Are you smarter than me?

Calculus Level 3

C o n s i d e r L = 2012 3 + 2013 3 + . . . . + 3011 3 R = 2013 3 + 2014 3 + . . . . + 3012 3 a n d I = ∫ 2012 3012 x 3 d x Consider\\ L=\sqrt [ 3 ]{ 2012 } +\sqrt [ 3 ]{ 2013 } +....+\sqrt [ 3 ]{ 3011 } \\ R=\sqrt [ 3 ]{ 2013 } +\sqrt [ 3 ]{ 2014 } +....+\sqrt [ 3 ]{ 3012 } \\ and\quad I=\int _{ 2012 }^{ 3012 }{ \sqrt[3]{ x } dx }

L + R = 2 I L+R=2I L + R > 2 I L+R>2I L R = I \sqrt { LR } =I L + R < 2 I \\ L+R<2I

Mehul Chaturvedi by Mehul Chaturvedi , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Sandeep Rathod
Nov 15, 2014

R = L + 3012 3 − 2012 3 R = L + \sqrt[3]{3012} - \sqrt[3]{2012}

I = 2 ( 3012 3 − 2012 3 I = 2(\sqrt[3]{3012} - \sqrt[3]{2012}

R = L + I 2 R = L + \frac{I}{2}

R + L = 2 L + I 2 R + L = 2L + \frac{I}{2}

R + L < 2 I R + L < 2I

1 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you get the second line of your answer, I = ... ?

Jamie Mackillop - 6 years, 6 months ago
展豪 張
Mar 23, 2016

x 3 \sqrt[3]{x} is concave, so L + R < 2 I L+R<2I

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...