Question

Evaluate: `int_0^oo xe^-x.dx`

Answer 1

`int_0^∞ xe^-x.dx`

= `[x int e^-x.dx]_0^∞ - int_0^∞[d/dx (x) int e^-x.dx].dx`

= `[x((e^-x)/-1)]_0^∞ - int^∞ 1.(e^-x)/((-1)).dx`

= `[- x.e^x]_0^∞ + int_0^∞ e^-x.dx`

= `[x.e^-x]_0^∞ + [e^-x/-1]_0^∞`

= `[x.e^-x]_0^∞ - [-e^x]_0^∞`

= `[∞.e^-∞-0.e^-0] - [e^-∞-e^-0]`

= `[0 - 0] - [0-1/e^0]`

= `- [0-1/1]`

= 1.       ...[∵ e0 = 1, e–x = 0, when x = ∞]