Question

What is the integral of e^-s^2ds from 0 to t

Answer 1

(e^-s^2)ds

Answer 2

The integral can not be written in terms of elementary functions.

Answer 3

Oh this is for differential equations

Answer 4

I need to find the derivative of y=(e^t^2 )(integral of e^-s^2ds from 0 to t)

Answer 5

Then you need to use the fundamental theorem of calculus

Answer 6

and the product rule.

Answer 7

yeah but how about at the integral of e^-s^2ds from 0 to t part?

Answer 8

The derivative of that is the integrand with t substituted for s

Answer 9

derivative with respect to t, that is

Answer 10

Okay so when I use product rule does the integral part when taking the derivative become e^t^2?

Answer 11

In the other summand, you will not get around writing out the integral :D

Answer 12

I think it'd be a lot more complicated than that?

Answer 13

e^-t^2*

Answer 14

it's 0 for some reason

Answer 15

Nope it would be \[ e^{-t^{-2}} \]

Answer 16

e^x is never 0. I don't know what you mean.

Answer 17

is it becomes Im integrating in terms of t and not s

Answer 18

Should I write out the whole derivative for you?

Answer 19

sure I'm slightly confused

Answer 20

Ok, first denote \[ f(t) = \int^t_0 e^{-1/s^2} ds \]

Answer 21

I typed it in wolfram and got 0 and I remember my TA saying something like that

Answer 22

okay

Answer 23

Then \[ f^\prime(t) = e^{-1/t^2} \]

Answer 24

by the fundamental theorem of calculus

Answer 25

Now set \[ g(t) = e^{t^2} f(t) \]

Answer 26

Is this the function you wanted to know the derivative of ?

Answer 27

Yeah but I don't see how it's e^-1/t^2?

Answer 28

Well the fundamental theorem of calculus says:
If \[ f(t) = \int^t_0 g(s) ds \] for some continous g, then
f is differentiable and
\[ f^\prime(t) = g(t) \]

Answer 29

okay thanks for your help I shall ask again, he said this was a tricky one where most people would just sub in t for s

Answer 30

just check out the fundamental theorem of calculus and use the product rule.. maybe the result is very simple, i didn't do the work

Answer 31

Nope I just did the question and the answer for that has to be 0 it has something to do with it being t and the integral being in terms of s.