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Consider $$ \begin{align} & L=\sqrt[3]{2012}+\sqrt[3]{2013}+\ldots +\sqrt[3]{3011} \\ & R=\sqrt[3]{2013}+\sqrt[3]{2014}+\ldots +\sqrt[3]{3012} \\ \end{align}\ $$ and $$ I=\int_{2012}^{3012}\sqrt[3]{x}\,dx $$ then which of the following holds? $$ \begin{align} & \text{(a) } L+R<2I\qquad \text{(b) }L+R>2I \\ & \text{(c) }L+R=2I\qquad \text{(d) }\sqrt{LR}=I \\ \end{align} $$

I have verified numerically that $\text{(a)}$ is indeed the answer, but I couldn't verify it analytically?

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Note that $$ L+R = \sum_{n = 2012}^{3011} (\sqrt[3]{n}+\sqrt[3]{n+1}) $$ and $$ \int_{2012}^{3012}\sqrt[3]{x}dx = \sum_{n = 2012}^{3011}\int_n^{n+1}\sqrt[3]{x}dx $$ So all you have to do is prove that $$ \sqrt[3]{n} + \sqrt[3]{n+1} < 2\int_n^{n+1}\sqrt[3]{x}dx $$ for any positive integer $n$. Can you manage that?

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