Friday

April 18, 2014

April 18, 2014

Posted by **MIKE9999** on Friday, March 8, 2013 at 9:41pm.

0

- calculus -
**Damon**, Friday, March 8, 2013 at 10:11pmI guess you mean integral from 0 to one

[ 3 e^x + x^3 ] at x = 1 - at x = 0

at x = 1

3 e^1 + 1 = 3e+1

at x = 0

3 e^0 + 0 = 3

so in the end

3 e - 2

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