Question

Integrate 0to 1 e^-4x

Answer 1

$\displaystyle\int_0^1e^{-4x}\,\mathrm dx=-\frac14\int_0^{-4}e^u\,\mathrm du$

where we let $u=-4x$, so that $-\dfrac{\mathrm du}4=\mathrm dx$.

$\displaystyle-\frac14\int_0^{-4}e^u\,\mathrm du=\frac14\int_{-4}^0e^u\,\mathrm du=\frac14e^u\bigg|_{u=-4}^{u=0}=\frac14\left(e^0-e^{-4}\right)=\frac{1-e^{-4}}4$