integration

Hi guys ,
I'm trying to integrate -4e^-t , isn't this in the form of y=e^f(x) hence dy/dx=f'(x)e^f(x) ? the answer comes out as +4e^-t but somehow i just can't get to that any help would be much appreciated

Reply 1

Notnek

Original post by Alen.m

What answer did you get? And could you post your working/thoughts to explain how you got it?

Reply 2

Alen.m

Original post by notnek

What answer did you get? And could you post your working/thoughts to explain how you got it?

first i diffrentiated -t which gave me -t^2/2 and then put in the form of dy/dx=f'(x)e^f(x) which again gave me -4(-t^2/2 e^-t) and this is just far away from what the text book said

Reply 3

Ayman!

Original post by Alen.m

first i diffrentiated -t which gave me -t^2/2 and then put in the form of dy/dx=f'(x)e^f(x) which again gave me -4(-t^2/2 e^-t) and this is just far away from what the text book said

Do you know that

\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left ( e^{-t} \right ) = -e^{-t}

? If you differentiate the result you obtained, it does not give your initial function.

Recall that

\displaystyle \int e^{x} \ \mathrm{d}x = e^{x} + \mathrm{C}

Reply 4

Notnek

Original post by Alen.m

first i diffrentiated -t which gave me -t^2/2 and then put in the form of dy/dx=f'(x)e^f(x) which again gave me -4(-t^2/2 e^-t) and this is just far away from what the text book said

You have integrated when you should have differentiated :smile:

\displaystyle \int -t \ dt = -\frac{1}{2}t^2 + c

Reply 5

Alexion

Original post by Alen.m

first i diffrentiated -t which gave me -t^2/2 and then put in the form of dy/dx=f'(x)e^f(x) which again gave me -4(-t^2/2 e^-t) and this is just far away from what the text book said

Differentiating -t with respect to t will give you -1.

Reply 6

Alen.m

Original post by aymanzayedmannan

Do you know that

\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left ( e^{-t} \right ) = -e^{-t}

? If you differentiate the result you obtained, it does not give your initial function.

Recall that

\displaystyle \int e^{x} \ \mathrm{d}x = e^{x} + \mathrm{C}

To be honest the text bok that im revising has got nothing about the formula you nust mentioned and that was why i kept trying wrong ways i guess but yours absoultely makes sense thanks

Reply 7

Notnek

Original post by Alen.m

To be honest the text bok that im revising has got nothing about the formula you nust mentioned and that was why i kept trying wrong ways i guess but yours absoultely makes sense thanks

This formula dy/dx=f'(x)e^f(x) does work.

You just made the mistake of integrating -t when you should have differentiated it to give f'(x) = -1.

Reply 8

Alen.m

Original post by notnek

This formula dy/dx=f'(x)e^f(x) does work.

You just made the mistake of integrating -t when you should have differentiated it to give f'(x) = -1.

Well actually the question is on M2 where you have to integrate certain expression to get to another expeesion for displacement, acceleration and velocity . Im half sure that i had to integrate certain expression which had -4e^-t in it as well to get to another expression. I'll send you the full question once i got home