# proving integral is negative (help!)

#### gomes

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

MHF Hall of Honor

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.

gomes and AllanCuz

#### gomes

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