Pick out the rings which are integral domains Pick out the rings which are integral domains:    
a. $\mathbb{R}[x]$, the ring of all polynomials in one variable with real coeffcients.  
b. $C^1[0, 1]$, the ring of continuously differentiable real-valued functions on
the interval [$0, 1$] (with respect to pointwise addition and pointwise multiplication). 
c. $M_n(\mathbb{R})$, the ring of all $n × n$ matrices with real entries.

(a) is an integral domain since $\mathbb{R}$ is a field and so $\mathbb{R}[x]$ is a Euclidean domain and hence a integral domain.
(c) is obviously not an integral domain.
(b) I am not sure.
Help me please.
 A: (For $n>1$) Multiply a matrix with zeros everywhere except a one in the top-right corner by itself.
(For $n=1$) These are just the real numbers, so it is integral.
For (b) use functions with support (the place there they are different from zero) are disjoint intervals of $[0,1]$. The function $e^{-1/x^2}e^{-1/(x-1/3)^2}$ in the interval $[0,1/3]$ and zero otherwise. And the function $e^{-1/(x-1/2)^2}e^{-1/(x-1)^2}$ in theinterval $[1/2,1]$ and zero otherwise. Multiply them.
A: I' assuming that the OP meant "b" instead of "c" in the second item of the second part of his question where he/she lists possible answers, so . . . 
$M_n(R)$ is not an integral domain for $n \ge 2$ since i.)  it contains nilpotent elements, e.g. the matrix
$N = [n_{ij}$ with $n_{ij} = 1$ if $j = i + 1$ and zeroes elsewhere, $1 \le i, j \le n$;  ii.)
it has other zero divisors as well, e.g. $A = [a_{ij}$ with $a_{ii} = 1$ for $1 \le i \le m < n \;$ and $a_{ij} = 0$ otherwise and $B = [b_{ij}$ with $b_{ii} = 1$ for $m < i \le n$ and $b_{ij} = 0$ elsewhere.  Then $AB = 0$ but $A \ne 0 \ne B$.  In fact, $A$ and $B$ are idempotent; $A^2 = A \ne 0$ and $B^2 = B \ne 0$.
*Edit:  This Just In!*In light of the OP's most recent edits to his question, we have:  $C^1[0, 1]$ is not an integral domain; it has zero divisors.  Consider the two functions
$f(x) = (x - 1/2)^2 \; \text{for} \; 0 \le x \le 1 / 2; f(x) = 0 \; \text{for} \; 1/2 \le x \le 1, \tag{1}$
and
$g(x) = 0  \; \text{for} \; 0 \le x \le 1 / 2; g(x) = (x - 1/2)^2  \; \text{for} \; 1/2 \le x \le 1; \tag{2}$
then $f(x), g(x) \in C^1[0, 1]$, $f(x)g(x) = 0$ but $f(x) \ne 0 \ne g(x)$!
Hope this helps.  Happy New Year,
and as always,
Fiat Lux!!!
A: To see $C^1[0,1]$ is not an integral domain, it may be easiest to draw a picture. Trying drawing two smooth curves where at least one of them is zero at any point in $[0,1]$.
This is a little handwavey, so you can use an explicit formula if you'd like. The function
$f\colon\mathbb{R}\to\mathbb{R}$ defined by
$$
f=\begin{cases}
e^{-1/x^2} & x>0,\\
0 & x\leq 0
\end{cases}
$$
is a well known smooth function. Then the function $g(x)=f(x-a)f(b-x)$ is also easily seen to be a smooth function, (since $f$ is smooth), which is positive on $(a,b)$ and zero elsewhere.
Using this, make two smooth functions and take the support to be $(1/5,2/5)$ for the first one, and $(3/5,4/5)$ as the support for the second, or something along those lines. These will be zero divisors. 
