Answer
$A=8$
Work Step by Step
$\Delta x=\dfrac {b-a}{n}=\dfrac {2-0}{n}=\dfrac {2}{n}\left( 1\right) $
$x_{k}=a+\kappa \Delta x\left( 2\right) $
$A=\sum ^{n}_{1}f\left( x_{k}\right) \Delta x\left( 3\right) $
$f\left( x_{k}\right) =3x^{2}_{k}=3\left( a+k\Delta x\right) ^{2}=3\left( a+k\left( \dfrac {b-a}{n}\right) \right) ^{2}\left( 4\right) $
From (4),(3),(1) we get
$A=\sum ^{n}_{k=1}3\left( a+k\times \dfrac {b-a}{n}\right) ^{2}\dfrac {b-a}{n}=3\sum ^{n}_{k=1}\left( 0+k\times \dfrac {2-0}{n}\right) ^{2}\times \dfrac {2-0}{n}=24\times \dfrac {1}{n^{3}}\sum ^{n}_{k=1}k^{2}=24\times \dfrac {1}{n^{3}}\times \dfrac {n\times \left( n+1\right) \left( 2n+1\right) }{6}=\dfrac {4}{n^{2}}\left( n+1\right) \left( 2n+1\right) $
$A=\lim _{n\rightarrow \infty }\dfrac {4}{n^{2}}\left( n+1\right) \left( 2n+1\right) =4\left( 1+\dfrac {1}{n}\right) \left( 2+\dfrac {1}{n}\right) =4\times 1\times 2=8 $
