How to differentiare/integrate e^t and the like .

[LinearAlgebra]

How to differentiare/integrate e^t and the like...

I'm taking linear algebra this semester and completely have forgotten how to integrate and differentiate e functions.

This is within the context of trying to learn how to integrate vector functions...

So how would you integrate t*e^t ? what do you do when there is a multiplicative factor?

How about if it were e^t2? or alternatively e^-2t??

Thanks for the help!

 

[D H]

Use the chain rule to integrate $f(t)e^t$. You can choose either $u = f(t), dv = e^t dt$ or $u = e^t, dv = f(t) dt$. With the former approach, $\int f(t)e^t = f(t)e^t - \int f^\prime(t) e^t dt$, and the latter, $\int f(t)e^t = F(t)e^t - \int F(t) e^t dt$. Use whichever approach simplifies the result.

There is no elementary function $f(t)$ such that $f(t) = \int e^{t^2} dt$. You can always define a special function that satisfies this equation. The reason no one has done so is because this integral does not come up very often.

On the other hand, $\int e^{-t^2} dt$ comes up all the time. A special function has been defined based on this integral, the error function $\text{erf}(x)$.

 

[LinearAlgebra]

DH, thanks a lot for your reply.

Can you please explain further why you use the chain rule?

Also, just to use an example - let's say you have to integrate t^2*e^t? How would you do that?

 

[D H]

You use integration by parts, which is the chain rule rewritten for integrals.

Choose $u = t^2, dv = e^t dt$. Then

$$\int t^2 e^t dt = t^2 e^t - 2\int t e^t dt$$

Use integration by parts again on the integral on the right hand side

$$\int t e^t dt = t e^t - \int e^t dt = (t-1)e^t$$

Now apply this result to the first result,

$$\int t^2 e^t dt = t^2 e^t - 2(t-1)e^t = (t^2-2t+2)e^t$$

Verify by differentiating:

$$\frac d {dt}\left((t^2-2t+2)e^t\right) = (2t-2)e^t + (t^2-2t+2)e^t = t^2e^t$$

 

[D H]

My bad! Integration by parts is the product rule rewritten for integrals.