Question

Consider
$L=\sqrt[3]{32012}+\sqrt[3]{32013}+.....+\sqrt[3]{33011}$
$R=\sqrt[3]{32013}+\sqrt[3]{32014}+.....+\sqrt[3]{33012}$
and $I={\int }_{2012}^{3012}\sqrt[3]{3x}dx$.

Then

• A. $L+R<2I$
• B. $L+R>2I$
• C. $L+R=2I$
• D. $\sqrt{LR}=I$

Answer 1

The correct option is B $L+R>2I$

$L=\sqrt[3]{32012}+\sqrt[3]{32013}+.........\sqrt[3]{33011}......\left(i\right)R=\sqrt[3]{32013}+\sqrt[3]{32014}+.........\sqrt[3]{33012}......\left(ii\right)$

$I={\int }_{2012}^{3012}\sqrt[3]{3x}dxf\left(x\right)=x^{\frac{1}{3}},h=1$

$n=\frac{b-a}{h}=\frac{3012-2012}{1}=1000$

$I=\frac{b-a}{n}\left[f\right(a)+f(a+h)+f(a+2h)+f(a+3h)..........f(a+(n-1)h\left)\right]$

$I=\frac{1000}{1000}\left[f\right(2012)+f(2013)+f(2014)........f(3011\left)\right]$

$I=\sqrt[3]{32012}+\sqrt[3]{32013}+\sqrt[3]{32014}..........\sqrt[3]{33011}$

$2I=2\sqrt[3]{32012}+2\sqrt[3]{32013}...............2\sqrt[3]{33011}$

$2I=\{\sqrt[3]{32012}+\sqrt[3]{32013}+\sqrt[3]{32014}..........\sqrt[3]{33011}\}+\{\sqrt[3]{32013}+\sqrt[3]{32014}..........\sqrt[3]{33012}\}+\sqrt[3]{32012}-\sqrt[3]{33012}$

$2I=L+R+\sqrt[3]{32012}-\sqrt[3]{33012}$

$\sqrt[3]{32012}-\sqrt[3]{33012}<0$

$\Rightarrow 2I<L+R$

So option $B$ is correct.