Question

Integration of 0.15e^(-0.15(x-0.5))

Answer 1

Substitute the stuff in exponent, 
$u=-0.15(x-0.5)\implies\,du=-0.15dx$

The integeral becomes
$-\int{e^u}\,du=-e^u+C=-e^{-0.15(x-0.5)}+C$