^1 40 \, dt = 40$
Now the line integral for $AB$ and $DA$ will be zero the way vector field is set up (you should check that). So the only other one you will have to find is along $CD$.
The other way is to apply Green's theorem which should make the job easier -
As it is a smooth simple closed curve, we can directly apply Green's theorem according to which, for a counterclockwise oriented curve,
$\int_C P \, dx + Q \, dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA$
Where $A$ is the area of the region. As our region is a square with $0 \leq x \leq 2, 0 \leq y \leq 2, \,$ and
$P = y^4, Q = 5x^2$
Our line integral is converted to
$\displaystyle \int_0^2 \int_0^2 (10x - 4y^3) \, dx \, dy$
which is easy to calculate.