Answer
$A=\dfrac {11}{6}$
Work Step by Step
$\Delta x=\dfrac {b-a}{n}\left( 1\right) $
$x_{k}=a+k\Delta x\left( 2\right) $
$A=\sum ^{n}_{k=1}f\left( x_{k}\right) \Delta x\left( 3\right) =\sum ^{n}_{k=1}f\left( a+k\Delta x\right) \times \Delta x=\sum ^{n}_{k=1}f\left( a+k\dfrac {b-a}{n}\right) \times \dfrac {b-a}{n} (3) $
$f\left( x\right) =x+x^{2}\Rightarrow f\left( a+k\dfrac {b-a}{n}\right) =a+k\dfrac {b-a}{n}+\left( a+\kappa \dfrac {b-a}{n}\right) ^{2}(4)$
From (3) and (4) we get
$A=\dfrac {b-a}{n}\sum ^{n}_{k=1}\left( a+k\dfrac {b-a}{n}+\left( a+k\dfrac {b-a}{n}\right) ^{2}\right) =\dfrac {b-a}{n}\sum ^{n}_{k=1}\left( a+a^{2}+3k\dfrac {b-a}{n}+k^{2}\dfrac {\left( b-a\right) ^{2}}{n^{2}}\right) =\dfrac {b-a}{n}\left( \left( a+a^{2}\right) \times n+\sum ^{n}_{1}3k\dfrac {b-a}{n}+\sum ^{n}_{1}k^{2}\dfrac {\left( b-a\right) ^{2}}{n^{2}}\right) =\dfrac {b-a}{n}\left( n\left( a+a^{2}\right) +3\dfrac {b-a}{n}\times \dfrac {n\times \left( n+1\right) }{2}+\dfrac {\left( b-a\right) ^{2}}{n^{2}}\times \dfrac {n\times \left( n+1\right) \left( 2n+1\right) }{6}\right) $
$b=1; a=0$
So we get $A=\dfrac {b-a}{n}\left( n\left( a+a^{2}\right) +3\dfrac {b-a}{n}\times \dfrac {n\times \left( n+1\right) }{2}+\dfrac {\left( b-a\right) ^{2}}{n^{2}}\times \dfrac {n\times \left( n+1\right) \left( 2n+1\right) }{6}\right) =\dfrac {1}{n}\left( \dfrac {3}{2}\left( n+1\right) +\dfrac {n}{3}+\dfrac {1}{2}+\dfrac {1}{6n}\right) =\dfrac {11}{6}+\dfrac {2}{n}+\dfrac {1}{6n^{2}}$
$\lim _{n\rightarrow \infty }\left( \dfrac {11}{6}+\dfrac {2}{n}+\dfrac {1}{6n^{2}}\right) =\dfrac {11}{6}$
