proving integral is negative (help!)

May 2010
39
0



I did the first part already, by letting
u = ax => dx = du/a, then just substituting in the values.
How do i explain why its negative?
 

Opalg

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I did the first part already, by letting
u = ax => dx = du/a, then just substituting in the values.
How do i explain why its negative?
For x > 0, \(\displaystyle e^x>1+x\). Take reciprocals to get \(\displaystyle e^{-x}<(1+x)^{-1}\). So the integrand is always negative and hence so is the integral.
 
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May 2010
39
0
For x > 0, \(\displaystyle e^x>1+x\). Take reciprocals to get \(\displaystyle e^{-x}<(1+x)^{-1}\). So the integrand is always negative and hence so is the integral.
Thanks :)