De Rham cohomology for $\mathbb{R^2}$ De Rham cohomology groups for $\mathbb{R^2}$.
$H^{0}_{dR}(\mathbb{R}^{2})=\mathbb{R}$ since  $Z^{0}(\mathbb{R}^{2})$ is the one dimensional space of locally constant functions on $\mathbb{R}^{2}$ and $B^{0}(\mathbb{R}^{2})=0$. Or one can use the fact $\mathbb{R}^{2}$ is connected. Can we also use Poincaré lemma in this case?
$H^{1}_{dR}(\mathbb{R}^{2})$  One can use Poincare lemma  and deduce that any closed form on $\mathbb{R}^{2}$ will be exact and then
$H^{1}_{dR}(\mathbb{R}^{2})=0$, but I would like to understand the following way of showing it explicitly:
-we choose an arbitrary closed $\omega=adx+bdy$ on  $\mathbb{R}^{2}$, if we are to show it is exact then we shall find $\Phi\in\Omega^{0}(\mathbb{R}^{2})$ such that $d\Phi=adx+bdy$ and here I have problems with understanding how to obtain the following:
$\Phi(x,y)=\displaystyle x\int_{0}^{1}a(tx,ty)dt+y\int_{0}^{1}b(tx,ty)dt$
What change of variables I have to make to get it? And how to verify that $\frac{\partial \Phi}{\partial x}=a(x,y)$ and $\frac{\partial \Phi}{\partial y}=b(x,y)$
Thank you 
 A: Suppose $a \, \mathrm{d}x + b \, \mathrm{d} y$ is closed, i.e.
$$\frac{\partial a}{\partial y} = \frac{\partial b}{\partial x}$$
and let $\Phi$ be as given. We just need to show that it works. For instance,
\begin{align*}
\frac{\partial \Phi}{\partial x} 
& = \int_0^1 a (t x, t y) \, d t + \int_0^1 t x \frac{\partial a}{\partial x} (t x, t y) \, d t + \int_0^1 t y \frac{\partial b}{\partial x} (t x, t y) \, d t \\
& = \int_0^1 a (t x, t y) \, d t + \int_0^1 t x \frac{\partial a}{\partial x} (t x, t y) \, d t + \int_0^1 t y \frac{\partial a}{\partial y} (t x, t y) \, d t \\
& = \int_0^1 a (t x, t y) \, d t + \int_0^1 t \frac{d}{d t} \left[ a (t x, t y) \right] \, d t \\
& = \int_0^1 \frac{d}{d t} \left[ t a (t x, t y) \right] \, dt \\
& = a (x, y)
\end{align*}
and a similar calculation works for the other partial derivative.
If you're wondering where $\Phi$ comes from: by the Stokes theorem, if $a \, \mathrm{d} x + b \, \mathrm{d} y = \mathrm{d} \Phi$ for some $\Phi$, then
$$\int_\gamma \left( a \, \mathrm{d} x + b \, \mathrm{d} y \right) = \Phi (x_1, y_1) - \Phi (x_0, y_0)$$
for all smooth paths $\gamma$ from $(x_0, y_0)$ to $(x_1, y_1)$; so for example,
$$\Phi (x, y) = \Phi (0, 0) + \int_0^1 \left( x a (t x, t y) + y b (t x, t y) \right) d t$$
by taking $\gamma$ to be the straight line from $(0, 0)$ to $(x, y)$. Clearly, $\Phi$ is only determined up to an additive constant, so we can take $\Phi (0, 0) = 0$ and obtain the given formula. 
