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?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}