Integrate....?

Question

how do you integrate t e^(-t^2) limits of integration is from 0 to infinity

Answer 1

One easy way is to make the substitution u = -t^2. Then du = -2t dt, and t dt = (-1/2) du.

Now integrate (-1/2) e^u du = (-1/2) e^u. Your limits on u are from 0 to infinity.

Answer 2

It is an improper integral


u=-t^2 and du = -2tdt

as 0<=t
The integral result -(e^(u))/2 and using the integrations limits you finally get -1/2

Answer 3

Integral {t*e^(-t^2) dt} from (0 to infinity)

Use u-substitution

u = -t^2
du = -2t dt ----> dt = -du/2t

Now substitute u and dt

Integral {t*e^(u) [-du/2t] } from (0 to infinity)
Integral {e^(u) [-t*du/2t] } from (0 to infinity)
Integral {e^(u) [-du/2] } from (0 to infinity)

= -e^(u) / 2
= -e^(-t^2) / 2
= -1 / 2e^(t^2) from 0 to infinity
[-1 / 2e^(infinity)] - [ -1 / 2e^(0) ]
= 0 + 1/2
=1/2

Answer 4

integration is (-1/2)*e^(-t^2)
result = 1/2

Answer 5

It is very simple.make the simple substitution u=-t/2 and change the limits