Deciding The region To Evaluate $\iint_{R}\left(x-y+1\right)dx dy$ 
QuestionEvaluate $$\iint_{R}\left(x-y+1\right)dxdy$$ where
  $R$ is region inside the unit square in which $ x +y\geq \dfrac{1}{2}$

My Approach
Figure 
I don't know how to plot graph In latex.I don't know anything
about latex coding.
Picture of region

In the figure There would be a unit square in first quadrant having
origin as one of its vertex .Straight line $\left(\frac{x}{\frac{1}{2}}+\frac{y}{\frac{1}{2}}=1\right)$divides
the unit square in two regions R$_{1}$and R$_{2}$. Where R$_{1}$region
is below the line and R$_{2}$region is above the line.
I think The region of integration $R= R_{2}$
$$\iint_{R_{2}}\left(x-y+1\right)dxdy=\iint_{R_{1}\cup R_{2}}\left(x-y+1\right)dxdy-\iint_{R_{1}}\left(x-y+1\right)dxdy$$
$\Longrightarrow$ 
This
is very easy to calculate.
Book Answer Books answer comes from This calculation
$$\iint_{R_{1}}\left(x-y+1\right)dxdy = \iint_{R_{1}\cup R_{2}}\left(x-y+1\right)dxdy - \iint_{R_{2}}\left(x-y+1\right)dxdy $$ $\Longrightarrow$
Region of integration is $R= R_{1}$
Am I wrong or is the book?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[15px,#ffc]{\ds{\int_{0}^{1}\int_{0}^{1}\pars{x - y + 1}\bracks{x + y \geq {1 \over 2}}\,\dd x\,\dd y}}\qquad\pars{~\bracks{\cdots}:\ Iverson\ Bracket~}
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}\pars{x - y + 1}\bracks{x \geq {1 \over 2} - y}
\,\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{1}\bracks{{1 \over 2} - y < 0}\int_{0}^{1}\pars{x - y + 1}
\,\dd x\,\dd y +
\int_{0}^{1}\bracks{0 < {1 \over 2} - y < 1}\int_{1/2 - y}^{1}\pars{x - y + 1}
\,\dd x\,\dd y
\\[1cm] = &\
\int_{0}^{1}\bracks{y > {1 \over 2}}\pars{{1 \over 2} - y + 1}\,\dd y
\\[2mm] + &\
\int_{0}^{1}\bracks{-\,{1 \over 2} < y < {1 \over 2}}\bracks{{1 \over 2}- {1 \over 2}\pars{{1 \over 2} - y}^{2} - \pars{y - 1}\pars{{1 \over 2} + y}}\,\dd y
\\[1cm] = &\
\int_{1/2}^{1}\pars{{3 \over 2} - y}\,\dd y +
\int_{0}^{1/2}\pars{{7 \over 8} + y - {3 \over 2}\,y^{2}}\,\dd y =
\bbx{7 \over 8} = 0.875
\end{align}
