If $\quad f(x)=\begin{cases} 2{ x }^{ 2 }+1,x\le 1 \\ 4{ x }^{ 2 }-1,x>1 \end{cases}$, then $\int _{ 0 }^{ 2 }{ f(x)dx }$ is

$\quad f(x)=\begin{cases} 2{ x }^{ 2 }+1,x\le 1 \\ 4{ x }^{ 2 }-1,x>1 \end{cases}$

from 0 to 1 we have $2x^2+1$ and from 1 to 2 we have $4x^2-1$

$=\int _{ 0 }^{ 1 }{ \left( 2{ x }^{ 2 }+1 \right) dx+\int _{ 1 }^{ 2 }{ \left( 4{ x }^{ 2 }-1 \right) dx } }$

$={ \left[ \frac { 2{ x }^{ 3 } }{ 3 } +x \right] }_{ 0 }^{ 1 }+{ \left[ \frac { 4{ x }^{ 3 } }{ 3 } -x \right] }_{ 1 }^{ 2 }$

$=\frac { 2 }{ 3 } +1-0-0+\frac { 4{ (2) }^{ 3 } }{ 3 } -2-\left( \frac { 4 }{ 3 } -1 \right)$

$=\frac { 2 }{ 3 } +1+\frac { 32 }{ 3 } -2-\frac { 4 }{ 3 } +1$

$=\frac { 2+32-4 }{ 3 }$

$=\dfrac{30}{3}$

=10.