Proving various relations are partial orders I am given these relations, in which I have to prove or disprove each and every one.
a. The relation $\trianglelefteq$ deﬁned on ℕ by a $\trianglelefteq$ b if a ≤ b²
b. The relation $\preceq$ deﬁned on ℤ by m $\preceq$ n if m ≤ n + 5.
c. The relation $\ll$ on the set of continuous functions deﬁned by f(x) $\ll$ g(x) if $\int^1_0f(x)dx \leqslant \int^1_0g(x)dx$
So, for a, I have the following: To prove reflexivity, it follows that a $\leqslant$ a², and therefore, a $\trianglelefteq$ a, so $\trianglelefteq$ is reflexive. For transitivity, for all a, b, and c in ℕ, if a ≤ b², and b ≤ c², then a ≤ c², and therefore the relation is transitive.
For b, I have that for reflexivity, m ≤ m+5, so the relation is reflexive. For transitivity,for all m,n, and p in ℤ, if m ≤ n + 5, and n ≤ p + 5, then m ≤ p + 5, so the relation is transitive.
For c, I wrote for reflexivity that since $\int^1_0f(x)dx \leqslant \int^1_0f(x)dx$, then it the relation is reflexive. For transitivity, if $\int^1_0f(x)dx \leqslant \int^1_0g(x)dx$, and $\int^1_0g(x)dx \leqslant \int^1_0h(x)dx$, then $\int^1_0f(x)dx \leqslant \int^1_0h(x)dx$, therefore, the relation is transitive.
As you can see, I left out antisymmetry, because I am not sure how to prove that. I think that my reflexive and transitive proofs are correct, but if anyone could help me out with these problems by helping me with the antisymmetry proofs and verifying/correcting my reflexive and transitive proofs, that would be great.

The first proof is not correct. $a\leq b^2$ and $b \leq c^2$ does not imply $a \leq c^2$. Consider $a=15, b=4, c=3$.
The second proof is also incorrect because again transitivity fails. Consider $m=10,n=5,p=0$.
Your problem is that you state, for example, that "if m ≤ n + 5, and n ≤ p + 5, then m ≤ p + 5" but do not give any justification. And the statement is simply false. If you work more carefully, you would do something like this:
If $m \leq n+5$ and $n \leq p+5$, then substituting $p+5$ for $n$ in the first inequality, we see $m \leq (p+5)+5 = p + 10$. 
You would then notice that this statement is not strong enough, and so you would start to look for a counterexample.
The last proof is correct, but you still need antisymmetry. To prove this property, you would assume $f \ll g$ and $g\ll f$ and then show $f = g$. However, it isn't true! Try $f(x) = 1-x$ and $g(x) = x$. Their integrals from $0$ to $1$ are equal, but the functions are not.
