Question

Question #da50f

Answer 1

`int_{-infty}^0 x5^{-x^2} dx`

by the substitution `u=-x^2`.
`Rightarrow {du}/{dx}=-2x Rightarrow dx={du}/{-2x}`
As `x` goes from `-infty` to `0`, `u` goes from `-infty` to `0`.

`=int_{-infty}^0x5^{u}{du}/{-2x}`

by cancelling out `x`'s and pulling `-1/2` out of the integral,

`=-1/2int_{-infty}^0 5^u du`

by the definition of improper integral,

`=-1/2lim_{t to -infty}int_t^0 5^u du`

by finding an anti-derivative,

`=-1/2lim_{t to -infty}[{5^u}/{ln5}]_t^0`

by pulling `1/{ln5}` out,

`=-1/{2ln5}lim_{t to -infty}(1-5^t)`

since `lim_{t to -infty}5^t=0`,

`=-1/{2ln5}`

Hence, the improper integral converges to `-1/{2ln5}`.

I hope that this was helpful.