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
`-\frac{\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`