# Integral help

• Feb 8th 2009, 01:27 PM
element
Integral help
Need to find the integral of e^[t^(2) - t]dt
• Feb 8th 2009, 01:39 PM
Moo
Hello,
Quote:

Originally Posted by element
Need to find the integral of e^[t^(2) - t]dt

This cannot be expressed in terms of simple functions.

Integrate[E^(x^2 - x), x] ==
(Sqrt[Pi]*Erfi[-1/2 + x])/(2*E^(1/4))
http://reference.wolfram.com/mathematica/ref/Erfi.html
• Feb 8th 2009, 01:48 PM
element
Quote:

Originally Posted by Moo
Hello,

This cannot be expressed in terms of simple functions.

Integrate[E^(x^2 - x), x] ==
(Sqrt[Pi]*Erfi[-1/2 + x])/(2*E^(1/4))
Erfi - Wolfram Mathematica

thanks. hmm. i probably did something wrong. here is the problem i am working on. we are supposed to use an integrating factor for these linear first order differential equations.

dP/dt + 2tP = P + 4t -2

i manipulated this into "standard form" ( dy/dx + P(x)y = f(x) )

dP/dt + (2t - 1)P = 4t - 2

then the integrating factor would be e^integral of [2t -1]dt, and that would be e^[t^(2) - t] correct?

now my differential equation looks like this

e^[t^(2) - t]y = integral of [ (4t-2)e^[t^(2) - t] ]dt

i separated the right side into
4 integral of te^[t^(2) - t]dt -2 integral of e^[t^(2) - t]dt

now im stuck on integrating that.
• Feb 8th 2009, 01:55 PM
HallsofIvy
Quote:

Originally Posted by element
thanks. hmm. i probably did something wrong. here is the problem i am working on. we are supposed to use an integrating factor for these linear first order differential equations.

dP/dt + 2tP = P + 4t -2

i manipulated this into "standard form" ( dy/dx + P(x)y = f(x) )

dP/dt + (2t - 1)P = 4t - 2

then the integrating factor would be e^integral of [2t -1]dt, and that would be e^[t^(2) - t] correct?

Yes, that is correct.

Quote:

now my differential equation looks like this

e^[t^(2) - t]y = integral of [ (4t-2)e^[t^(2) - t] ]dt

i separated the right side into
4 integral of te^[t^(2) - t]dt -2 integral of e^[t^(2) - t]dt

now im stuck on integrating that.
Don't separate! Let u= t^2- t. The integral becomes $\displaystyle 2\int e^u du$
• Feb 8th 2009, 02:00 PM
element
Quote:

Originally Posted by HallsofIvy
Yes, that is correct.

Don't separate! Let u= t^2- t. The integral becomes $\displaystyle 2\int e^u du$

thank you very much! i cant believe i didn't see that. thanks.